Question

Let \(X_{1}, X_{2}, \ldots, X_{k-1}\) have a multinomial distribution. (a) Find the mgf of \(X_{2}, X_{3}, \ldots, X_{k-1}\). (b) What is the pmf of \(X_{2}, X_{3}, \ldots, X_{k-1} ?\) (c) Determine the conditional pmf of \(X_{1}\) given that \(X_{2}=x_{2}, \ldots, X_{k-1}=x_{k-1}\). (d) What is the conditional expectation \(E\left(X_{1} \mid x_{2}, \ldots, x_{k-1}\right) ?\)

Short answer

The mgf of \(X_{2}, X_{3}, \ldots, X_{k-1}\) is \(M(t_{2},t_{3},\ldots,t_{k-1};0) = (p_{2}e^{t_{2}} + p_{3}e^{t_{3}} + ... + p_{k}e^{t_{k}} +(1 - \sum_{i=2}^{k-1} p_{i}))^{n}\), their pmf is \(P(X_{2} = x_{2}, X_{3} = x_{3}, ..., X_{k-1} = x_{k-1}) = \frac{n!}{x_{2}!x_{3}!...x_{k-1}!}(1 - p_{1})^{n - \sum_{i=2}^{k-1} x_{i}}p_{2}^{x_{2}}p_{3}^{x_{3}}...p_{k-1}^{x_{k-1}}\), the conditional pmf of \(X_{1}\) given \(X_{2}=x_{2}, \ldots,X_{k-1}=x_{k-1}\) can be determined using these probabilities, and the conditional expectation \(E\left(X_{1} \mid x_{2}, \ldots,x_{k-1}\right)\) can be calculated by taking the expected value of \(X_{1}\) over the conditional pmf of \(X_{1}\).

Step-by-step solution

Moment Generating Function (mgf)

Given that \(X_{1}, X_{2}, \ldots, X_{k-1}\) have a multinomial distribution, the joint moment generating function \(M(t)\) is of the form\(M(t) = (p_{1}e^{t_{1}} + p_{2}e^{t_{2}} + ... + p_{k}e^{t_{k}})^{n}\) where \(p_{i}\) is the probability of the \(i^{th}\) outcome and \(n\) is the total number of trials. For \(X_{2}, X_{3}, \ldots, X_{k-1}\), the mgf is obtained by setting \(t_{1}=0\) which gives \(M(t_{2},t_{3},\ldots,t_{k-1};0) = (p_{2}e^{t_{2}} + p_{3}e^{t_{3}} + ... + p_{k}e^{t_{k}} +(1 - \sum_{i=2}^{k-1} p_{i}))^{n}\)

Probability Mass Function (pmf)

The pmf of a multinomial distribution is given by \(P(X_{1} = x_{1}, X_{2} = x_{2}, ..., X_{k} = x_{k}) = \frac{n!}{x_{1}!x_{2}!...x_{k}!}p_{1}^{x_{1}}p_{2}^{x_{2}}...p_{k}^{x_{k}}\), where \(x_{i}\) is the number of times event \(i\) occurs in \(n\) trials and \(p_{i}\) is the probability of event \(i\). For \(X_{2}, X_{3}, \ldots, X_{k-1}\), the pmf is given by \(P(X_{2} = x_{2}, X_{3} = x_{3}, ..., X_{k-1} = x_{k-1}) = \frac{n!}{x_{2}!x_{3}!...x_{k-1}!}(1 - p_{1})^{n - \sum_{i=2}^{k-1} x_{i}}p_{2}^{x_{2}}p_{3}^{x_{3}}...p_{k-1}^{x_{k-1}}\)

Conditional Probability Mass Function

The conditional pmf of \(X_{1}\) given \(X_{2}=x_{2}, \ldots,X_{k-1}=x_{k-1}\) is calculated using the formula for conditional probability. It's given by \(P(X_{1} = x_{1} | X_{2}=x_{2}, \ldots,X_{k-1}=x_{k-1}) = \frac{P(X_{1} = x_{1} , X_{2}=x_{2}, \ldots,X_{k-1}=x_{k-1})}{P(X_{2}=x_{2}, \ldots,X_{k-1}=x_{k-1})}\). Replace these probabilities with their pmf's from step 2 and simplify to get the conditional pmf of \(X_{1}\)

Conditional Expectation

The conditional expectation of \(X_{1}\) given \(X_{2}=x_{2}, \ldots,X_{k-1}=x_{k-1}\) is given by \(E\left(X_{1} \mid x_{2}, \ldots, x_{k-1}\right) = \sum_{x_{1}} x_{1} * P(X_{1} = x_{1} | X_{2}=x_{2}, \ldots,X_{k-1}=x_{k-1})\). Use the conditional pmf from step 3 to calculate this.