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diff --git a/usth/MATH1.5/homework/cursived.pdf b/usth/MATH1.5/cursived.pdf Binary files differindex c7a2337..c7a2337 100644 --- a/usth/MATH1.5/homework/cursived.pdf +++ b/usth/MATH1.5/cursived.pdf diff --git a/usth/MATH1.5/homework/cursived.tex b/usth/MATH1.5/cursived.tex index 59a3ac7..59a3ac7 100644 --- a/usth/MATH1.5/homework/cursived.tex +++ b/usth/MATH1.5/cursived.tex diff --git a/usth/MATH1.5/homework/review.pdf b/usth/MATH1.5/review.pdf Binary files differindex 10f4b74..10f4b74 100644 --- a/usth/MATH1.5/homework/review.pdf +++ b/usth/MATH1.5/review.pdf diff --git a/usth/MATH1.5/homework/review.tex b/usth/MATH1.5/review.tex index 55ad0a0..55ad0a0 100644 --- a/usth/MATH1.5/homework/review.tex +++ b/usth/MATH1.5/review.tex diff --git a/usth/MATH2.1/problems.pdf b/usth/MATH2.1/problems.pdf Binary files differnew file mode 100644 index 0000000..4889f5c --- /dev/null +++ b/usth/MATH2.1/problems.pdf diff --git a/usth/MATH2.1/solutions.pdf b/usth/MATH2.1/solutions.pdf Binary files differnew file mode 100644 index 0000000..12d4b19 --- /dev/null +++ b/usth/MATH2.1/solutions.pdf diff --git a/usth/MATH2.1/solutions.tex b/usth/MATH2.1/solutions.tex new file mode 100644 index 0000000..46c6915 --- /dev/null +++ b/usth/MATH2.1/solutions.tex @@ -0,0 +1,1633 @@ +\documentclass[a4paper,12pt]{article} +\usepackage[english,vietnamese]{babel} +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{enumerate} +\usepackage{lmodern} +\usepackage{mathtools} +\usepackage{pgfplots} +\usepackage{venndiagram} + +\newcommand{\E}{\mathbf E} +\newcommand{\C}{\mathrm C} +\newcommand{\N}{\mathcal N} +\newcommand{\ud}{\,\mathrm{d}} +\newcommand{\var}{\mathrm{var}} +\newcommand{\cov}{\mathrm{cov}} +\newcommand{\exercise}[1]{\noindent\textbf{#1.}} +\renewcommand{\thesubsection}{\Alph{subsection}} +\renewcommand{\thefootnote}{\fnsymbol{footnote}} + +\title{Probability Homework} +\author{Nguyễn Gia Phong} +\date{Fall 2019} + +\begin{document} +\maketitle +\section{Basic Probability 1} + +\exercise 1 Problems regarding de Morgan's law +\begin{enumerate}[(a)] + \item Consider rolling a six-sided die, where + \begin{align*} + &\begin{cases} + A = \{2, 4, 6\}\iff A^\C = \{1, 3, 5\}\\ + B = \{4, 5, 6\}\iff B^\C = \{1, 2, 3\} + \end{cases}\\ + \Longrightarrow &\begin{cases} + (A \cup B)^\C = \{2, 4, 5, 6\}^\C = \{1, 3\} = A^\C \cap B^\C\\ + (A \cap B)^\C = \{4, 6\}^\C = \{1, 2, 3, 5\} = A^\C \cup B^\C\\ + \end{cases} + \end{align*} + \item By de Morgan's law, + \begin{align*} + P\left(A^\C \cap B^\C\right) &= P\left((A \cup B)^\C\right)\\ + &= 1 - P\left(A \cup \left(A^\C \cap B\right)\right)\\ + &= 1 - P(A) - P\left(A^\C \cap B\right) + \tag{since $A \cap \left(A^\C \cap B\right) = \varnothing$}\\ + &= 1 -P(A) -P\left((A\cap B)\cup\left(A^\C\cap B\right)\right) +P(A\cap B) + \tag{since $(A\cap B)\cap\left(A^\C\cap B\right) = \varnothing$}\\ + &= 1 - P(A) - P(B) + P(A\cap B) + \end{align*}\label{1.b} + \item Consider events A and B such that $P(A) = 1/2$, $P(A\cup B) = 3/4$, + $P\left(B^\C\right) = 5/8$. + + \begin{align*} + & P\left(A^\C\cap B\right) = P(A\cup B) - P(A) + = \frac 3 4 - \frac 1 2 = \frac 1 4\\ + & P\left(A^\C\cap B^\C\right) = P\left((A\cup B)^\C\right) + = P(\Omega) - P(A\cup B) = 1 - \frac 3 4 = \frac 1 4\\ + & P\left(A\cap B^\C\right) = P\left(B^\C\right)-P\left((A\cup B)^\C\right) + = \frac 5 8 - \frac 1 4 = \frac 3 8\\ + & P(A\cap B) = P(A) - P\left(A\cap B^\C\right) + = \frac 1 2 - \frac 3 8 = \frac 1 8\\ + & P\left(A^\C\cup B^\C\right) = P\left((A\cap B)^\C\right) + = P(\Omega) - P(A\cap B) = 1 - \frac 1 8 = \frac 7 8 + \end{align*} +\end{enumerate} + +\exercise 2 A four-sided die is rolled repeatedly, +until the first time (if ever) that an even number is obtained. +What is the sample space for this experiment? + +Let the outcome be a n-dimensional vector, whose elements are values +of each roll in chronological order. The sample space would then be +\[\Omega = \{v \in \{1, 3\}^m\times\{2, 4\} \mid m \in \mathbb N\}\] + +\exercise 3 A ball is drawn at random from a box containing 6 red balls, +4 white balls, and 5 blue balls. + +Let $\Omega$ be the sample space then $n(\Omega) = 6 + 4 + 5 = 15$. +Let the R, W and B be the event where a red, white and blue ball is drawn +respectively, each of these events are mutually exclusive. Suppose each ball +is equally likely to be drawn, we get +\[\begin{cases} + n(R) = 6\\ + n(W) = 4\\ + n(B) = 5 +\end{cases} +\Longrightarrow\begin{dcases} + P(R) = \frac{n(R)}{n(\Omega)} = \frac{2}{5}\\ + P(W) = \frac{n(W)}{n(\Omega)} = \frac{4}{15}\\ + P(B) = \frac{n(B)}{n(\Omega)} = \frac{1}{3} +\end{dcases}\] + +\begin{enumerate}[(a)] + \item For a ball that is not red to be drawn, the probability is + \[P\left(R^\C\right) = P(\Omega) - P(R) = 1 - \frac 2 5 = \frac 3 5\] + \item For a ball that is either red or white to be drawn, the probability is + \[P(R\cup W) = P(R) + P(W) = \frac{2}{5} + \frac{4}{15} = \frac 2 3\] +\end{enumerate} + +\exercise 4 Given $P(C_a) = 0.8$, $P(C_b) = 0.6$ and $P(C_a\cap C_b) = 0.5$. + +We can easily prove that $P(C_a\cup C_b) = P(C_a) + P(C_b) - P(C_a\cap C_b)$ +(similar to what we did in exercise 1.b). Thus the probability that the student +will get at least one offer from these two companies is $0.8 + 0.6 - 0.5 = 0.9$. + +\exercise 5 Let G and C be the events that the selected student is a genius and +is a chocolate lover, respectively, then $P(G) = 0.6$, $P(C) = 0.7$ and +$P(G\cap C) = 0.4$. The probability that a randomly selected student is +neither a genius nor a chocolate lover is +\[P\left((G\cup C)^\C\right) = 1 - P(G) - P(C) + P(G\cap C) += 1 - 0.6 - 0.7 + 0.4 = 0.1\] + +\exercise 6 First, consider Rick's choice of entrance. We denote +the outcome that he chooses each gate as $R_A$, $R_B$, $R_C$ and $R_D$, +then $P\{R_A\} = 1/3$ and $P\{R_B\} = P\{R_C\} = P\{R_D\} = 2/9$. +The sample space is $\Omega_R = \{R_A, R_B, R_C, R_D\}$. + +Similarly, denote Brenda's and Ali's choices as $B_Y$ and $A_X$ respectively, +where $X$, $Y$ (and later $Z$) are one of the four entrances +$\omega = \{A, B, C, D\}$, we get +\[\begin{dcases} + P\{B_A\} = P\{B_B\} = P\{B_C\} = P\{B_D\} = \frac 1 4\\ + P\{A_A\} = P\{A_B\} = \frac{2}{35}\\ + P\{A_C\} = \frac 2 7\\ + P\{A_D\} = \frac 3 5 +\end{dcases}\] +The sample spaces of these two models are $\Omega_B = \{B_A, B_B, B_C, B_D\}$ +and $\Omega_A = \{A_A, A_B, A_C, A_D\}$. + +Now consider the probability model of the choices of the three friends. +The sample space is $\Omega = \Omega_R\times\Omega_B\times\Omega_A$. +Since the three friends chooses their entrance independently, +for all $\mathbf v = \langle R_Z, B_Y, A_X\rangle$ in $\Omega$, +\[P\{\mathbf v\} = P\{R_Z\} \cdot P\{B_Y\} \cdot P\{A_X\}\] + +\begin{enumerate}[(a)] + \item The event that at least two friends choose entrance B is + \[a = (\Omega_R \times \{B_B\} \times \{A_B\}) + \cup (\{R_B\} \times \Omega_B \times \{A_B\}) + \cup (\{R_B\} \times \{B_B\} \times \Omega_A)\] + + Notice that + \begin{align*} + &(\Omega_R \times \{B_B\} \times \{A_B\}) + \cap (\{R_B\} \times \Omega_B \times \{A_B\})\\ + =\,&(\Omega_R \times \{B_B\} \times \{A_B\}) + \cap (\{R_B\} \times \Omega_B \times \{A_B\}) + \cap (\{R_B\} \times \{B_B\} \times \Omega_A)\\ + =\,&\{R_B, B_B, A_B\} + \end{align*} + + Therefore the probability of this event is + \begin{align*} + P(a) &= P(\Omega_A \times \{B_B\} \times \{A_B\})\\ + &+ P(\{R_B\} \times \Omega_B \times \{A_B\})\\ + &- P\{R_B, B_B, A_B\}\\ + &+ P(\{R_B\} \times \{B_B\} \times \Omega_A)\\ + &- P\{R_B, B_B, A_B\}\\ + &= P\{B_B\} \cdot P\{A_B\}\\ + &+ P\{R_B\} \cdot P\{A_B\}\\ + &+ P\{R_B\} \cdot P\{B_B\}\\ + &- 2 \cdot P\{R_B\} \cdot P\{B_B\} \cdot P\{A_B\}\\ + &= \frac{1}{4} \cdot \frac{2}{35} + + \frac{2}{9} \cdot \frac{2}{35} + + \frac{2}{9} \cdot \frac{1}{4} + - \frac{2}{9} \cdot \frac{1}{4} \cdot \frac{2}{35} + = \frac{8}{105} + \end{align*} + \item The only four cases where all friends choose the same entrance are + $\{\langle R_X, B_X, A_X\rangle \mid X = \omega\}$. + Hence the probability of this event is + \begin{align*} + P(b) &= \sum_{X \in \omega} P\{R_X\}\cdot P\{B_X\}\cdot P\{A_X\}\\ + &= \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{2}{35} + + \frac{2}{9} \cdot \frac{1}{4} \cdot \frac{2}{35} + + \frac{2}{9} \cdot \frac{1}{4} \cdot \frac{2}{7} + + \frac{2}{9} \cdot \frac{1}{4} \cdot \frac{3}{5} + = \frac{2}{35} + \end{align*} +\end{enumerate} + +\exercise 7 We roll two fair six-sided dice. +\begin{enumerate}[(a)] + \item The event that doubles are rolled has six outcomes, thus its probability + is 6/36 = 1/6. + \item Among the six outcomes where the result is four or less + (\{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)\}), + there are two that are doubles, hence the probability would then be 1/3. + \item Let $\omega = \{1, 2, 3, 4, 5, 6\}$, + the sample space is $\Omega = \omega^2$. For one die roll is a six, + the event is $C = (\{6\}\times\omega) \cup (\omega\times\{6\})$. + Since $(\{6\}\times\omega) \cap (\omega\times\{6\}) = \{(6, 6)\}$, + \begin{align*} + P(C) &= P(\{6\}\times\omega) + P(\omega\times\{6\}) - P\{(6, 6)\}\\ + &= \frac{n(\{6\}\times\omega) + n(\omega\times\{6\}) - n\{(6, 6)\}} + {n(\Omega)}\\ + &= \frac{6 + 6 - 1}{36} = \frac{11}{36} + \end{align*} +\end{enumerate} + +\exercise 8 A baby rolls two six-sided dice. Assumed that the dice are fair. +Let $\omega = \{1, 2, 3, 4, 5, 6\}$, the sample space is $\Omega = \omega^2$. +\begin{enumerate}[(a)] + \item There are six outcomes where the result of seven: + \[A = \{(m, 7 - m) \mid m \in \omega\}\] + hence this event's probability is + \[P(A) = \frac{n(A)}{n(\Omega)} = \frac{6}{36} = \frac{1}{6}\] + \item There are two outcomes where the result of eleven: + $B = \{(5, 6), (6, 5)\}$, thus $P(B) = 1/18$. + As $A$ and $B$ are disjoint, the probability of not getting + a sum of seven or eleven is + \begin{align*} + P\left((A\cup B)^\C\right) &= P(\Omega) - P(A\cup B)\\ + &= 1 - (P(A) + P(B))\\ + &= 1 - \frac{1}{6} - \frac{1}{18} = \frac{7}{9} + \end{align*} +\end{enumerate} + +\exercise 9 Given $n(\Omega) = 25$, $n(C) = 9$, $n(D) = 8$ +and $n\left((C\cup D)^\C\right) = 10$. + +By the Venn diagram, $a = C\cap D^\C$, $b = C\cap D$ and $c = C^\C\cap D$. +Let $E = C\cup D$ and $d = E^\C$, $n(d) = 10$ and $n(E) = n(\Omega)-n(d) = 15$. +Assume that the boy fairly randomly selected, +\begin{align*} + P(E) &= \frac{n(E)}{n(\Omega)} + = \frac{15}{25} + = \frac{3}{5}\\ + P(a) &= P\left((D\cap d)^\C\right) + = 1 - P(D\cap d) + = 1 - P(D) - P(d)\\ + &= 1 - \frac{n(D) + n(d)}{n(\Omega)} + = 1 - \frac{8 + 10}{25} + = \frac{7}{25}\\ + P(c) &= P\left((C\cap d)^\C\right) + = 1 - P(C\cap d) + = 1 - P(C) - P(d)\\ + &= 1 - \frac{n(C) + n(d)}{n(\Omega)} + = 1 - \frac{9 + 10}{25} + = \frac{6}{25}\\ + P(b) &= \frac{n(b)}{n(\Omega)} + = \frac{n(E) - n(a) - n(c)}{n(\Omega)} + = P(E) - P(a) - P(c)\\ + &= \frac{3}{5} - \frac{7}{25} - \frac{6}{25} + = \frac{2}{25} +\end{align*} +\pagebreak + +\exercise{10} +\begin{enumerate}[(a)] + \item Venn diagram: + \begin{venndiagram3sets}[ + labelOnlyB={5}, + labelABC={2}, + labelNotABC={0}, + overlap=1cm] + \setpostvennhook{ + \draw[-stealth] (labelA) -- ++(135:2.5cm) node[left]{15}; + \draw[-stealth] (labelB) -- ++(45:2.5cm) node[right]{8}; + \draw[-stealth] (labelC) -- ++(-90:2cm) node[below]{12}; + \draw[-stealth] (0,0) -- ++(-135:1.5cm) node[below]{27}; + \draw[-stealth] (1.4,1.9) -- ++(180:2.5cm) node[left]{4}; + \draw[-stealth] (2.4,3.6) -- ++(-157:3.8cm); + \draw[-stealth] (4,3) -- ++(-70:4cm) node[below]{21}; + \draw[-stealth] (1,2.5) -- ++(-40:5.3cm); + \draw[-stealth] (3.3,1.5) -- ++(-50:3cm); + } + \end{venndiagram3sets} + \item The number of tourists who had not visited Burundi: + \[n\left(B^\C\right) = n(\Omega) - n(B) = 27 - 8 = 19\] + \item The number of tourists who had not visited Cameroon + unless they had visited all three countries: + \[n\left((A\cap B)\cup C^\C\right) + = n(\Omega) - n(C) + n(A\cap B\cap C) + = 27 - 12 + 2 = 17\] + \item For the randomly selected tourist to have visited at least + two countries, that person must not visited only one country. + Thus the event can be denoted as + \[d = \Omega\setminus((A\setminus B\setminus C)\cup + (B\setminus C\setminus A)\cup(C\setminus A\setminus B))\] + Since the selection is random, the event's probability can be calculated as + \[P(d) = 1 - \frac{n((A\setminus B\setminus C)\cup + (B\setminus C\setminus A)\cup(C\setminus A\setminus B))}{n(\Omega)} + = 1 - \frac{21}{27} = \frac{2}{9}\] +\end{enumerate} +\pagebreak + +\section{Basic Probability 2} +\exercise 1 Let $A$ be the event that the chosen transistor is defective, +$B$ be the event that the chosen one is partially defective +and $C$ be the event that the chosen one is acceptable. +$A$, $B$ and $C$ are disjoint and $A\cup B\cup C = \Omega$, thus +\[n(\Omega) = n(A) + n(B) + n(C) = 5 + 10 + 25 = 40\] + +The probability that the chosen transistor does not immediately fail is +$P\left(A^\C\right) = 1 - P(A) = 1 - n(A)/n(\Omega) = 1 - 5/40 = 7/8$. + +Given this condition, the probability the chosen transistor is acceptable is +\[P\left(C|A^\C\right) = \frac{P\left(C\cap A^\C\right)}{P\left(A^\C\right)} += \frac{P(C)}{7/8} += \frac{8n(C)}{7n(\Omega)} = \frac{8\cdot 25}{7\cdot 40} = \frac 5 7\] + +\exercise 2 Denote the outcomes of tossing a coin as $H$ (head) and $T$ (tail). +\begin{enumerate}[(a)] + \item Consider tossing a coin $n$ times, the sample space is + $\Omega = \{H, T\}^n$. Let $A$ be the event of getting at least a head, + $A^\C$ would then be getting all tails ($\{T\}^n$). Suppose the chance of getting + head and tail are equal, + \[P\left(A^\C\right) = \frac{n\left(A^\C\right)}{n(\Omega)} = \frac{1}{2^n} + \Longrightarrow P(A) = 1 - P\left(A^\C\right) = \frac{2^n - 1}{2^n}\] + \item For $n = 4$, $P(A) = \dfrac{2^4 - 1}{2^4} = \dfrac{15}{16}$. + \item Consider rolling a six-sided die $n$ times, the sample space is + \[\Omega = \{1, 2, 3, 4, 5, 6\}^n\] + Let $B$ be the event of getting a six, $B^\C = \{1, 2, 3, 4, 5\}^n$. + Therefore the probability of $B$ is + \[P(B) = 1 - P\left(B^\C\right) + = 1 - \frac{n\left(B^\C\right)}{n(\Omega)} = 1 - \frac{5^n}{6^n}\] + For $n = 4$, $P(B) = 671/1296$. + \item For $P(B) = 0.99$, + \[1 - \left(\frac{5}{6}\right)^n = 0.99 + \iff \left(\frac{5}{6}\right)^n = 0.01 + \iff n = \log_{5/6}0.01 \approx 25\] +\end{enumerate} + +\exercise 3 Let $B$ be the event that the woman rides the bicycle to work, +$B^\C$ would be that she ride the scooter. Let $L$ be that she is late, +\begin{align*} + P(B) &= 0.7\\ + P\left(B^\C\right) &= 0.3\\ + P(L|B) &= 0.03\\ + P\left(L|B^\C\right) &= 0.02 +\end{align*} +\begin{enumerate}[(a)] + \item By Total Probability Theorem, the probability the woman + is late for work is + \[P(L) = P(B)\cdot P(L|B) + P\left(B^\C\right)\cdot P\left(L|B^\C\right) + = 0.7\cdot 0.03 + 0.3\cdot 0.02 = 0.027\] + \item The probability she is not late for work is + \[P\left(L^\C\right) = 1 - P(L) = 1 - 0.027 = 0.973\] + Since the woman is expected to be on time roughly 223 days a year, + she goes to work $223 / P\left(L^\C\right) \approx 229$ days a year. +\end{enumerate} + +\exercise 4 Consider flipping the coin twice, the sample space is +\[\Omega = \{(H, H), (H, T), (T, H), (T, T)\}\] +where $H$ stands for head and $T$ stands for tail. + +Denote getting a head from the first flip as $H_1$ and getting a head +from the second one as $H_2$. Assume that $P(H_1) = P(H_2) = 0.6$. +It is obvious that these two events are independent, or in other words +\[P(H_1\cap H_2) = P(H_1)\cdot P(H_2)\] + +Similarly, +\begin{align*} + P\{(H, T)\} &= P\left(H_1\cap H_2^\C\right) + = P\left(H_1\right)\cdot\left(1 - P\left(H_2\right)\right) = 0.24\\ + P\{(T, H)\} &= P\left(H_1^\C\cap H_2\right) + = \left(1 - P\left(H_1\right)\right)\cdot P\left(H_2\right) = 0.24 +\end{align*} + +Therefore if Minh and Nam flip the coin twice for both head and tail +and choose K-pop when they get a head first and US music otherwise, +the genre would be chosen equally even. + +\exercise 5 Place three maths, two history and four biology book on a shelf. +\begin{enumerate}[(a)] + \item There would be $(3 + 2 + 4)! = 362880$ ways to do it + without any further restriction. + \item If each subject needs to stay together, + there are $3! 