Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a Poisson distribution with parameter \(\theta>0\). From Remark 7.6.1, we know that \(E\left[(-1)^{X_{1}}\right]=e^{-2 \theta}\). (a) Show that \(E\left[(-1)^{X_{1}} \mid Y_{1}=y_{1}\right]=(1-2 / n)^{y_{1}}\), where \(Y_{1}=X_{1}+X_{2}+\cdots+X_{n}\). Hint: First show that the conditional pdf of \(X_{1}, X_{2}, \ldots, X_{n-1}\), given \(Y_{1}=y_{1}\), is multinomial, and hence that of \(X_{1}\), given \(Y_{1}=y_{1}\), is \(b\left(y_{1}, 1 / n\right)\). (b) Show that the mle of \(e^{-2 \theta}\) is \(e^{-2 \bar{X}}\). (c) Since \(y_{1}=n \bar{x}\), show that \((1-2 / n)^{y_{1}}\) is approximately equal to \(e^{-2 \pi}\) when \(n\) is large.

Short answer

Step 1 shows the conditional expectation based on multinomial distribution for random sample. In Step 2, the MLE of \(e^{-2 \theta}\) is derived to be \(e^{-2 \bar{X}}\). Lastly in Step 3, the estimation is taken to the scenario where \(n\) is large, showing \((1-2 / n)^{y_{1}}\) approximates to \(e^{-2 \bar{X}}\) when \(n\) is large.

Step-by-step solution

Derive the Conditional Probability

Use the definition of Poisson and Multinomial Distribution given, with the condition \(Y_{1}=y_{1}\), to demonstrate that \(E\left[(-1)^{X_{1}} \mid Y_{1}=y_{1}\right]=(1-2 / n)^{y_{1}}\). This is accomplished by using the hint, formulating the multinomial distribution, and then expressing the conditional expectation accordingly.

Compute Maximum Likelihood Estimation (MLE)

Implement the method of maximum likelihood estimation (MLE) to show that the MLE of \(e^{-2 \theta}\) is \(e^{-2 \bar{X}}\). This is done by taking the log of the joint likelihood, differentiating with respect to \(\theta\), and setting to zero to find the extremum.

Apply the result to Large Sample

Now, apply the result from Step 1 to large samples (say, when n is very large) using the Standard Limit Theorems {(Law of Large Numbers and Central Limit Theorem). Show that \((1-2 / n)^{y_{1}}\) approximates to \(e^{-2 \bar{X}}\) when n is large. This is done by taking the limit \(n\) to infinity and comparing two approaches to estimate the underlying population parameter.