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Chapter 3: Problem 94

Let \(N\) be a hypergeometric random variable having the distribution of the number of white balls in a random sample of size \(r\) from a set of \(w\) white and \(b\) blue balls. That is, $$ P\\{N=n\\}=\frac{\left(\begin{array}{c} w \\ n \end{array}\right)\left(\begin{array}{c} b \\ r-n \end{array}\right)}{\left(\begin{array}{c} w+b \\ r \end{array}\right)} $$ where we use the convention that \(\left(\begin{array}{c}m \\\ j\end{array}\right)=0\) if either \(j<0\) or \(j>m\). Now, consider a compound random variable \(S_{N}=\sum_{i=1}^{N} X_{i}\), where the \(X_{i}\) are positive integer valued random variables with \(\alpha_{j}=P\left\\{X_{i}=j\right\\}\) (a) With \(M\) as defined as in Section \(3.7\), find the distribution of \(M-1\). (b) Suppressing its dependence on \(b\), let \(P_{w, r}(k)=P\left\\{S_{N}=k\right\\}\), and derive a recursion equation for \(P_{w, r}(k)\). (c) Use the recursion of (b) to find \(P_{w, r}(2)\).

Short Answer

Expert verified
\(P_{w, r}(2) = \frac{b(\, _{w-1}C_{0}\, _{b}C_{r})(\, _{w-1}C_{0}\, _{b}C_{r-1})}{(\, _{w+b}C_{r})(\, _{w+b-1}C_{r-1})} + \frac{w(\, _{w}C_{1}\, _{b}C_{r-1})(\, _{w}C_{1}\, _{b}C_{r-2})}{(\, _{w+b}C_{r})(\, _{w+b}C_{r-1})}\).

Step by step solution

01

Review the definition of \(M\)

Recall that \(M\) is the moment generating function of the random variable \(X\), defined as \(M(t) = E[e^{tX}] = \sum_{i=1}^{+\infty}\alpha_i e^{tX_i}\). The moment generating function of \(S_N\) (denoted \(M_N(t)\)) can be obtained from the moment generating functions of the \(X_i\) by multiplying them together; that is, \(M_N(t) = \prod_{i=1}^{N} M(t)\).
02

Determine the moment generating function of \(M-1\)

Let \(Y = X_i - 1\). The moment generating function of \(Y\) is: \(M_Y(t) = E[e^{t(Y)}] = E[e^{t(X_i-1)}] = e^{-t} M(t)\). Since \(S_N = \sum_{i=1}^{N} X_i\) and \(S'_N = \sum_{i=1}^{N} (X_i - 1) = S_N - N\), we have \(M_{S'_N}(t) = \left(e^{-t} M(t)\right)^N\). (b) Find a recursion equation for \(P_{w, r}(k) = P\\{S_N = k\\}\)
03

Derive a recursion equation

Note that \(P_{w, r}(k) = P\\{S_N = k\\} = P\\{S_{N-1} = k-N, N-1 \text{ white balls}\\} + P\\{S_N = k, N \text{ white balls}\\}\). Alternatively, we can state this as: \(P_{w, r}(k) = P\\{ S_N = k | N-1 \text{ white balls} \\}P\\{ N-1 \text{ white balls}\\} + P\\{S_N = k | N \text{ white balls}\\}P\\{ N \text{ white balls}\\}\). Using the probability mass function of \(N\), we can write this as: \(P_{w, r}(k) = \frac{\left(\begin{array}{c}w-1\\n-1\end{array}\right)\left(\begin{array}{c}b\\r-n+1\end{array}\right)}{\left(\begin{array}{c}w+b\\r\end{array}\right)}P_{w-1, r-1}(k-n+1) + \frac{\left(\begin{array}{c}w\\n\end{array}\right)\left(\begin{array}{c}b\\r-n\end{array}\right)}{\left(\begin{array}{c}w+b\\r\end{array}\right)}P_{w, r-1}(k-n)\). (c) Use the recursion equation from (b) to find \(P_{w, r}(2)\)
04

Compute \(P_{w, r}(2)\)

We are given \(S_N = 2\). Using the recursion equation from part (b), we have: \(P_{w, r}(2) = \frac{\left(\begin{array}{c}w-1\\0\end{array}\right)\left(\begin{array}{c}b\\r\end{array}\right)}{\left(\begin{array}{c}w+b\\r\end{array}\right)}P_{w-1, r-1}(1) + \frac{\left(\begin{array}{c}w\\1\end{array}\right)\left(\begin{array}{c}b\\r-1\end{array}\right)}{\left(\begin{array}{c}w+b\\r\end{array}\right)}P_{w, r-1}(1)\). Now, \(P_{w-1, r-1}(1) = \frac{\left(\begin{array}{c}w-1\\0\end{array}\right)\left(\begin{array}{c}b\\r-1\end{array}\right)}{\left(\begin{array}{c}w+b-1\\r-1\end{array}\right)}\) and \(P_{w, r-1}(1) = \frac{\left(\begin{array}{c}w\\1\end{array}\right)\left(\begin{array}{c}b\\r-2\end{array}\right)}{\left(\begin{array}{c}w+b\\r-1\end{array}\right)}\). Substituting these values into the recursion equation, we get: \(P_{w, r}(2) = \left(\frac{\left(\begin{array}{c}w-1\\0\end{array}\right)\left(\begin{array}{c}b\\r\end{array}\right)}{\left(\begin{array}{c}w+b\\r\end{array}\right)}\right)\left(\frac{\left(\begin{array}{c}w-1\\0\end{array}\right)\left(\begin{array}{c}b\\r-1\end{array}\right)}{\left(\begin{array}{c}w+b-1\\r-1\end{array}\right)}\right) + \left(\frac{\left(\begin{array}{c}w\\1\end{array}\right)\left(\begin{array}{c}b\\r-1\end{array}\right)}{\left(\begin{array}{c}w+b\\r\end{array}\right)}\right)\left(\frac{\left(\begin{array}{c}w\\1\end{array}\right)\left(\begin{array}{c}b\\r-2\end{array}\right)}{\left(\begin{array}{c}w+b\\r-1\end{array}\right)}\right)\). Simplifying the expression will give us the probability \(P_{w, r}(2)\).

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Most popular questions from this chapter

Let \(p_{0}=P[X=0\\}\) and suppose that \(0

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