2! 4! 3! = 1728$ ways. + \item If only biology books must stay together, + we can do it in $4!(3 + 2 + 1)!$ or 17280 ways. +\end{enumerate} + +\exercise 6 Seat six people around a table. +\begin{enumerate}[(a)] + \item If they can sit anywhere, there are $6!/6 = 120$ arrangements. + \item If two particular people must sit next to each other, + there are $2\cdot 5!/5$ or 48 arrangements; thus if those two cannot + sit side-by-side, the figure is $120 - 48 = 72$. +\end{enumerate} + +\exercise 7 Let $\omega$ be the set of cards in a standard 52-card deck. +Shuffle the deck an draw seven cards, +the sample space of this probability model is $\Omega = \binom{\omega}{7}$, +$n(\Omega) = \binom{52}{7}$. +\begin{enumerate}[(a)] + \item Let $A$ be the event that exactly three of the drawn ones are aces, + \[n(A) = \binom{4}{3}\binom{48}{4} + \Longrightarrow P(A) = \frac{n(A)}{n(\Omega)} = \frac{9}{1547}\] + \item Let $K$ be the event that exactly two of the drawn ones are kings, + \[n(K) = \binom{4}{2}\binom{48}{5} + \Longrightarrow P(K) = \frac{n(K)}{n(\Omega)} = \frac{594}{7735}\] + \item The probability that exactly three aces and two kings are drawn is + \[n(A\cap K) = \binom{4}{3}\binom{4}{2}\binom{44}{2} + \Longrightarrow P(A\cap K) = \frac{n(A\cap K)}{n(\Omega)} + = \frac{1419}{8361535}\] + + Thus probability that either exactly three aces or two kings are drawn is + \[P(A\cup K) = P(A) + P(K) - P(A\cap K) = \frac{137868}{1672307}\] +\end{enumerate} + +\exercise 8 Let $M$ be the event that a red marble is picked +and $C$ be the event of getting head from tossing the coin, we have +\begin{align*} + P(M|C) &= 0.6\\ + P\left(M|C^\C\right) &= 0.2\\ + P(C) = P\left(C^\C\right) &= 0.5 +\end{align*} + +\begin{enumerate}[(a)] + \item By Total Probability Theorem, the probability a red marble is picked is + \[P(M) = P(C)\cdot P(M|C) + P\left(C^\C\right)\cdot P\left(M|C^\C\right) + = 0.4\] + \item The probability that a blue marble is picked is + \[P\left(M^\C\right) = 1 - P(M) = 0.6\] + \item The probability of getting a head if the red marble is picked is + \[P(C|M) = \frac{P(C\cap M)}{P(M)} = \frac{P(C)\cdot P(M|C)}{P(M)} + = \frac{0.5\cdot 0.6}{0.4} = 0.75\] +\end{enumerate} + +\exercise 9 Consider $n$ random people and their birthdays, assuming +that all 366 birthdays are equally likely\footnote{If you are wondering +how one could be equally likely to be born on the leap day, then well, +the distribution of birthdays on other days is in fact not uniform either. +\textit{Don't complicate it, don't drive yourself insane!}}. +The size of the sample space is $n(\Omega_n) = 366^n$. + +Let $A_n$ be the event that no two of these $n$ people to celebrate +their birthday on the same day, $n(A_n) = \prod_{i=0}^{n-1}(366-i)$. +Thus the probability of this is +\[P(A_n) = \frac{n(A_n)}{n(\Omega_n)} = \prod_{i=1}^{n-1}\frac{366 - i}{366}\] + +Since $P(A_{23}) < 0.5 < P(A_{22})$, $n$ needs to be at least 23 +for the probability to be less than 0.5. + +\exercise{10} The reasoning is not correct because: +\begin{itemize} + \item If he is not to be released, the answer from the guard will be + both of other prisoners, and everyones' fate will be known. + \item Otherwise, in case the guard only gives one name, + our protagonist will sure be released. +\end{itemize} +\pagebreak + +\section{Discrete Random Variable 1} +\subsection{Discrete Random Variable and PMF} +\exercise 1 Consider a fair coin. +\begin{enumerate}[(a)] + \item Toss it twice and let $X$ be the number of heads, + $X$ would be a binomial random variable + \begin{align*} + X\colon \Omega &\to \{0, 1, 2\}\\ + \omega &\mapsto x + \end{align*} + whose probability mass function is + \[p_X(x) = \binom{2}{x}\cdot 0.5^x\cdot 0.5^{2-x} + = \frac{1}{2x!(2-x)!}\] + + Therefore PMF of $X$ for each case is + \begin{align*} + p_X(0) = p_X(2) &= \frac{1}{2\cdot 0!2!} = \frac{1}{4}\\ + p_X(1) &= \frac{1}{2\cdot 1!1!} = \frac{1}{2} + \end{align*} + \item Toss it thrice and let $Y$ be the number of heads, + $Y$ would be a binomial random variable + \begin{align*} + Y\colon \Omega &\to \{0, 1, 2, 3\}\\ + \omega &\mapsto y + \end{align*} + whose probability mass function is + \[p_Y(y) = \binom{3}{y}\cdot 0.5^y\cdot 0.5^{3-y} + = \frac{3}{4y!(3-y)!}\] + + Therefore PMF of $Y$ for each case is + \begin{align*} + p_Y(0) = p_Y(3) &= \frac{3}{4\cdot 0!3!} = \frac{1}{8}\\ + p_Y(1) = p_Y(2) &= \frac{3}{4\cdot 1!2!} = \frac{3}{8} + \end{align*} +\end{enumerate} + +\exercise 2 Toss a pair of fair siz-sided dice +and let $X$ be the sum of the points +\begin{align*} + X\colon \Omega &\to [2, 12]\cap\mathbb Z\\ + \omega &\mapsto x +\end{align*} +with $\Omega = S^2 = \{1, 2, 3, 4, 5, 6\}^2 \Longrightarrow n(\Omega) = 36$. + +\begin{enumerate}[(a)] + \item $X$ is a random variable whose PMF is + \begin{align*} + p_X(2) &= P\{(1,1)\} = \frac{1}{36}\\ + p_X(3) &= P\{(1,2), (2,1)\} = \frac{1}{18}\\ + p_X(4) &= P\{(1,3), (2,2), (3,1)\} = \frac{1}{12}\\ + p_X(5) &= P\{(1,4), (2,3), (3,2), (4,1)\} = \frac{1}{9}\\ + p_X(6) &= P\{(1,5), (2,4), (3,3), (4,2), (5,1)\} = \frac{5}{36}\\ + p_X(7) &= P\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\} = \frac{1}{6}\\ + p_X(8) &= P\{(2,6), (3,5), (4,4), (5,3), (6,2)\} = \frac{5}{36}\\ + p_X(9) &= P\{(3,6), (4,5), (5,4), (6,3)\} = \frac{1}{9}\\ + p_X(10) &= P\{(4,6), (5,5), (6,4)\} = \frac{1}{12}\\ + p_X(11) &= P\{(5,6), (6,5)\} = \frac{1}{18}\\ + p_X(12) &= P\{(6,6)\} = \frac{1}{36} + \end{align*} + \item The graph of $p_X(x)$: + + \begin{tikzpicture} + \begin{axis}[xlabel={$x$}, ylabel={$p_X(x)$}] + \addplot[ycomb, samples at={2,3,...,12}]{1/6-abs(x-7)/36}; + \end{axis} + \end{tikzpicture} +\end{enumerate} +\pagebreak + +\exercise 3 Denote the event of winning, tying and losing the first game as +$A_2$, $A_1$ and $A_0$ respectively, we get $P(A_2) = P(A_1) = 0.2$ +and $P(A_0) = 0.6$. Similarly, let $B_2$, $B_1$ and $B_0$ in that order +be the event MIT soccer team winning, tying and losing the second game, +we get $P(B_2) = P(B_1) = 0.35$ and $P(B_0) = 0.3$. + +Let $A$ and $B$ be the random variable satisfying +\begin{align*} + A &= \begin{cases} + 2\text{ if }A_2\\ + 1\text{ if }A_1\\ + 0\text{ if }A_0 + \end{cases} + \Longrightarrow\begin{cases} + p_A(2) = p_A(1) = 0.2\\ + p_A(0) = 0.6 + \end{cases}\\ + B &= \begin{cases} + 2\text{ if }B_2\\ + 1\text{ if }B_1\\ + 0\text{ if }B_0 + \end{cases} + \Longrightarrow\begin{cases} + p_B(2) = p_B(1) = 0.35\\ + p_B(0) = 0.3 + \end{cases} +\end{align*} +then the number of points the team earns over the weekend is $X = A + B$. + +Since the outcome of the two games are independent, +\begin{align*} + p_X(0) &= p_A(0)\cdot p_B(0) = 0.18\\ + p_X(1) &= p_A(0)\cdot p_B(1) + p_A(1)\cdot p_B(0) = 0.27\\ + p_X(2) &= p_A(0)\cdot p_B(2) + p_A(1)\cdot p_B(1) + p_A(2)\cdot p_B(0) + = 0.34\\ + p_X(3) &= p_A(1)\cdot p_B(2) + p_A(2)\cdot p_B(1) = 0.14\\ + p_X(4) &= p_A(2)\cdot p_B(2) = 0.07 +\end{align*} + +\subsection{Expectation of Random Variables} +\exercise 4 Given a random variable +\[X = \begin{cases} + -2&\text{ with probability of 1/3}\\ + 3&\text{ with probability of 1/2}\\ + 1&\text{ with probability of 1/6} +\end{cases}\] +\begin{align*} + \E[X] &= \sum_{x\in\{-2, 1, 3\}}xp_X(x) = 1\\ + \E[2X+5] &= \sum_{x\in\{-2, 1, 3\}}(2x + 5)p_X(x) = 7\\ + \E\left[X^2\right] &= \sum_{x\in\{-2, 1, 3\}}x^2 p_X(x) = 6 +\end{align*} + +\exercise 5 Consider the genders of the three children, and assume +that both genders\footnote{You SJWs really need to calm down. +This is just a mathematical problem.