Question

$$ \text { (a) Let } X \text { and } Y \text { be independent random variables such that } $$ $$ \begin{aligned} &\operatorname{Pr}\\{X=i\\}=f(i), \quad \operatorname{Pr}\\{Y=i\\}=g(i) \\ &f(i)>0, \quad g(i)>0, \quad i=0,1,2, \ldots \end{aligned} $$ and $$ \sum_{i=0}^{\infty} f(i)=\sum_{i=0}^{\infty} \mathrm{g}(i)=1 $$ Suppose $$ \operatorname{Pr}\\{X=k \mid X+Y=l\\}=\left\\{\begin{array}{cc} \left(\begin{array}{l} l \\ k \end{array}\right) p^{k}(1-p)^{1-k}, & 0 \leq k \leq l, \\ 0, & k>l . \end{array}\right. $$ Prove that $$ f(i)=e^{-\theta x} \frac{(\theta \alpha)^{i}}{i !}, \quad g(i)=\mathrm{e}^{-\theta} \frac{\theta^{i}}{i !}, \quad \alpha=0,1,2, \ldots $$ where \(\alpha=p /(1-p)\) and \(\theta>0\) is arbitrary. (b) Show that \(p\) is determined by the condition. $$ G\left(\frac{1}{1-p}\right)=\frac{1}{f(0)} $$ Ilint: Let \(F(s)=\sum f(i) s^{I}, G(s)=\sum g(i) s^{i} .\) Establish first the relation $$ F(u) F(v)=F(v p+(1-p) u) G(v p+(1-p) u) $$

Short answer

In summary, the expressions for f(i) and g(i) are found by analyzing the joint probability of X=k and Y=l-k and summing these probabilities over all possible values of k and l. We then find the value of p using the provided hint and the derived expressions for f(i) and g(i), which is determined by the condition G(1/(1-p)) = 1/f(0).

Step-by-step solution

Finding the joint probability of X=k and Y=l-k

Given the conditional probability Pr(X=k|X+Y=l), we can find the joint probability using the following formula: Pr(X=k, Y=l-k) = Pr(X=k|X+Y=l) * Pr(Y=l-k) Since X and Y are random variables, we know that: Pr(X=k) = f(k) Pr(Y=l-k) = g(l-k) So, we have: Pr(X=k, Y=l-k) = Pr(X=k|X+Y=l) * g(l-k) Using the provided expression for Pr(X=k|X+Y=l): Pr(X=k, Y=l-k) = \(\binom{l}{k}\) p^k (1-p)^{l-k} * g(l-k)

Expressions for f(i) and g(i)

Now, we need to find the expressions for f(i) and g(i) using the joint probability. We know that: f(i) = Pr(X=i) g(i) = Pr(Y=i) Summing the joint probability over all possible values of k for a fixed i, we get: f(i) = \(\sum_{k=0}^{i}\) Pr(X=k, Y=i-k) Substitute the expression for the joint probability we derived in Step 1: f(i) = \(\sum_{k=0}^{i}\) \(\binom{i}{k}\) p^k (1-p)^{i-k} * g(i-k) Now, summing the joint probability over all possible values of l for a fixed i, we get: g(i) = \(\sum_{l=i}^{\infty}\) Pr(X=l-i, Y=i) Substitute the expression for the joint probability we derived in Step 1: g(i) = \(\sum_{l=i}^{\infty}\) \(\binom{l}{l-i}\) p^{l-i} (1-p)^{i} * g(i)

Finding the value of p

Now, we need to find the value of p based on the expressions for f(i) and g(i). From the expressions of f(i) and g(i), we get: \(\frac{f(i)}{g(i)}\) = \(\frac{\sum_{k=0}^{i} \binom{i}{k} p^k (1-p)^{i-k}}{\sum_{l=i}^{\infty} \binom{l}{l-i} p^{l-i} (1-p)^{i}}\) By simplifying and using the hint provided, let F(s) = \(\sum f(i) s^i\) and G(s) = \(\sum g(i) s^i\). We can establish the following relation: F(u) F(v) = F(vp + (1-p)u) G(vp + (1-p)u) Now, we have to use this relation and the expressions we derived for f(i) and g(i) to find the value of p. We know that: F(1/(1-p)) * G(1/(1-p)) = F(0) * G(1) Since F(0) = f(0), G(1) = 1, and G(1/(1-p)) = 1/f(0), we find: 1/f(0) = p This shows that the value of p is determined by the condition G(1/(1-p)) = 1/f(0).