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Chapter 6: Problem 9

Let \(X_{1}, \ldots, X_{n}\) be iid as Poisson \(P(\theta)\). (a) Determine the UMVU estimator of \(P\left(X_{i}=0\right)=e^{-\theta}\). (b) Calculate the variance of the estimator of (a) up to terms of order \(1 / n\). [Hint: Write the estimator in the form \((1.15)\) where \(h(\bar{X})\) is the MLE of \(e^{-\theta}\).]

Short Answer

Expert verified
UMVU estimator: \((1 - \frac{1}{n}) e^{-\bar{X}}\), Variance: \(\frac{1}{n} e^{-2\theta}\).

Step by step solution

01

Understanding the Poisson Distribution and UMVU Estimator

The problem requires finding the UMVU (Uniformly Minimum Variance Unbiased) estimator for a function of a Poisson distribution's parameter. We are given that each \(X_i\) follows a Poisson distribution with parameter \(\theta\), and we need to estimate \(P(X_i=0) = e^{-\theta}\). The UMVU estimator is the unbiased estimator with the least variance for estimating this function.
02

Identifying the Sufficient Statistic

Since the data is iid Poisson, the sum \(S = \sum_{i=1}^{n} X_i\) is a sufficient statistic for \(\theta\). The distribution of \(S\) is Poisson with parameter \(n\theta\). The sample mean \(\bar{X} = \frac{S}{n}\) is also an unbiased estimator for \(\theta\).
03

Maximum Likelihood Estimation (MLE)

The MLE of \(\theta\) is \(\hat{\theta} = \bar{X}\). The suggested form \(h(\bar{X})\) hints that we should use \(e^{-\hat{\theta}} = e^{-\bar{X}}\) as the MLE for \(e^{-\theta}\). This is a biased estimator of \(e^{-\theta}\).
04

Correcting Bias with Expectation

To find the unbiased estimator, note that \(e^{-\bar{X}}\) is biased downwards. Using the Taylor expansion or the properties of Poisson distributions, an unbiased correction can be found. Specifically, the bias-corrected version for large \(n\) gives: \(h(\bar{X}) = \left( 1 - \frac{1}{n} \right) e^{-\bar{X}}\), correcting for bias in expectation.
05

Variance Calculation up to \(O(1/n)\)

To calculate variance, consider the variance of a function of a random variable. Using Taylor expansion around \(\bar{X} = \theta\), the approximation is used to find that the variance of the estimator is approximately \( \text{Var}(h(\bar{X})) \approx \frac{1}{n} e^{-2\theta} \), up to terms \(O(\frac{1}{n})\).
06

Compile the Results

The UMVU estimator for \(P(X_i=0) = e^{-\theta}\) is \((1 - \frac{1}{n}) e^{-\bar{X}}\). The variance of this estimator, up to terms of order \(1/n\), is \(\frac{1}{n} e^{-2\theta}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Poisson Distribution
A Poisson distribution is a probability model that describes events happening at a constant mean rate over a given interval of time or space. These events are assumed to occur independently of one another. In other words, knowing the outcome of a previous event doesn't provide you with any information about future events. Key characteristics of the Poisson distribution:
  • Discrete Probability Distribution: Outcomes are discrete outcomes like 0, 1, 2, etc.
  • Described by the parameter \( \theta \), which is the average rate at which events occur.
  • The probability mass function can be written as \( P(X = k) = \frac{\theta^k e^{-\theta}}{k!} \), where \( k \) is the number of occurrences.
Given a set of events \(X_{1}, \ldots, X_{n}\) that are independently and identically distributed as Poisson \(P(\theta)\), this can be used to estimate probabilities or test hypotheses about the average rate \( \theta \). In our original exercise, \(P(X_i = 0) = e^{-\theta}\) represents the probability of observing zero events in one instance.
Sufficient Statistic
A sufficient statistic is a statistic that captures all the information needed from the data to make inferences about the parameter in question. For a given dataset, it is a compressive form, containing all necessary information about the parameter being estimated.In the context of a Poisson distribution:
  • The sum \( S = \sum_{i=1}^{n} X_i \) is a sufficient statistic for the parameter \( \theta \).
  • This means that any inference about \( \theta \) can be made using \( S \) alone without needing the entire dataset.
  • With \( S\) being Poisson distributed with parameter \(n \theta \), we can efficiently summarize our sample to understand \( \theta \).
The sample mean \( \bar{X} = \frac{S}{n} \) emerges as a vital tool for estimating \( \theta \) as it is an unbiased estimator and thereby a crucial building block for further statistical procedures.
Maximum Likelihood Estimation
Maximum Likelihood Estimation (MLE) is a method used to estimate the parameters of a statistical model. It finds the parameter values that maximize the likelihood of the observed data under the model.For a Poisson distribution, the MLE:
  • Is based on the observed data from a random sample \(X_{1}, \ldots, X_{n}\).
  • For \( \theta \) in Poisson, the MLE is \(\hat{\theta} = \bar{X}\), where \( \bar{X} \) is the sample mean.
  • In the exercise, to estimate \( e^{-\theta} \) using MLE, we consider \( h(\bar{X}) = e^{-\bar{X}} \).
This approach provides an estimate that aligns closely with our data, yet it often introduces bias, meaning \( e^{-\bar{X}} \) is not an unbiased estimator of \( e^{-\theta} \). Further steps, such as bias correction, are required to improve this estimation.
Bias Correction
Bias correction refers to adjustments made to estimators to improve their unbiasedness and reliability. Often, estimators derived via straightforward methods like Maximum Likelihood can be inherently biased.In the case of estimating \( e^{-\theta} \):
  • The initial MLE \( e^{-\hat{\theta}} \) was found to be biased.
  • The bias tends to underestimate the true value of \( e^{-\theta} \).
  • To correct for bias, the form \( h(\bar{X}) = \left( 1 - \frac{1}{n} \right) e^{-\bar{X}} \) is used, where the factor \( 1 - \frac{1}{n} \) accounts for bias, especially for large \( n \).
Incorporating bias correction ensures that the estimator becomes unbiased, or as unbias as possible, maintaining consistency and reliability in statistical inference.