} are equally likely. + +Let $X$ be the number of girls, $X$ is a binomial random variable +whose probability mass function is +\begin{multline*} + p_X(x) = \binom{3}{x}\cdot 0.5^x\cdot 0.5^{3-x} = \frac{3}{4x!(3-x)!}\\ + \Longrightarrow \E[X] = \sum_{x=0}^3\frac{3x}{4x!(3-x)!} = \frac 3 2 +\end{multline*} + +\exercise 6 Consider rolling a fair six-sided die, the sample space is +$\Omega = \{1, 2, 3, 4, 5, 6\}$. Let $X$ be a random variable given by +\begin{multline*} + X(\omega) = \begin{cases} + -1&\text{ if }\omega \in \{1, 2, 3\}\\ + 2&\text{ if }\omega \in \{4, 5\}\\ + 8&\text{ if }\omega = 6 + \end{cases} + \Longrightarrow\begin{dcases} + p_X(-1) = \frac{3}{6} = \frac{1}{2}\\ + p_X(2) = \frac{2}{6} = \frac{1}{3}\\ + p_X(8) = \frac{1}{6} + \end{dcases}\\ + \Longrightarrow \E[X] = \frac{-1}{2} + \frac{2}{3} + \frac{4}{3} = \frac{3}{2} +\end{multline*} + +Practically, this means that at the end of the day, +it is very unlikely that the house will win. + +\exercise 7 Let $X$ be the prize in dollars on a randomly chosen +lottery ticket, its PMF is +\begin{align*} + p_X(100) &= \frac{5}{10\,000} = \frac{1}{2\,000}\\ + p_X(25) &= \frac{20}{10\,000} = \frac{1}{5\,000}\\ + p_X(5) &= \frac{200}{10\,000} = \frac{1}{500}\\ + p_X(0) &= \frac{10\,000-200-20-5}{10\,000} = \frac{391}{400} +\end{align*} + +Thus the expected value for a ticket's value in dollars is +\[\E[X] = \sum_x x\cdot p_X(x) = \frac{13}{200}\] +or 6.5 cents. + +\exercise 8 Let $X$ be the prize in dollars on a randomly chosen +raffle ticket, its PMF is +\begin{align*} + & p_X(1998) = p_X(999) = \frac{1}{5000}\\ + & p_X(498) = \frac{2}{5000} = \frac{1}{2500}\\ + & p_X(98) = \frac{5}{5000} = \frac{1}{1000}\\ + & p_X(-2) = \frac{5000-5-2-1-1}{5000} = \frac{4991}{5000} +\end{align*} + +Thus the expected value in dollars to get when buying a ticket is +\[\E[X] = \sum_x x\cdot p_X(x) = \frac{-11}{10}\] +or to lose \$1.1. + +\subsection{Variance and Standard Deviation} +\exercise{9} Given the outcome $X$ from rolling a fair six-sided die. +\begin{align*} + \E[X] &= \frac{1+2+3+4+5+6}{6} = \frac{7}{2}\\ + \Longrightarrow \var(X) &= \E\left[(X - \E[X])^2\right] + = \sum_x \frac{(x - \frac{7}{2})^2}{6} = \frac{35}{24}\\ + \Longrightarrow \sigma_X &= \sqrt{\var(X)} = \sqrt\frac{35}{24} +\end{align*} + +\exercise{10} Based on the result of exercise 2, +\[E[X] = 7,\qquad\var(X) = \frac{35}{6},\qquad\sigma_X = \sqrt\frac{35}{6}\] + +\exercise{11} Given the integral random variable $X$ with PMF +\[p_X(x) = \begin{dcases} + \frac{1}{9} &\text{ if } x \in [-4, 4]\\ + 0 &\text{ otherwise} +\end{dcases}\] + +Let $S = \{-4, -3, -2, -1, 0, 1, 2, 3, 4\}$, +\begin{multline*} + \E[X] = \sum_{x\in\mathbb Z}x\cdot p_X(x) + = \sum_{x\in S}\frac{x}{9} + \sum_{x\in\mathbb Z\setminus S}x\cdot 0 = 0\\ + \Longrightarrow \var(X) = \E\left[X^2\right] + = \sum_{x\in S}\frac{x^2}{9} = \frac{20}{3} +\end{multline*} + +\exercise{12} Given the integral random variable $X$ with PMF +\[p_X(x) = \begin{dcases} + \frac{x^2}{a} &\text{ if } x \in [-3, 3]\\ + 0 &\text{ otherwise} +\end{dcases}\] +\begin{enumerate}[(a)] + \item Let $S = \{-3, -2, -1, 0, 1, 2, 3\}$. Since + \begin{multline*} + \sum_{x\in\mathbb Z}p_X(x) = 1 + \iff \sum_{x\in S}\frac{x^2}{a} + \sum_{x\in\mathbb Z\setminus S}0 = 1 + \iff a = \sum_{x\in S}x^2 = 28\\ + \Longrightarrow \E[X] = \sum_{x\in S}\frac{x^3}{28} = 0 + \end{multline*} + \item Let $Z = (X - \E[X])^2 = X^2$, + the range of $Z$ is $\{z^2\mid z\in\mathbb Z\}$. + For all $z > 9$, it is trivial that $p_Z(z) = 0$. Otherwise, + \begin{align*} + p_Z(0) &= P(Z = 0) = P(X=0) = p_X(0) = \frac{0^2}{28} = 0\\ + p_Z(1) &= P(X = \pm 1) = p_X(-1) + p_X(1) + = \frac{(-1)^2}{28} + \frac{1^2}{28} = \frac{1}{14}\\ + p_Z(4) &= P(X = \pm 2) = p_X(-2) + p_X(2) + = \frac{(-2)^2}{28} + \frac{2^2}{28} = \frac{2}{7}\\ + p_Z(9) &= P(X = \pm 3) = p_X(-3) + p_X(3) + = \frac{(-3)^2}{28} + \frac{3^2}{28} = \frac{9}{14} + \end{align*} + \item The variance of $X$ is + \[\var(X) = \E\left[(X - \E[X])^2\right] + = \E[Z] = 1\cdot\frac{1}{14} + 4\cdot\frac{2}{7} + 9\cdot\frac{9}{14} = 7\] +\end{enumerate} +\pagebreak + +\section{Discrete Random Variable 2} +\subsection{Conditional PMF and Expectation} +\exercise 1 Compute conditional PMF: +\begin{enumerate}[(a)] + \item Let $X$ be the roll if a fair six-sided die and $A$ be the event that + the roll is an number greater or equal to 4, we have $A = \{X \ge 4\}$ + and $P(A) = 0.5$, thus + \[p_{X|A}(x) = \frac{P(\{X = x\}\cap\{X \ge 4\})}{P(A)}\] + + For $x \in \{1, 2, 3\}$, $\{X = x\}\cap\{X \ge 4\} = \varnothing$ + so $p_{X|A}(x) = 0/0.5 = 0$. + + For $x \in \{4, 5, 6\}$, $\{X = x\}\cap\{X \ge 4\} = \{x\}$, + \[p_{X|A}(x) = \frac{1/6}{0.5} = \frac{1}{3}\] + \item Let $X$ represent number of heads from the three-time toss + of a fair coin and $B = \{X \ge 2\}$, + $P(B) = \binom{3}{2}0.5^3 + \binom{3}{3}0.5^3 = 0.5$. + \begin{multline*} + p_{X|B}(x) = \frac{P(\{X = x\}\cap B)}{P(B)} + = \frac{P(\{X = x\}\cap\{X \ge 2\})}{0.5}\\ + \Longrightarrow\begin{dcases} + p_{X|B}(0) = p_{X|B}(1) = 0\\ + p_{X|B}(2) = \frac{\binom{3}{2}0.5^3}{0.5} = \frac{3}{4}\\ + p_{X|B}(3) = \frac{\binom{3}{3}0.5^3}{0.5} = \frac{1}{4} + \end{dcases} + \end{multline*} + \item Let $X$ be the roll of a pair of fair dice and $C = \{X = 7\}$. + As shown in the previous section, $P(C) = 1/6$ and thus + \[p_{X|C}(x) = \begin{cases} + 1\text{ if }x = 7\\ + 0\text{ if }x \neq 7 + \end{cases}\] +\end{enumerate} + +\exercise 2 Consider the destination of the message and denote the event +it arrives at Liberty City, Chicago and San Fierro as $B$, $C$ and $F$ +respectively, we have $B\cup C\cup F = \Omega$. The expected transit time is +\begin{align*} + \E[X] &= P(B)\E[X|B] + P(C)\E[X|C] + P(F)\E[X|F]\\ + &= 0.5\cdot 0.05 + 0.3\cdot 0.1 + 0.2\cdot 0.3 = 0.115 +\end{align*} + +\exercise 3 Let $V$ and $T$ be respectively the speed (in mph) and time +(in hours) Alyssa get to class. Denote the event she walk to class as $W$, +$P(W) = 0.6$, +\[\begin{cases} + \E[V|W] = 5\\ + \E\left[V|W^\C\right] = 30 +\end{cases} +\Longrightarrow\begin{dcases} + \E[T|W] = \frac{2}{5}\\ +\E\left[T|W^\C\right] = \frac{1}{15} +\end{dcases}\] +\begin{enumerate}[(a)] + \item The expected value of Alyssa's speed is + \[\E[V] = P(W)\E[V|W] + P\left(W^\C\right)\E\left[V|W^\C\right] + = 0.6\cdot 5 + (1 - 0.6)30 = 15\] + \item The expected value of the time Alyssa she takes to get to class is + \[\E[T] = P(W)\E[T|W] + P\left(W^\C\right)\E\left[T|W^\C\right] + = 0.6\cdot\frac{2}{5} + (1 - 0.6)\frac{1}{15} = \frac{4}{15}\] +\end{enumerate} + +\exercise 4 With $X$ being the number of tries until the program works +correctly and $p$ being the probability each try succeed, +we have $\mathrm{range}(X) = \mathbb N^*$ and $p_X(x) = (1 - p)^{x - 1}p$. +The mean of $X$ is +\begin{align*} + \E[X] &= \sum_{x\in\mathbb N^*}x\cdot p_X(x)\\ + &= \sum_{x\in\mathbb N^*}x(1 - p)^{x - 1}p\\ + &= -p\sum_{x\in\mathbb N^*}\frac{\ud (1 - p)^x}{\ud p}\\ + &= -p\frac{\ud}{\ud p}\left(\sum_{x\in\mathbb N}(1 - p)^{x} - 1\right)\\ + &= -p\frac{\ud}{\ud p}\left(\frac{1}{p} - 1\right)\\ + &= \frac{1}{p} +\end{align*} + +Similarly, +\begin{align*} + \E\left[X^2\right] &= p\sum_{x\in\mathbb N^*}x^2(1 - p)^{x - 1}\\ + &= -p\frac{\ud}{\ud p} + \left(\frac{1 - p}{p}\sum_{x\in\mathbb N^*}x(1 - p)^{x - 1}p\right)\\ + &= -p\frac{\ud}{\ud p}\left(\frac{1}{p^2} - \frac{1}{p}\right)\\ + &= \frac{2}{p^2} - \frac{1}{p} +\end{align*} + +Therefore the variance of $X$ is +\[\var(X) = \E\left[X^2\right] - (\E[X])^2 = \frac{1}{p^2} - \frac{1}{p}\] + +\subsection{Joint PMF and independent variables} +\exercise 5 Consider two independent coin tosses, +each with a 3/4 probability of a head, +and let $X$ be the number of heads obtained, +$X$ is a binomial random variable. +\[\E[X] = 0p_X(0) + 1p_X(1) + 2p_X(2) += \binom{2}{1}\frac{3}{4}\left(1 - \frac{3}{4}\right) ++ 2\binom{2}{2}\left(\frac{3}{4}\right)^2 += \frac{3}{2}\] + +\exercise 6 Let $X$ be the number of red traffic lights Alyssa encounters, +$X$ is a binomial random variable whose PMF is +\[p_X(x) = \binom{4}{x}0.5^4\] + +The mean of $X$ is +\[\E[X] = \frac{1}{16}\sum_{x=0}^4 x\binom{4}{x} = 2\] + +The variance of $X$ is +\[\var(X) = \E\left[X^2\right] - (\E[X])^2 += \frac{1}{16}\sum_{x=0}^4 x^2\binom{4}{x} - 4 = 1\] + +\exercise 7 Let $X_i$ be 1 if the $i$th person gets his or her own hat and 0 +otherwise, then for all positive integer $i \le n$ +\[\E[X_i] = p_{X_i}(1) = \frac{(n - 1)!