--- title: "Sample Homework Solution" author: "Marek Rychlik (derived from a student's homework; some solutions are incorrect!!!)" date: "11/02/2021" output: html_document: toc: yes pdf_document: toc: yes editor_options: markdown: wrap: 72 chunk_output_type: console --- Setup of some 'knitr' parameters ```{r setup, include=FALSE} knitr::opts_chunk$set(echo = FALSE) ``` **Install required packages** You may omit this R "chunk" after the packages are installed. ```{r install.packages("bookdown");install.packages("markdown")} ``` **Exercise 4.5** Let $0 < p = 1 - q < 1$ From Example 4.3: The sequence given by $u_n = \binom{N}{n}$ if $n=0,1,2,...,N$ $u_n = 0$ otherwise has generating function $U(s) = \sum_{n=0}^{N} \binom{N}{n}s^n = (1+s)^N$ generating function: $U(s) = \sqrt{1-4pgs^2} = (1 - 4pqs^2)^{\frac{1}{2}}$ sequence: $u_n = \binom{-\frac{1}{2}}{n}$ if $n=0,1,2,...,N$ $u_n = 0$ otherwise \newpage **Exercise 4.18** $X$ is a random variable with probability generating function $G_{X}(s)$ $k$ is a positive integer $Y =kX$ and $Z = X + k$ have probability generating functions $G_{Y}(s) = G_{X}(s^k)$, $G_{Z}(s) = s^k G_{X}(s)$ \newpage **Exercise 4.19** $X$ is uniformly distributed on {$0,1,2,...,a$} in that $P(X=k) = \frac{1}{a+1}$ for $k=0,1,2,...,a$ $X$ has probability generating function $G_{X}(s) = \frac{1-s^{a+1}}{(a+1)(1-s)}$ \newpage **Exercise 4.30** From Example 4.16: If $X$ has the Poisson distribution with parameter $\lambda$, then $$G_{X}(s) = \sum_{k=0}^{\infty} \frac{1}{k!} \lambda^{k} e^{- \lambda} s^{k} = e^{\lambda(s-1)}$$ From Proof 4.25: $E(X) = G_{X}'(s)$ $$ \begin{aligned} E(X) = G_{X}'(s) &= \frac{d}{ds} e^{\lambda(s-1)} \\ &= e^{\lambda(s-1)} \frac{d}{ds} {\lambda}(s-1)\\ &= e^{\lambda(s-1)} \lambda\left[\frac{d}{ds}(s) - \frac{d}{ds}(1)\right]\\ &= e^{\lambda(s-1)} \lambda[1-0]\\ &= e^{\lambda(s-1)} \lambda \end{aligned} $$ From Equation 4.26: $E(X) = G_{X}'(1)$ $$ \begin{aligned} E(X) = G_{X}'(1) &= e^{\lambda[(1)-1]} \lambda\\ &= e^{\lambda(0)} \lambda\\ &= e^{0} \lambda\\ &= \lambda \end{aligned} $$ From Equation 4.28: $var(X) = G_{X}''(1) + G_{X}'(1) - (G_{X}'(1))^2$ $$ \begin{aligned} G_{X}''(s) &= \frac{d}{ds} \lambda e^{\lambda(s-1)}\\ & = \lambda \frac{d}{ds} e^{\lambda(s-1)}\\ & = \lambda e^{\lambda(s-1)} \frac{d}{ds} \lambda(s-1)\\ & = \lambda e^{\lambda(s-1)} \lambda(1-0)\\ & = \lambda e^{\lambda(s-1)} \lambda\\ & = \lambda^2 e^{\lambda(s-1)}\\ G_{X}''(1) &= \lambda^2 e^{\lambda[(1)-1]}\\ & = \lambda^2 e^{\lambda(0)}\\ & = \lambda^2\\ var(X) &= \lambda^2 + \lambda - (\lambda)^2\\ var(X) &= \lambda^2 + \lambda - \lambda^2\\ var(X) &= \lambda \end{aligned} $$ A random variable having the Poisson distribution with parameter $\lambda$ has both mean and variance equal to $\lambda$. \newpage **Exercise 4.31** $X$ has the negative binomial distribution with parameters $n$ and $p$ From Example 4.17: If $X$ has the negative binomial distribution with parameters $n$ and $p$, then $$G_{X}(s) = \sum_{k=n}^{\infty} \binom{k-1}{n-1} p^{n}q^{k-n}s^{k} = (\frac{ps}{1-qs})^n$$ if $|s| < q^{-1}$ \$G\_{X}'(s) = \$ $E(X) = \frac{n}{p}$, $var(X) = \frac{nq}{p^2}$ where $q = 1-p$ \newpage **Exercise 4.41** Find distribution of $X + Y$, where $X$ and $Y$ are independent random variables, $X$ having the binomial distribution with parameters $m$ and $p$, and Y having the binomial distribution with parameters $n$ and $p$. Deduce that the sum of $n$ independent random variables, each having the Bernoulli distribution with parameter $p$, has the binomial distribution with parameters $n$ and $p$. \newpage **Exercise 4.42** Egg cracks with probability of $p$ Number of eggs laid today by the hen has the Poisson