Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Poisson distribution with parameter \(\theta>0\) (a) Find the MVUE of \(P(X \leq 1)=(1+\theta) e^{-\theta}\). Hint: \(\quad\) Let \(u\left(x_{1}\right)=1, x_{1} \leq 1\), zero elsewhere, and find \(E\left[u\left(X_{1}\right) \mid Y=y\right]\), where \(Y=\sum_{1}^{n} X_{i}\). (b) Express the MVUE as a function of the mle. (c) Determine the asymptotic distribution of the mle.

Short answer

The MVUE of \(P(X ≤ 1)=(1+\theta) e^{-\theta}\) is \((1+y)e^{-y}/n\), which can be expressed as \((1+\frac{Y}{n})e^{-Y/n}\) in terms of MLE. The asymptotic distribution of MLE is \(N(\theta,\frac{\theta}{n})\).

Step-by-step solution

Finding the MVUE

Using the given hint, let \(u(X_1)\) be an indicator function such that it equals to 1 if \(x_1 ≤ 1\) and 0 otherwise. Considering the given random sample from Poisson distribution with parameter \(\theta\), the expected value \(E[u(X_1)|Y=y]\) where \(Y = \sum_{i=1}^{n} X_i\) can be computed as follows which is the MVUE:\n\[E[u(X_1)|Y=y] = P(X_1 ≤ 1|Y=y) = P(X_1 =0|Y=y) + P(X_1=1|Y=y) = (1+y)e^{-y}/n\]

Expressing the MVUE as a function of the MLE

The maximum likelihood estimator (MLE) for a Poisson distribution is the sample mean. Let \(Y/n\) be the MLE, we know that \(Y\) follows also poisson distribution with parameter \(n\theta \). Thus, the MVUE can be expressed as a function of the MLE as follow:\n \[(1+\frac{Y}{n})e^{-Y/n}\]

Determine the asymptotic distribution of MLE

The asymptotic distribution of the MLE can be found using the central limit theorem for large n, which suggests that the distribution of MLE tends to a normal distribution. To be clear, when n tends to infinity, the distribution of \(\sqrt{n}(\theta^{MLE} - \theta )\to N(0, \frac{\theta}{n})\). Considering the MLE for \(\theta\) is \(Y/n\), the final asymptotic distribution will be \(N(\theta,\frac{\theta}{n})\).

Key concepts

Poisson Distribution

The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring within a fixed interval of time or space. These events must occur independently from one another. One of the key characteristics of the Poisson distribution is that the events happen with a known constant mean rate, which is denoted as \( \theta \). In mathematical terms:

• The probability of observing \( x \) events is given by the formula: \( P(X = x) = \frac{e^{-\theta} \theta^x}{x!} \)
• \( \theta \) is the average rate of occurrence; it is greater than zero and can be any positive number.

Given its nature, Poisson distribution is often used to model rare events, like the number of typos in a book, or the number of cars passing through a checkpoint in an hour.When dealing with samples from a Poisson distribution, it is useful to find estimators like MVUE and MLE for better statistical analysis and predictions.

MVUE (Minimum Variance Unbiased Estimator)

The Minimum Variance Unbiased Estimator or MVUE is an important concept in statistics as it provides the best unbiased estimation of a parameter with the minimum variance among all unbiased estimators. For example, in the exercise given, we aim to find the MVUE of \( P(X \leq 1) \). The concept of unbiased estimation is crucial because it ensures that, on average, the estimator will yield the correct parameter.To find the MVUE, the exercise uses an indicator function \( u(X_1) \), which equals 1 if \( x_1 \leq 1 \) and 0 otherwise. The expected value of this function given the sample sum \( Y \) is calculated: \[E[u(X_1)|Y=y] = P(X_1 \leq 1|Y=y) = P(X_1 =0|Y=y) + P(X_1=1|Y=y) = \frac{(1+y)e^{-y}}{n}\]This result effectively provides the MVUE for the given scenario.The method outlines how to systematically break down the problem and apply the principle of conditioning to isolate our estimate.

MLE (Maximum Likelihood Estimation)

Maximum Likelihood Estimation, or MLE, is a method used in statistics to estimate the parameters of a probability distribution by maximizing a likelihood function. In the context of the Poisson distribution, the MLE for the parameter \( \theta \) is the sample mean, which is \( \frac{Y}{n} \), where \( Y \) is the sum of the sample data.To express the MVUE as a function of the MLE, we used the sample mean \( \frac{Y}{n} \) as our estimate for \( \theta \). Consequently, the MVUE can be written as:\[(1+\frac{Y}{n})e^{-Y/n}\]When considering the asymptotic distribution of the MLE, the Central Limit Theorem helps. As the sample size \( n \) becomes larger, the distribution of the MLE approximates a normal distribution. Specifically, for the Poisson MLE \( \theta^{MLE} = \frac{Y}{n} \), it follows:\[\sqrt{n}(\theta^{MLE} - \theta) \rightarrow N(0, \frac{\theta}{n})\]This suggests that the MLE is consistent and asymptotically normal, meaning it becomes more accurate as the sample size grows.