}{n!} = \frac{1}{n}\] +since if we fix one hat to its owner, there are $(n - 1)!$ arrangements +for the rest. Due to the linearity property of expectation, +\[\E[X] = \E\left[\sum_{i=1}^n X_i\right] = \sum_{i=1}^n\E[X_i] = 1\] + +\exercise 8 Consider four independent rolls of a six-sided die. +Let $X$ and $Y$ be the number of ones and twos obtained respectively, +both are binomial random variables: +\[p_X(k) = p_Y(k) += \binom{4}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{4 - k} += \binom{4}{k}\frac{5^{4 - k}}{1296}\] + +Given $Y = y$, $X$ is the number of ones in the remaining $4 - y$ rolls, +each of which can take the values other than two equally likely: +\[p_{X|Y}(x|y) += \binom{4 - y}{x}\left(\frac{1}{5}\right)^x\left(\frac{4}{5}\right)^{4 - y - x} += \binom{4 - y}{k}\frac{4^{4 - y - x}}{625}\] + +Thus the joint PMF of $X$ and $Y$ is +\[p_{X,Y}(x, y) = p_Y(y)p_{X|Y}(x|y) += \binom{4}{x}\binom{4 - y}{k}\frac{5^{4 - x}4^{4 - y - x}}{810\,000}\] + +\exercise 9 Given the joint PMF of two discrete random variables $X$ and $Y$ +\[p_{X,Y}(x, y) = \begin{cases} + c(2x + y) &\text{where }(x, y)\in\{0, 1, 2\}\times\{0, 1, 2, 3\}\\ + 0 &\text{otherwise} +\end{cases}\] +\begin{enumerate}[(a)] + \item Consider all cases: + \begin{align*} + \sum_x\sum_y p_{X,Y}(x, y) = 1 + &\iff \sum_{x=0}^2\sum_{y=0}^3 c(2x + y) = 1\\ + &\iff \sum_{x=0}^2 c(8x + 6) = 1\\ + &\iff c(24 + 18) = 1\\ + &\iff c = \frac{1}{42} + \end{align*} + \item $P(X = 2, Y = 1) = (2\cdot 2 + 1)/42 = 5/42$. + \item Similarly, $P(X \ge 1, Y \le 2) = 4/7$. + \item The marginal PMF of $X$: + \[p_X(x) = \sum_y p_{X,Y}(x, y) + = \sum_{y=0}^3\frac{2x + y}{42} = \frac{4x + 3}{21}\] + \item The marginal PMF of $Y$: + \[p_Y(y) = \sum_x p_{X,Y}(x, y) + = \sum_{x=0}^2\frac{2x + y}{42} = \frac{2 + y}{14}\] + \item Since $p_X(2)p_Y(1) \neq p_{X,Y}(2, 1)$, + the two variables are dependent. + \item Given $X = 2$, + \[p_{Y|X}(y|2) = \frac{p_{X,Y}(2, y)}{p_X(2)} = \frac{4 + y}{22} + \Longrightarrow p_{Y|X}(1|2) = \frac{5}{22}\] + \item Given $Y = 2$, + \[p_{X|Y}(x|2) = \frac{p_{X,Y}(x, 2)}{p_Y(2)} = \frac{x + 1}{6} + \Longrightarrow p_{X|Y}(3|2) = \frac{2}{3}\] +\end{enumerate} + +\exercise{10} Given the joint PMF of two discrete random variables $X$ and $Y$ +\[p_{X,Y}(x, y) = \begin{cases} + cxy &\text{where }(x, y)\in\{1, 2, 3\}\times\{1, 2, 3\}\\ + 0 &\text{otherwise} +\end{cases}\] +\begin{enumerate}[(a)] + \item Consider all cases: + \begin{align*} + \sum_x\sum_y p_{X,Y}(x, y) = 1 + &\iff \sum_{x=1}^3\sum_{y=1}^3 cxy = 1\\ + &\iff 36c = 1\\ + &\iff c = \frac{1}{36} + \end{align*} + \item $P(X = 2, Y = 3) = 1/6$. + \item Similarly, $P(1\le X\le 2, Y\le 2) = 1/4$. + \item By the result of (e), $P(X\ge 2) = 5/6$, + $P(Y < 2) = P(Y = 1) = P(X = 1) = 1/6$ and $P(Y = 3) = 1/2$. + \item The marginal PMF of $X$: + \[p_X(x) = \sum_y p_{X,Y}(x, y) + = \sum_{y=1}^3\frac{xy}{36} = \frac{x}{6}\] + The marginal PMF of $Y$: + \[p_Y(y) = \sum_x p_{X,Y}(x, y) + = \sum_{x=1}^3\frac{xy}{36} = \frac{y}{6}\] + \item Since $p_{X,Y}(x, y) = p_X(x)p_Y(y)$, $X$ and $Y$ are independent. +\end{enumerate} +\pagebreak + +\section{Continuous Random Variable 1} +\subsection{PDF and CDF} +\exercise 1 Given a PDF such that +\[f_X(x) = \begin{cases} + cx^2 &\text{if }0 < x < 3\\ + 0 &\text{otherwise} +\end{cases}\] +\begin{enumerate}[(a)] + \item By the normalization property, + \[1 = \int_{-\infty}^{\infty}f_X(x)\ud x = \int_0^3 cx^2\ud x = 9c + \Longrightarrow c = \frac{1}{9}\] + \item $\displaystyle P(1 < X < 2) = \int_1^2\frac{x^2}{9}\ud x = \frac{7}{27}$ + \item The CDF of $X$: + \[F_X(x) = \begin{dcases} + 0 &\text{if }x \le 0\\ + 1 &\text{if }x \ge 3\\ + \int_0^x\frac{t^2}{9}\ud t = \frac{x^3}{27} &\text{otherwise} + \end{dcases}\] + \item $P(1 < X \le 2) = F_X(2) - F_X(1) + = \dfrac{8}{27} - \dfrac{1}{27} = \dfrac{7}{27}$ +\end{enumerate} + +\exercise 2 Denote the event that the day is sunny as $A$, $P(A) = 2/3$. +\begin{align*} + f_{X|A}(x) &= \begin{cases} + b &\text{if }15 \le x \le 20\\ + 0 &\text{otherwise} + \end{cases} + \Longrightarrow \int_{15}^{20}b\ud x = 1 + \iff b = \frac{1}{5}\\ + f_{X|A^\C}(x) &= \begin{cases} + c &\text{if }20 \le x \le 25\\ + 0 &\text{otherwise} + \end{cases} + \Longrightarrow \int_{20}^{25}c\ud x = 1 + \iff c = \frac{1}{5} +\end{align*} + +By the total probability theorem, +\[f_X(x) = P(A)f_{X|A}(x) + P\left(A^\C\right)f_{X|A^\C}(x) + = \begin{dcases} + \frac{2}{15} &\text{if }15 \le x < 20\\ + \frac{1}{15} &\text{if }20 \le x < 25\\ + 0 &\text{otherwise} + \end{dcases}\] +\pagebreak + +\exercise 3 Given a random variable $X$ with the PDF +$f_X(x) = \dfrac{c}{x^2 + 1}$. +\begin{enumerate}[(a)] + \item By the normalization property, + \[\int_{-\infty}^\infty\frac{c}{x^2 + 1}\ud x = 1 + \iff \left.c\arctan x\right|_{-\infty}^\infty = 1 + \iff c\pi = 1 \iff c = \frac{1}{\pi}\] + \item The probability that $X^2$ to lie between 1/3 and 1 is + \begin{align*} + P\left(\frac{1}{3} < X^2 < 1\right) + &= P\left(-1 < X < \frac{-1}{\sqrt 3}\right) + + P\left(\frac{1}{\sqrt 3} < X < 1\right)\\ + &= \left.\frac{\arctan x}{\pi}\right|_{-1}^{\frac{-1}{\sqrt 3}} + + \left.\frac{\arctan x}{\pi}\right|_{\frac{1}{\sqrt 3}}^{1} + = \frac{1}{6} + \end{align*} + \item The CDF of $X$: + \[F_X(x) = \int_{-\infty}^{x}\frac{1}{\pi\left(x^2 + 1\right)}\ud t + = \frac{\arctan x}{\pi} + \frac{1}{2}\] +\end{enumerate} + +\exercise 4 Given a random variable $X$ with the CDF +\[F_X(x) = \begin{cases} + 1 - \exp(-2x) &\text{if } x \ge 0\\ + 0 &\text{otherwise} + \end{cases}\] +\begin{enumerate}[(a)] + \item Let $g$ be the antiderivative of $f_X$, $g$ is a constant function + in $(-\infty, 0)$ and for all positive $x$ + \[g(x) = g(0) + 1 - \exp(-2x) + \Longrightarrow f_X(x) = 2\exp(-2x)\] + \item The probability that $X > 2$ is + \[P(X > 2) = P(\Omega) - P(X \le 2) = 1 - F_X(2) = \frac{1}{e^4}\] + \item The probability that $-3 < X \le 4$ is + \[P(-3 < X \le 4) = F_X(4) - F_X(-3) = 1 - \frac{1}{e^8}\] +\end{enumerate} + +\exercise 5 Given random variable with the following PDF: +\[f_X(x) = \begin{dcases} + \frac{10}{x^2} &\text{if }x > 10\\ + 0 &\text{otherwise} +\end{dcases} +\Longrightarrow F_X(x) = \begin{dcases} + 1 - \frac{10}{x} &\text{if }x > 10\\ + 0 &\text{otherwise} +\end{dcases}\tag{b}\] + +\[P(X > 20) = P(\Omega) - F_X(20) = 1 - 1 + \frac{10}{20} = \frac{1}{2}\tag{a}\] + +Let $Y$ be the number out of six devices that will function for at least +15 hours, $Y$ is a binomial random variable whose PMF is +\begin{align*} + p_Y(y) &= \binom{6}{y} P^y(X \ge 15) P^{6-y}(X < 15)\\ + &= \binom{6}{y} \left(1 - F_X(15)\right)^y F_X^{6-y}(15)\\ + &= \binom{6}{y} \left(\frac{10}{15}\right)^y + \left(1 - \frac{10}{15}\right)^{6-y}\\ + &= \binom{6}{y} \frac{2^y}{3^6} +\end{align*} + +Denote $A$ as the event that at least three out of six devices will function +for at least 15 hours, +\[P(A) = \sum_{y=3}^6 p_Y(y) = \frac{656}{729}\tag{c}\] + +\subsection{Expectation, Variance and STD\protect\footnote{No, not that STD.