distribution, parameter $\lambda$ Number of uncracked ages has the Poisson distribution with parameter $\lambda(1-p)$ $G_{N}(s) = E(s^N) = (p + ps)^{\lambda}$ $G_{X}(s) = e^{\lambda (1-p)(s-1)}$ $S = X_1 + X_2 + ... + X_N$ $G_{S}(s) = G_{N}(G_{X}(s)) = (p + p(e^{\lambda (1-p)(s-1)}))^{\lambda}$ \newpage **Exercise C4.4.2** A random variable $X$ has generating function $G_{X}(s) = (\frac{1}{2} + \frac{1}{2} e^{3(s-1)})^{20}$ \newpage **Problem 1** Let $X$ have probability generating function $G_{X}(s)$ and let $u_n = P(X > n)$ The generating function $U(s)$ of the sequence $u_0, u1, ...$ satisfies $(1-s)U(s) = 1 - G_{X}(s)$ whenever the series defining these generating function coverage \newpage **Problem 2** Symmetrical die thrown independently 7 times. $$ P(X_j = k) = \begin{cases} \frac{1}{6}, &k \in \{1,2,3,4,5,6\}\\ 0 &\text{otherwise.} \end{cases} $$ $$G_{X_j}(s) = \sum_{k=1}^6 \frac{1}{6} s^k = \frac{1}{6} s \frac{1-s^6}{1-s}$$ $$G_{X}(s) = \left(G_{X_1}(s)\right)^6 = \left(\frac{1}{6} s \frac{1-s^6}{1-s}\right)^6$$ $$ \begin{aligned} P(X = 14) &= [s^{14}]\left(\frac{1}{6} s \frac{1-s^6}{1-s}\right)^6 \quad \text{($g = 14-6 = 8$)}\\ &= \frac{1}{6^6} [s^g] \left(\frac{1-s^6}{1-s}\right)^6\\ &= \frac{1}{6^6} [s^g] (1-s^6)^6 (1-s)^{-6}\\ &= \frac{1}{6^6} [s^g] \left[\sum_{k=0}^{6} \binom{6}{k} (-s^6)^k\right] \left[\sum_{l=0}^{\infty} \binom{-6}{l} (-s)^l\right]\\ &= \frac{1}{6^6} [s^g] \sum_{k=0}^{6} \sum_{l=0}^{\infty} \binom{6}{k} (-1)^{k+l} \binom{-6}{l} s^{6k+l}\\ &= \frac{1}{6^6} \sum_{k \in {0,1,...,6}} \binom{6}{k} \binom{-6}{l} (-1)^{k+l}\\ &= \frac{1}{6^6} \left[- \binom{6}{0} \binom{-6}{8} + \binom{6}{1} \binom{-6}{3}\right]\\ &= \frac{1}{46656} \left[-(1) \binom{-6}{8} + (6) \binom{-6}{3}\right] \end{aligned} $$ \`\`\` \newpage **Problem 3** 3 players throw a perfect die in turn independently in the order A, B, C, A, ... until one wins by throwing a 5 or 6. Probability generating function $F(s)$ for the random variable $X$ which takes the value $r$ if the game ends on the $r$th throw can be written as: $$F(s) = \frac{9s}{27-8s^3} + \frac{6s^2}{27-8s^3} + \frac{4s^3}{27-8s^3}$$ \newpage **Problem 5** Tree of a particular type flowers once each year. Probability a tree has $n$ flowers is $(1-p)p^n, n=0,1,2,...$ where 0 \< $p$ \< 1 Each flower has probability $\frac{1}{2}$ of producing a ripe fruit, independently of all other flowers. a) $P(X = r)$ b) $P(X = n | X = r)$ \newpage **Problem 7** Let $X$ and $Y$ be independent random variables having Poisson distributions with parameters $\lambda$ and $\mu$ respectively. $X + Y$ has a Poisson distribution $var(X + Y) = var(X) + var(Y)$ conditional probability: $P(X = k | X + Y = n)$ for $0 \leq k \leq n$ conditional expectation of $X$ given that $X + Y = n$: $$E(X|X + Y =n) = \sum_{k=0}^{\infty} k P(X = k | X + Y = n)=\frac{n \lambda}{\lambda + \mu}$$ \newpage **Problem 10** Probability generating function $\phi$ If $\phi(s)$ has the form $\frac{p(s)}{q(s)}$, the mean value is $\frac{(p'(1) - q'(1))}{q(1)}$ \newpage **Problem 11** A random number $N$ of foreign objects in soup, with mean $\mu$ and finite variance. Each object is a fly with probability $p$, and otherwise spider. Different objects have independent types. Let $F$ be the number of flies and $S$ the number of spiders. a) $G_{F}(s) = G_N(ps + 1 -p)$ b) $N$ has the Poisson distribution with parameter $\mu$. $F$ has the Poisson distribution with parameter $\mu p$. $F$ and $S$ are independent. c) Let $p = \frac{1}{2}$ and suppose $F$ and $S$ are independent. $G_{N}(s) = G_{N}\left(\frac{1}{2} [1 + s]\right)^2$ $\left[1 + (\frac{x}{n}) + o(n^{-1})\right]^n \rightarrow e^x$ as $n \rightarrow \infty$