Setup of some ‘knitr’ parameters
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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
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 \(G_{Y}(s) = G_{X}(s^k)\), \(G_{Z}(s) = s^k G_{X}(s)\)
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 \(G_{X}(s) = \frac{1-s^{a+1}}{(a+1)(1-s)}\)
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 \(\lambda\).
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 \(q = 1-p\)
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 \(p\).
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 \(\lambda(1-p)\)
\(G_{N}(s) = E(s^N) = (p + ps)^{\lambda}\)
\(G_{X}(s) = e^{\lambda (1-p)(s-1)}\)
\(S = X_1 + X_2 + ... + X_N\)
\(G_{S}(s) = G_{N}(G_{X}(s)) = (p + p(e^{\lambda (1-p)(s-1)}))^{\lambda}\)
Exercise C4.4.2
A random variable \(X\) has generating function \(G_{X}(s) = (\frac{1}{2} + \frac{1}{2} e^{3(s-1)})^{20}\)
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
Problem 2
Symmetrical die thrown independently 7 times. \[ P(X_j = k) = \begin{cases} \frac{1}{6}, &k \in \{1,2,3,4,5,6\}\\ 0 &\text{otherwise.} \end{cases} \] \[G_{X_j}(s) = \sum_{k=1}^6 \frac{1}{6} s^k = \frac{1}{6} s \frac{1-s^6}{1-s}\] \[G_{X}(s) = \left(G_{X_1}(s)\right)^6 = \left(\frac{1}{6} s \frac{1-s^6}{1-s}\right)^6\]
\[ \begin{aligned} P(X = 14) &= [s^{14}]\left(\frac{1}{6} s \frac{1-s^6}{1-s}\right)^6 \quad \text{($g = 14-6 = 8$)}\\ &= \frac{1}{6^6} [s^g] \left(\frac{1-s^6}{1-s}\right)^6\\ &= \frac{1}{6^6} [s^g] (1-s^6)^6 (1-s)^{-6}\\ &= \frac{1}{6^6} [s^g] \left[\sum_{k=0}^{6} \binom{6}{k} (-s^6)^k\right] \left[\sum_{l=0}^{\infty} \binom{-6}{l} (-s)^l\right]\\ &= \frac{1}{6^6} [s^g] \sum_{k=0}^{6} \sum_{l=0}^{\infty} \binom{6}{k} (-1)^{k+l} \binom{-6}{l} s^{6k+l}\\ &= \frac{1}{6^6} \sum_{k \in {0,1,...,6}} \binom{6}{k} \binom{-6}{l} (-1)^{k+l}\\ &= \frac{1}{6^6} \left[- \binom{6}{0} \binom{-6}{8} + \binom{6}{1} \binom{-6}{3}\right]\\ &= \frac{1}{46656} \left[-(1) \binom{-6}{8} + (6) \binom{-6}{3}\right] \end{aligned} \] ```
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:
\[F(s) = \frac{9s}{27-8s^3} + \frac{6s^2}{27-8s^3} + \frac{4s^3}{27-8s^3}\]
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.
\(P(X = r)\)
\(P(X = n | X = r)\)
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 \(X + Y = n\):
\[E(X|X + Y =n) = \sum_{k=0}^{\infty} k P(X = k | X + Y = n)=\frac{n \lambda}{\lambda + \mu}\]
Problem 10
Probability generating function \(\phi\)
If \(\phi(s)\) has the form \(\frac{p(s)}{q(s)}\), the mean value is \(\frac{(p'(1) - q'(1))}{q(1)}\)
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.
\(G_{F}(s) = G_N(ps + 1 -p)\)
\(N\) has the Poisson distribution with parameter \(\mu\). \(F\) has the Poisson distribution with parameter \(\mu p\). \(F\) and \(S\) are independent.
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 \(n \rightarrow \infty\)