}} +\exercise 6 Given a random variable $X$ with the PDF +$f_X(x) = \frac{\lambda}{2}\exp(-\lambda|x|)$. +\begin{align*} + \int_{-\infty}^\infty f_X(x)\ud x + &= \int_{-\infty}^0 \frac{\lambda}{2}\exp(\lambda x)\ud x + + \int_0^\infty \frac{\lambda}{2}\exp(-\lambda x)\ud x\\ + &= \frac{1}{2}\left(\int_{-\infty}^0\ud\exp(\lambda x) + - \int_0^\infty\ud\exp(-\lambda x)\right)\\ + &= \frac{1 - (-1)}{2} = 1 +\end{align*} + +Let $g(x) = x f_X(x)$, the mean of $X$ is +$\E[X] = \displaystyle\int_{-\infty}^\infty g(x)\ud x$. Since $g(-x) = -g(x)$ +for all $x$, $\E[X] = 0$. + +The variance of $X$ can be calculated as +$\var(X) = \E\left[X^2\right] - \E^2[X] = \E\left[X^2\right]$ +or $\var(X) = \displaystyle\int_{-\infty}^\infty x^2 f_X(x)\ud x$. + +Let $h(x) = x^2 f_X(x)$, we have $h(-x) = h(x)$ for all $x$ and thus +\begin{align*} + \E\left[X^2\right] &= 2\int_{-\infty}^\infty h(x)\ud x\\ + &= 2\int_0^\infty\frac{\lambda x^2}{2}\exp(-\lambda x)\ud x\\ + &= 2\int_0^\infty\frac{x^2}{-2}\ud\exp(-\lambda x)\\ + &= 2\int_0^\infty\exp(-\lambda x)\ud\frac{x^2}{2} + - \int_0^\infty\ud x^2\exp(-\lambda x)\\ + &= 2\int_0^\infty x\exp(-\lambda x)\ud x\\ + &= \frac{-2}{\lambda}\int_0^\infty x\ud\exp(-\lambda x)\\ + &= \frac{2}{\lambda}\left(\int_0^\infty\exp(-\lambda x)\ud x + - \int_0^\infty\ud x\exp(-\lambda x)\right)\\ + &= \frac{-2}{\lambda^2}\int_0^\infty\ud\exp(-\lambda x) + = \frac{2}{\lambda^2} +\end{align*} + +\exercise 7 Given a random variable $X$ with PDF +\[f_X(x) = \begin{cases} + 2\exp(-2 x) &\text{if }x > 0\\ + 0 &\text{otherwise} +\end{cases}\] + +Since $X$ is exponentially distributed with the parameter $\lambda = 2$, +\[\E[X] = \sigma_X = \frac{1}{\lambda} = \frac{1}{2},\, +\var(X) = \frac{1}{\lambda^2} = \frac{1}{4}\] + +Thus $\E\left[X^2\right] = \var(X) + \E^2[X] = 1/2$. + +\exercise 8 Given a random variable $X$ with PDF +\[f_X(x) = \begin{cases} + a + bx^2 &\text{if }0 \le x \le 1\\ + 0 &\text{otherwise} +\end{cases}\] + +By the normalization probability, +\[\int_0^1\left(a + bx^2\right)\ud x = 1 \iff a + \frac{b}{3} = 1\] + +Since $\E[X] = 3/5$, +$\displaystyle\int_0^1\left(ax + 3x^3\right)\ud x = \frac{3}{5}$ +or $a/2 + b/4 = 0.6$ +and thus $a = 0.6$ and $b = 1.2$. + +\exercise 9 The expectation is +\begin{align*} + \E[X] &= \int_0^\infty\frac{x^2}{e^x}\ud x\\ + &= -\int_0^\infty x^2\ud e^{-x}\\ + &= \int_0^\infty e^{-x}\ud x^2 - \int_0^\infty\ud\frac{x^2}{e^x}\\ + &= \int_0^\infty\frac{2x}{e^x}\ud x\\ + &= -2\int_0^\infty x\ud e^{-x}\\ + &= 2\int_0^\infty e^{-x}\ud x - 2\int_0^\infty\ud\frac{x}{e^x}\\ + &= -2\int_0^\infty\ud e^{-x} = 2 +\end{align*} + +\exercise{10} The $X$ is exponentially distributed by the PDF +$f_X(x) = \dfrac{1}{3}\exp\dfrac{-x}{3}$. +\begin{enumerate}[(a)] + \item $\E[X] = \sigma_X = 3$ and $\var(X) = 9$. + \item The CDF of $X$ is $F_X(x) = 1 - \exp(-x/3)$ and thus + \[P(2 < X \le 4) = F_X(4) - F_X(2) + = \exp\frac{-2}{3} - \exp\frac{-4}{3}\] +\end{enumerate} +\pagebreak + +\section{Continuous Random Variable 2} +\exercise 1 Let $X$ be the time to repair the machine, +\begin{align*} + f_X(x) &= \begin{dcases} + \frac{1}{2}\exp\frac{-x}{2} &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{dcases}\\ + \Longrightarrow F_X(x) &= \begin{dcases} + 1 - \exp\frac{-x}{2} &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{dcases} +\end{align*} +\begin{enumerate}[(a)] + \item $P(X > 2) = 1 - F_X(2) = \exp\dfrac{-2}{2} = \dfrac{1}{e}$ + \item Let $A = \{X > 10\}$ and $B = \{X > 8\}$, we have + $P(B) = \exp\dfrac{-8}{2} = \dfrac{1}{e^4}$ and + $P(A\cap B) = P(A) = \exp\dfrac{-10}{2} = \dfrac{1}{e^5}$. + Hence the conditional probability that a repair exceeding eight hours + takes at least 10 hours is + \[P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{1}{e}\] +\end{enumerate} + +\exercise 2 Denote the event that the day is sunny as $A$, $P(A) = 2/3$. +\begin{align*} + f_{X|A}(x) &= \begin{cases} + b &\text{if }15 \le x \le 23\\ + 0 &\text{otherwise} + \end{cases} + \Longrightarrow \int_{15}^{20}b\ud x = 1 + \iff b = \frac{1}{8}\\ + f_{X|A^\C}(x) &= \begin{cases} + c &\text{if }20 \le x \le 25\\ + 0 &\text{otherwise} + \end{cases} + \Longrightarrow \int_{20}^{25}c\ud x = 1 + \iff c = \frac{1}{5} +\end{align*} + +By the total probability theorem, +\[f_X(x) = P(A)f_{X|A}(x) + P\left(A^\C\right)f_{X|A^\C}(x) + = \begin{dcases} + \frac{1}{12} &\text{if }15 \le x < 20\\ + \frac{3}{20} &\text{if }20 \le x < 23\\ + \frac{1}{15} &\text{if }23 \le x < 25\\ + 0 &\text{otherwise} + \end{dcases}\] + +\exercise 3 Let $X$ be the waiting time +and $A$ be the event that one arrives at the station before 7:15, +we have $P(A) = 0.25$, $P\left(A^\C\right) = 0.75$. + +\begin{align*} + f_{X|A}(x) &= \begin{dcases} + \frac{1}{5} &\text{if }0 \le x < 5\\ + 0 &\text{otherwise} + \end{dcases}\\ + f_{X|A^\C}(x) &= \begin{dcases} + \frac{1}{15} &\text{if }5 \le x \le 15\\ + 0 &\text{otherwise} + \end{dcases} +\end{align*} + +By the total probability theorem, +\[f_X(x) = P(A)f_{X|A}(x) + P\left(A^\C\right)f_{X|A^\C}(x) + = \begin{dcases} + \frac{1}{10} &\text{if }0 \le x < 5\\ + \frac{1}{20} &\text{if }5 \le x < 15\\ + 0 &\text{otherwise} + \end{dcases}\] + +\exercise 4 Let $X$ be a random variable with PDF +\[f_X(x) = \begin{dcases} + \frac{x}{4} &\text{if }1 < x \le 3\\ + 0 &\text{otherwise} +\end{dcases}\] +and $A = \{X \ge 2\}$. +\begin{enumerate}[(a)] + \item $X$ has the mean of + $\displaystyle\E[X] = \int_1^3\frac{x^2}{4}\ud x = \frac{13}{6}$. + The CDF of $X$ is + \[F_X(x) = \begin{dcases} + 0 &\text{if }x \le 1\\ + \frac{x^2 - 1}{8} &\text{if }1 < x \le 3\\ + 1 &\text{otherwise} + \end{dcases}\] + thus $P(A) = P(\Omega) - P(X < 2) = 1 - F_X(2) = \dfrac{5}{8}$. + \[f_{X|A}(x) = \frac{P(\{X = x\}\cap A)}{P(A)} + = \frac{8}{5}P(\{X = x\}\cap A)\] + It is trivial that $\{X = x\}\cap A = \varnothing$ if $x < 2$ + and $P(\{X = x\}\cap A) = f_X(x)$ otherwise, so + \[f_{X|A}(x) = \begin{dcases} + \frac{2x}{5} &\text{if }2 \le x \le 3\\ + 0 &\text{otherwise} + \end{dcases} + \Longrightarrow \E[X|A] = \int_2^3\frac{2x^2}{5}\ud x = \frac{38}{15}\] + \item Let $Y = X^2$, the $Y$ has the expectation of + \[\E[Y] = \E\left[X^2\right] = \int_1^3\frac{x^3}{4}\ud x = 5\] + + The variance of $Y$ is + \[\var(Y) = \E\left[Y^2\right] - \E^2[Y] = \E\left[X^4\right] - 5^2 + = \int_1^3\frac{x^5}{4}\ud x - 25 = \frac{16}{3}\] +\end{enumerate} + +\exercise 5 Let $X$ be Alyssa's waiting time and $A$ be the event there is +a customer ahead, then $P(A) = P\left(A^\C\right) = 0.5$ and +\begin{align*} + f_{X|A}(x) &= \begin{cases} + \lambda\exp(-\lambda x) &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{cases}\\ + \Longrightarrow F_{X|A}(x) &= \begin{cases} + 1 - \exp(-\lambda x) &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{cases}\\ + p_{X|A^\C}(x) &= \begin{cases} + 1 &\text{if }x = 0\\ + 0 &\text{otherwise} + \end{cases}\\ + \Longrightarrow F_{X|A^\C}(x) &= \begin{cases} + 1 &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{cases} +\end{align*} + +Therefore +\begin{align*} + F_X(x) &= P(A)F_{X|A}(x) + P\left(A^\C\right)F_{X^\C}(x)\\ + &= \begin{dcases} + 1 - 0.5\exp(-\lambda x) &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{dcases} +\end{align*} + +\exercise 6 Given the joint PDF of $X$ and $Y$ +\[f_{X,Y}(x, y) = \begin{cases} + cxy &\text{if }0 < x < 4\text{ and }1 < y < 5\\ + 0 &\text{otherwise} +\end{cases}\] +\begin{enumerate}[(a)] + \item By the normalization probability + \[\int_0^4\int_1^5 cxy\ud y\ud x = 1 \iff 96c = 1 \iff c = \frac{1}{96}\] + \item $\displaystyle P(1 < X < 2, 2 < Y < 