Question

Let \(X\) be a nonnegative integer-valued random variable with probability generating function \(f(s)=\sum_{n=0}^{\infty} a_{n} s^{n}\), After observing \(X\), then conduct \(X\) binomial trials with probability \(p\) of success. Let \(Y\) denote the resulting number of successes. (a) Determine the probability generating function of \(Y\). (b) Determine the probability generating funetion of \(X\) given that \(Y=X\).

Short answer

To summarize, the probability generating function of \(Y\) is given by: \(g(s)=\sum_{n=0}^{\infty} s^n \sum_{m=0}^{\infty} \binom{m}{n} p^n (1-p)^{m-n} a_m\) And the probability generating function of \(X\) given that \(Y=X\) is: \(P(X=k|Y=k)=\sum_{k=0}^{\infty} \frac{a_k p^k}{g(s)}s^k\)

Step-by-step solution

Probability generating functions and conditional probability

Recall that the probability generating function of a random variable is given by the series \(\sum_{n=0}^{\infty} a_{n} s^{n}\), where \(a_n\) is the probability that the random variable takes the value \(n\). The probability generating function of a binomial distribution with parameters \(n\) and \(p\) is given by \((1-p+ps)^n\). Lastly, recall the conditional probability: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\).

Determine the probability generating function of \(Y\)

To determine the probability generating function of \(Y\), we need to relate the probability generating function of a binomial distribution with the probability generating function of \(X\). Since the number of trials in each binomial experiment depends on the value of \(X\), we can write the probability generating function of \(Y\) in the following way: \[g(s) = E\left[(1-p+ps)^X\right]\] As \(X\) is a non-negative random variable, we can expand and use the property of linearity of expectation: \begin{align*} g(s) &= E\left[\sum_{n=0}^{\infty} \binom{X}{n} p^n (1-p)^{X-n} s^n\right] \\ &= \sum_{n=0}^{\infty} s^n E\left[ \binom{X}{n} p^n (1-p)^{X-n}\right] \end{align*} Now, we can use the fact that \(E[h(X)]=\sum_{n=0}^{\infty} h(n) P(X=n)\): \[g(s)=\sum_{n=0}^{\infty} s^n \sum_{m=0}^{\infty} \binom{m}{n} p^n (1-p)^{m-n} P(X=m)\] Since the generating function of \(X\) is \(f(s)=\sum_{n=0}^{\infty} a_n s^n\), we obtain: \[g(s)=\sum_{n=0}^{\infty} s^n \sum_{m=0}^{\infty} \binom{m}{n} p^n (1-p)^{m-n} a_m\]

Determine the probability generating function of \(X\) given that \(Y=X\)

We want to calculate the probability generating function of \(X\) given that \(Y=X\). To do this, we need to find the probability generating function of the joint event \((X=k, Y=k)\) and divide it by the probability generating function of the event \(Y=k\). First, calculate the probability generating function of the joint event, \(h(s) = P(X=k, Y=k)\). Since \(Y=k\) is the outcome of \(k\) successes in a binomial trial with parameters \(X\) and \(p\), we can write: \[h(s) = \sum_{k=0}^{\infty} a_k \binom{k}{k} p^k (1-p)^{0} s^k = \sum_{k=0}^{\infty} a_k p^k s^k\] Now, to find the conditional probability generating function of \(X\) given that \(Y=X\), we need to calculate \(P(X=k|Y=k)\), which can be done using the relation \(P(A|B) = \frac{P(A \cap B)}{P(B)}\): \[P(X=k|Y=k) = \frac{P(X=k, Y=k)}{P(Y=k)} = \frac{a_k p^k}{g(s)}\] Thus \[P(X=k|Y=k)=\sum_{k=0}^{\infty} \frac{a_k p^k}{g(s)}s^k\]