3) + = \int_1^2\int_2^3\frac{xy}{96}\ud y\ud x = \frac{5}{128}$ + \item $\displaystyle P(X \ge 3, Y \le 2) + = \int_3^4\int_1^2\frac{xy}{96}\ud y\ud x = \frac{7}{128}$ + \item The marginal PDF of $X$ is $\displaystyle f_X(x) + = \int_1^5\frac{xy}{96}\ud y = \frac{x}{8}$ and that of $Y$ is + $\displaystyle f_Y(y) = \int_0^4\frac{xy}{96}\ud x = \frac{y}{12}$. + \item The region with nonzero probability where $X + Y < 3$ is + $\{(x, y) \in \mathbb R^2 \mid 0 < x < 2, 1 < y < 3 - x\}$, thus + \[P(X + Y < 3) = \int_0^2\int_1^{3-x}\frac{xy}{96}\ud y\ud x + = \int_0^2\frac{x^3 - 6x^2 + 8x}{192}\ud x = \frac{1}{48}\] + \item Let $C_u$ be the line $X + 2Y = u$, then the PDF of $U = X + 2Y$ is + \[f_U(u) = P(X + 2Y = u) = \int_{C_u}f_{X,Y}(x, y)\ud s\] + where $\ud s$ is the infinitesimal length of $C_u$. + + Let $t$ satisfy $x = u - 2t$ and $y = t$ we get + \begin{align*} + f_U(u) &= \int_{-\infty}^\infty f_{X,Y}(u - 2t, t) + \sqrt{\left(\frac{\ud x}{\ud t}\right)^2 + + \left(\frac{\ud y}{\ud t}\right)^2}\ud t\\ + &= \int_{-\infty}^\infty f_{X,Y}(u - 2t, t)\sqrt 5\ud t + \end{align*} + + For $u < 2$, with $x \in (0, 4)$, + \[0 < u - 2t < 4 \Longrightarrow 2t < u < 2 \iff y = t < 1\] + and thus $f_{X,Y}(u - 2t, t) = 0$. Similarly, the joint PDF of $X$ and $Y$ + also equals zero when $u > 15$, so $f_U(u) = 0$ + for $u \in \mathbb R \setminus [2, 14]$. + + For $u \in [2, 6]$, + \begin{align*} + \begin{cases} + 0 < x < 4\\ + 1 < y < 5 + \end{cases} + &\iff \begin{cases} + 0 < u - 2t < 4\\ + 1 < t < 5 + \end{cases}\\ + &\iff \frac{u}{2} - 2 \le 1 < t < \frac{u}{2} < 5\\ + &\iff 1 < t < \frac{u}{2} + \end{align*} + so $\displaystyle f_U(u) = \frac{\sqrt 5}{96}\int_1^{u/2}(ut - 2t^2)\ud t + = \frac{\sqrt 5}{2304}(u^3 - 12u + 16)$. + + For $u \in (6, 10)$, + \begin{align*} + \begin{cases} + 0 < x < 4\\ + 1 < y < 5 + \end{cases} + &\iff 1 < \frac{u}{2} - 2 < t < \frac{u}{2} < 5\\ + &\iff \frac{u}{2} - 2 < t < \frac{u}{2} + \end{align*} + so $\displaystyle f_U(u) = \frac{\sqrt 5}{96} + \int_{u/2-2}^{u/2}(ut - 2t^2)\ud t + = \frac{\sqrt 5}{288}(u^2 - 2u)$. + + For $u \in [10, 14]$, + \begin{align*} + \begin{cases} + 0 < x < 4\\ + 1 < y < 5 + \end{cases} + &\iff 1 < \frac{u}{2} - 2 < t < 5 \le \frac{u}{2}\\ + &\iff \frac{u}{2} - 2 < t < 5 + \end{align*} + so $\displaystyle f_U(u) = \frac{\sqrt 5}{96}\int_{u/2-2}^5(ut - 2t^2)\ud t + = \frac{\sqrt 5}{2304}(348u - u^3 - 2128)$. +\end{enumerate} + +\exercise 7 Given the joint PDF of $X$ and $Y$ +\[f_{X,Y}(x, y) = \begin{cases} + 8xy &\text{if }0 \le y \le x \le 1\\ + 0 &\text{otherwise} +\end{cases}\] +\begin{enumerate}[(a)] + \item The marginal PDFs are + \begin{align*} + f_X(x) &= \begin{dcases} + \int_0^x 8xy\ud y = 4x^3 &\text{if }0 \le x \le 1\\ + 0 &\text{otherwise} + \end{dcases}\\ + f_Y(y) &= \begin{dcases} + \int_y^1 8xy\ud x = 4y - 4y^3 &\text{if }0 \le y \le 1\\ + 0 &\text{otherwise} + \end{dcases} + \end{align*} + \item The conditional PDFs are + \begin{align*} + f_{X|Y}(x|y) = \frac{f_{X,Y}(x, y)}{f_Y(y)} &= \begin{dcases} + \frac{2x}{1 - y^2} &\text{if }0 \le y \le x \le 1\\ + 0 &\text{otherwise} + \end{dcases}\\ + f_{Y|X}(y|x) = \frac{f_{X,Y}(x, y)}{f_X(x)} &= \begin{dcases} + \frac{2y}{x^2} &\text{if }0 \le y \le x \le 1\\ + 0 &\text{otherwise} + \end{dcases} + \end{align*} + \item The conditional expectations are + \begin{align*} + \E[X|Y=y] &= \int_{-\infty}^\infty xf_{X|Y}(x|y)\ud x\\ + &= \begin{dcases} + \int_y^1\frac{2x^2}{1-y^2}\ud x = \frac{2x^2 + 2x + 2}{3x + 3} + &\text{if }0 \le y \le 1\\ + 0 &\text{otherwise} + \end{dcases}\\ + \E[Y|X=x] &= \int_{-\infty}^\infty yf_{Y|X}(y|x)\ud y\\ + &= \begin{dcases} + \int_0^x\frac{2y^2}{x^2}\ud y = \frac{2x}{3} &\text{if }0 \le x \le 1\\ + 0 &\text{otherwise} + \end{dcases} + \end{align*} + \item It is trivial that given $X = x \in \mathbb R \setminus [0, 1]$, + $\var(Y|X=x) = 0$. Otherwise, + \begin{align*} + \var(Y|X=x) &= \E\left[Y^2|X=x\right] - \E^2[Y|X=x]\\ + &= \int_0^x\frac{2y^3}{x^2}\ud y - \left(\frac{2x}{3}\right)^2 + = \frac{x^2}{2} - \frac{4x^2}{9} = \frac{x^2}{16} + \end{align*} +\end{enumerate} + +\exercise 8 Given the joint PDF of $X$ and $Y$ +\[f_{X,Y}(x, y) = \begin{cases} + \exp(-x-y) &\text{if }x \ge 0\text{ and }y \ge 0\\ + 0 &\text{otherwise} +\end{cases}\] + +The marginal PDFs are +\begin{align*} + f_X(x) &= \begin{dcases} + \int_0^\infty\exp(-x-y)\ud y = \exp(-x) &\text{if }x \ge 0\\ + 0 &\text{otherwise} + \end{dcases}\\ + f_Y(y) &= \begin{dcases} + \int_0^\infty\exp(-x-y)\ud x = \exp(-y) &\text{if }y \ge 0\\ + 0 &\text{otherwise} + \end{dcases} +\end{align*} + +It is noticeable that $X$ and $Y$ are independent, +thus $f_{X|Y}(x|y) = f_X(x)$ and $f_{Y|X}(y|x) = f_Y(y)$. +\pagebreak + +\section{Continuous Random Variable 3} +\subsection{PDF and CDF} +\exercise 1 Given $Z \sim \N(0, 1)$. +\begin{enumerate}[(a)] + \item $P(Z > 1.2) = 1 - \Phi(1.2) = 1 - 0.8849 = 0.1101$ + \item $P(-2 < Z < 2) = \Phi(2) - \Phi(-2) = 2\Phi(2) - 1 + = 2\cdot 0.9772 - 1 = 0.9544$ + \item $P(-1.2 < Z < 1) = \Phi(1) + \Phi(1.2) - 1 + = 0.8413 + 0.8849 - 1 = 0.7262$ +\end{enumerate} + +\exercise 2 Given $X \sim \N(4, 9)$. Let $Y = \dfrac{X - 4}{3}$, +$Y$ is a standard normal random variable +and $F_X(x) = \Phi\left(\dfrac{x - 4}{3}\right)$. +\begin{enumerate}[(a)] + \item $P(X > 6) = 1 - F_X(6) = 1 - \Phi\left(\dfrac{6 - 4}{3}\right) + \approx 1 - \Phi(0.67) = 0.2514$ + \item $P(X > 1) = 1 - F_X(6) = 1 - \Phi\left(\dfrac{1 - 4}{3}\right) + = 1 - \Phi(-1) = \Phi(1) = 0.8413$ +\end{enumerate} + +\exercise 3 Let $X$ be the annual snowfall in inches +and $Y = \dfrac{X - 60}{20}$, we have $Y \sim \N(0, 1)$ +and $F_X(x) = \Phi\left(\dfrac{x - 60}{20}\right)$. +The probability that snowfall will be at least 80 inches is +\[P(X \ge 80) = 1 - F_X(80) = 1 - \Phi\left(\frac{80 - 60}{20}\right) += 1 - \Phi(1) = 1 - 0.8413 = 0.1587\] + +\exercise 4 Let $X$ be the number of customers arriving during an one-hour +period, $X$ is a Poisson random variable whose PMF is +\begin{multline*} + p_X(x) = \frac{24^x}{e^{24}x!},\qquad x \in \mathbb N\\ + \Longrightarrow P(X < 15) = \sum_{x=0}^{14}p_X(x) + = \sum_{x=0}^{14}\frac{24^x}{e^{24}x!} \approx 0.019825332823463673 +\end{multline*} + +\subsection{Covariance and Correlation Coefficient} +\exercise 5 Given the joint PMF of $X$ and $Y$ +\[p_{X,Y}(x, y) = \begin{cases} + c(2x + y) &\text{where }x \in \{0, 1, 2\}\text{ and }y \in \{0, 1, 2, 3\}\\ + 0 &\text{otherwise} +\end{cases}\] + +By the normalization property, +\[\sum_{x=0}^2\sum_{y=0}^3 c(2x + y) = 1 \iff c = \frac{1}{42}\] +\begin{enumerate}[(a)] + \item The maginal PMFs are + \begin{align*} + p_X(x) &= \sum_{y=0}^3 p_{X,Y}(x, y) = \begin{dcases} + \frac{4x + 3}{21} &\text{if }x \in \{0, 1, 2\}\\ + 0 &\text{otherwise} + \end{dcases}\\ + p_Y(y) &= \sum_{x=0}^2 p_{X,Y}(x, y) = \begin{dcases} + \frac{y + 2}{14} &\text{if }y \in \{0, 1, 2, 3\}\\ + 0 &\text{otherwise} + \end{dcases} + \end{align*} + + Therefore we can compute these expectations: + \begin{align*} + \E[X] &= \sum_{x=0}^2 x p_X(x) + = \sum_{x=0}^2\frac{4x^2 + 3x}{21} = \frac{29}{21}\\ + \E[Y] &= \sum_{y=0}^3 y p_Y(y) + = \sum_{y=0}^3\frac{y^2 + 2y}{14} = \frac{13}{7}\\ + \E[XY] &= \sum_{x=0}^2\sum_{y=0}^3 xy p_{X,Y}(x,y) + = \sum_{x=0}^2\sum_{y=0}^3\frac{2x^2 y + xy^2}{42} = \frac{17}{7} + \end{align*} + \item The variances of these variables can be calculated as + \begin{align*} + \var(X) &= \E\left[X^2\right] - \E^2[X] = \frac{230}{21}\\ + \var(Y) &= \E\left[Y^2\right] - \E^2[Y] = \frac{25}{7} + \end{align*} + where + \begin{align*} + \E\left[X^2\right] &= \sum_{x=0}^2 x^2 p_X(x) + = \sum_{x=0}^2\frac{4x^3 + 3x^2}{21} = \frac{17}{7}\\ + \E\left[Y^2\right] &= \sum_{y=0}^3 y^2 p_Y(y) + = \sum_{y=0}^3\frac{y^3 + 2y^2}{14} = \frac{32}{7} + \end{align*} + \item $\cov(X, Y) = \E[XY] - \E[X]\E[Y] = \dfrac{-20}{147}$ so + \[\rho(X, Y) = \dfrac{\cov(X, Y)}{\sqrt{\var(X)\var(Y)}} \approx -0.027\] +\end{enumerate} + +\exercise 6 Given the joint PDF of $X$ and $Y$ as followed +\[f_{X,Y}(x, y) = \begin{cases} + c(2x + y) &\text{where }(x, y) \in (2, 6) \times (0, 5)\\ + 0 &\text{otherwise} +\end{cases}\] + +By the normalization property, +\[\int_2^6\int_0^5 c(2x + y)\ud y\ud x = 1 \iff c = \frac{1}{210}\] +\begin{enumerate}[(a)] + \item The maginal PMFs are + \begin{align*} + f_X(x) &= \int_0^5 f_{X,Y}(x, y)\ud y = \begin{dcases} + \frac{4x + 5}{84} &\text{if }2 < x < 6\\ + 0 &\text{otherwise} + \end{dcases}\\ + f_Y(y) &= \int_2^6 f_{X,Y}(x, y)\ud x = \begin{dcases} + \frac{2y + 16}{105} &\text{if }0 < y < 5\\ + 0 &\text{otherwise} + \end{dcases} + \end{align*} + + Therefore we can compute these expectations: + \begin{align*} + \E[X] &= \int_2^6 x f_X(x)\ud x + = \int_1^6\frac{4x^2 + 5x}{84}\ud x = \frac{268}{63}\\ + \E[Y] &= \int_0^5 y f_Y(y)\ud y + = \int_0^5\frac{2y^2 + 16y}{105}\ud y = \frac{170}{63}\\ + \E[XY] &= \int_2^6\int_0^5 xy f_{X,Y}(x,y)\ud y\ud x + = \int_2^6\int_0^5\frac{2x^2 y + xy^2}{210}\ud y\ud x = \frac{80}{7} + \end{align*} + \item The variances of these variables can be calculated as + \begin{align*} + \var(X) &= \E\left[X^2\right] - \E^2[X] = \frac{5036}{3969}\\ + \var(Y) &= \E\left[Y^2\right] - \E^2[Y] = \frac{16225}{7938} + \end{align*} + where + \begin{align*} + \E\left[X^2\right] &= \int_2^6 x^2 f_X(x) + = \int_2^6\frac{4x^3 + 5x^2}{84} = \frac{1220}{63}\\ + \E\left[Y^2\right] &= \int_0^5 y^2 f_Y(y) + = \int_0^5\frac{2y^3 + 16y^2}{105} = \frac{1175}{126} + \end{align*} + \item $\cov(X, Y) = \E[XY] - \E[X]\E[Y] = \dfrac{-200}{3969}$ so + \[\rho(X, Y) = \dfrac{\cov(X, Y)}{\sqrt{\var(X)\var(Y)}} \approx -0.0313\] +\end{enumerate} + +\subsection{Derived Distribution} +\exercise 7 With $X$ being uniform on $[0, 1]$, by the normalization property, +$f_X(x) = 1$ $\Longrightarrow F_X(x) = x$ on this interval. Given $Y = \sqrt X$, +\[F_Y(y) = P(Y \le y) = P\left(\sqrt X \le y\right) = F_X\left(y^2\right) = y^2 +\Longrightarrow f_y(y) = \frac{\ud F_Y}{\ud y} = 2y\] +if $0 < Y < 1$, otherwise $f_Y(y) = 0$. + +\exercise 8 Let $X$ be the speed in miles per hour, +\[f_X(x) = \begin{dcases} + \frac{1}{30} &\text{if }30 \le x \le 60\\ + 0 &\text{otherwise} +\end{dcases} +\Longrightarrow F_X(x) = \begin{dcases} + 0 &\text{if }x < 30\\ + \frac{x}{30} - 1 &\text{if }30 \le x < 60\\ + 1 &\text{otherwise} +\end{dcases}\] +then the duration of the trip is $Y = 180/X$. Where $3 \le Y \le 6$, +\begin{multline*} + F_Y(y) = P\left(\frac{180}{X} \le y\right) + = P\left(X \ge \frac{180}{y}\right) + = 1 - F_X\left(\frac{180}{y}\right) = 2 - \frac{6}{y}\\ + \Longrightarrow f_Y(y) = \frac{\ud F_Y}{\ud y} = \frac{6}{y^2} +\end{multline*} + +Otherwise, it is obvious that $f_Y(y) = 0$. + +\exercise{9}\footnote{IMHO this is a really poor example to demonstrate +the usefulness of this method.} Let $A$ be the event that one arrives at +the station before 7:15, we have $P(A) = 0.25$ and $P\left(A^\C\right) = 0.75$. +The probability that $X = x$ is 0.05 for all $x \in [0, 20)$ +and is 0 otherwise, thus +\begin{align*} + F_{X|A}(x) &= \begin{dcases} + 0 &\text{if }x < 0\\ + \frac{\int_0^x 0.05\ud t}{0.25} = \frac{0.05x}{0.25} + &\text{if }0 \le x < 5\\ + 1 &\text{otherwise} + \end{dcases}\\ + F_{X|A^\C}(x) &= \begin{dcases} + 0 &\text{if }x < 5\\ + \frac{\int_5^x 0.05\ud t}{0.25} = \frac{0.05x - 0.25}{0.75} + &\text{if }5 \le x < 20\\ + 1 &\text{otherwise} + \end{dcases} +\end{align*} + +With $Y = 5 - X$ if $A$ and $Y = 20 - X$ otherwise, +\begin{multline*} + \begin{aligned} + F_Y(y) &= P(Y \le y)\\ + &= 1 - P(Y > y)\\ + &= P(A)P(5 - X > y|A) + P\left(A^\C\right)P\left(20 - X > y|A^\C\right)\\ + &= 1 - 0.25P(X < 5 - y|A) - 0.75P\left(X < 20 - y|A^\C\right)\\ + &= 1 - 0.25F_{X|A}(5 - y) - 0.75F_{X|A^\C}(20 - y)\\ + &= \begin{dcases} + 0 &\text{if }y < 0\\ + 1 - 0.05(5 - y) - 0.05(20 - y) + 0.25 = 0.1y &\text{if }0 \le y < 5\\ + 1 - 0.05(20 - y) + 0.25 = 0.05y + 0.25 &\text{if }5 \le y < 15\\ + 1 &\text{otherwise} + \end{dcases} + \end{aligned}\\ + \Longrightarrow f_Y(y) = \frac{\ud F_Y}{\ud y} = \begin{dcases} + 0.1 &\text{if }0 \le y < 5\\ + 0.05 &\text{if }5 \le y < 15\\ + 0 &\text{otherwise} + \end{dcases} +\end{multline*} +\pagebreak + +\setcounter{section}{8} +\section{Limit Theorem} +\exercise 1 Let $S_{100} = \sum_{i=1}^{100}X_i$, $M_{100} = S_{100}/100$ +is the sample mean. We have +\[Z_{100} = \frac{S_{100} - 100\cdot 10}{4\sqrt{100}} = 2.5M_{100} - 25\] + +Since 100 is large, we can use the approximation +$P(Z_{100} \le z) \approx \Phi(z)$: +\begin{multline*} + P(S_{100} \le 900) = P(M_{100} \le 9) = P(2.5M_{100} - 25 \le 22.5 - 25)\\ + = P(Z_n \le -2.5) \approx \Phi(-2.5) = 1 - \Phi(2.5) = 0.0062 +\end{multline*} + +\exercise 2 Let $X$ be the weight of a box in lbs, and denote +\[Z_{49} = \frac{\sum_{i=1}^{49}X_i - 49\cdot 205}{15\sqrt{49}} += \frac{S_{49}}{105} - \frac{287}{3}\] + +Since 49 is large, we can use the following approximation +to compute the probability that all 49 boxes can be +safely loaded onto the freight elevator and transported +\[P(S_{49} \le 9800) = P\left(Z_{49} \le \frac{-7}{3}\right) +\approx 1 - \Phi\left(\frac 7 3\right) = 0.0099\] + +\exercise 3 Let $X$ be the number of tickets to be purchased by a student, +and denote +\[Z_{100} = \frac{\sum_{i=1}^{100}X_i - 100\cdot 2.4}{2\sqrt{100}} += \frac{S_{100}}{20} - 12\] + +Since 100 is large, we can use the following approximation +to compute the probability that all 100 students will be able to purchase +the tickets they desire from the 250 that is left: +\[P(S_{100} \le 250) = P(Z_{100} \le 0.5) +\approx \Phi(0.5) = 0.6915\] + +\exercise 4 Let $X$ be the time in minutes to complete one problem, +and denote +\[Z_{40} = \frac{\sum_{i=1}^{40}X_i - 40\cdot 5}{2\sqrt{40}} += \frac{S_{40} - 200}{4\sqrt{10}}\] + +Since 40 is large, we can use the following approximation +to compute the probability that all 40 problems within 3 hours +\[P(S_{40} \le 180) = P\left(Z_{49} \le -\sqrt\frac{5}{2}\right) +\approx 1 - \Phi(1.58) = 0.0571\] + +\exercise 5 Let $X$ be the size in MB of an image and denote +\[Z_{80} = \frac{\sum_{i=1}^{80}X_i - 80\cdot 0.6}{0.4\sqrt{80}} += \frac{5S_{40} - 240}{8\sqrt{5}}\] + +Since 80 is large, we can use the following formula to approximate +the probability that the total size is between 47 and 56 MB +\[P(47 \le S_{80} \le 56) += P\left(\frac{\sqrt 5}{-8} \le Z_{49} \le \sqrt 5\right) +\approx \Phi(2.23) - 1 + \Phi(0.28) = 0.5974\] + +\exercise 6 After 11 weeks, the station is supplied +\[74\,000 + 47\,000 \cdot 11 = 591\,000\text{ (gallons)}\] + +Let $X$ be the gasoline consumption in gallons a week and denote +\[Z_{11} = \frac{\sum_{i=1}^{11}X_i - 11\cdot 50\,000}{10\,000\sqrt{11}} += \frac{S_{11} - 550\,000}{10\,000\sqrt{11}}\] + +While 11 is not exactly large, for the ease of calculation, +we still use the approximation $P(Z_{11} \le z) \approx \Phi(z)$. + +\begin{enumerate}[(a)] + \item The probability that the remain will be below 20\,000 gallons is + \begin{multline*} + P(591\,000 - S_{11} < 20\,000) + = P(S_{11} > 571\,000) + = P\left(Z_{11} > \frac{21}{10\sqrt{11}}\right)\\ + \approx 1 - \Phi(0.63) = 0.2643 + \end{multline*} + \item Let $w$ be the weekly delivery satisfying the probability + that below 20\,000 gallons will be remained is 0.5\%, we have + \begin{align*} + &P(74\,000 + 11w - S_{11} < 20\,000) = 0.005\\ + \iff &P(S_{11} > 54\,000 + 11w) = 0.005\\ + \iff &P\left(Z_{11}>\frac{11w-496\,000}{10\,000\sqrt{11}}\right) = 0.005\\ + \iff &1 - \Phi\left(\frac{11w-496\,000}{10\,000\sqrt{11}}\right) = 0.005\\ + \iff &\Phi\left(\frac{11w-496\,000}{10\,000\sqrt{11}}\right) = 0.995\\ + \iff &\frac{11w-496\,000}{10\,000\sqrt{11}} = 2.57\\ + \iff &w = 52840 + \end{align*} +\end{enumerate} +\end{document} |