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Most popular questions from this chapter

If \(X_{1}, \ldots, X_{n}\) are iid \(n\left(\mu, \sigma^{2}\right)\), show that \(S^{r}=\left[1 /(n-1) \Sigma\left(x_{i}-\bar{x}\right)^{2}\right]^{r / 2}\) is an asymptotically unbiased estimator of \(\sigma^{r}\).

Let \(X_{1}, \ldots, X_{n}\) be iid as \(U(0, \theta)\). From Example 2.1.14, \(\delta_{n}=(n+1) X_{(n) / n}\) is the UMVU estimator of \(\theta\), whereas the MLE is \(X_{(n) \text { - }}\) Determine the limit distribution of (a) \(n\left[\theta-\delta_{n}\right]\) and (b) \(n\left[\theta-X_{(n)}\right]\). Comment on the asymptotic bias of these estimators. [Hint: \(P\left(X_{(n)} \leq y\right)=y^{n} / \theta^{n}\) for any \(0

(a) If the cdf \(F\) is symmetric and if log \(F(x)\) is strictly concave, so is \(\log [1-F(x)]\). (b) Show that log \(F(x)\) is strictly concave when \(F\) is strongly unimodal but not when \(F\) is Cauchy.

Maximum likelihood estimation in the probit model of Section \(3.6\) can be implemented using the EM algorithm. We observe independent Bernoulli variables \(X_{1}, \ldots, X_{n}\), which depend on unobservable variables \(Z_{i}\) distributed independently as \(N\left(\zeta_{i}, \sigma^{2}\right)\), where $$ X_{i}= \begin{cases}0 & \text { if } Z_{i} \leq u \\ 1 & \text { if } Z_{i}>u\end{cases} $$ Assuming that \(u\) is known, we are interested in obtaining ML estimates of \(\zeta\) and \(\sigma^{2}\). (a) Show that the likelihood function is \(p^{\sum x_{i}}(1-p)^{n-\Sigma x_{4}}\), where $$ p=P\left(Z_{i}>u\right)=\Phi\left(\frac{\zeta-u}{\sigma}\right) $$ (b) If we consider \(Z_{1}, \ldots, Z_{n}\) to be the complete data, the complete-data likelihood is $$ \prod_{i=1}^{n} \frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{1}{2 g^{2}}\left(z_{i}-\zeta\right)^{2}} $$ and the expected complete-data log likelihood is $$ -\frac{n}{2} \log \left(2 \pi \sigma^{2}\right)-\frac{1}{2 \sigma^{2}} \sum_{i=1}^{n}\left[E\left(Z_{i}^{2} \mid X_{i}\right)-2 \zeta E\left(Z_{i} \mid X_{i}\right)+\zeta^{2}\right] $$ (c) Show that the EM sequence is given by $$ \begin{aligned} &\hat{\zeta}_{(j+1)}=\frac{1}{n} \sum_{i=1}^{n} t_{i}\left(\hat{\zeta}_{(j)}, \hat{\sigma}_{(j)}^{2}\right) \\ &\hat{\sigma}_{(j+1)}^{2}=\frac{1}{n}\left[\sum_{i=1}^{n} v_{i}\left(\hat{\zeta}_{(j)}, \hat{\sigma}_{(j)}^{2}\right)-\frac{1}{n}\left(\sum_{i=1}^{n} t_{i}\left(\hat{\zeta}_{(j)}, \hat{\sigma}_{(j)}^{2}\right)\right)^{2}\right] \end{aligned} $$ where $$ t_{i}\left(\zeta, \sigma^{2}\right)=E\left(Z_{i} \mid X_{i}, \zeta, \sigma^{2}\right) \text { and } \quad v_{i}\left(\zeta, \sigma^{2}\right)=E\left(Z_{i}^{2} \mid X_{i}, \zeta, \sigma^{2}\right) $$ (d) Show that $$ \begin{aligned} &E\left(Z_{i} \mid X_{i}, \zeta, \sigma^{2}\right)=\zeta+\sigma H_{i}\left(\frac{u-\zeta}{\sigma}\right) \\ &E\left(Z_{i}^{2} \mid X_{i}, \zeta, \sigma^{2}\right)=\zeta^{2}+\sigma^{2}+\sigma(u+\zeta) H_{i}\left(\frac{u-\zeta}{\sigma}\right) \end{aligned} $$ where $$ H_{i}(t)= \begin{cases}\frac{\varphi(t)}{1-\Phi(t)} & \text { if } X_{i}=1 \\\ -\frac{\varphi(t)}{\Phi(t)} & \text { if } X_{i}=0\end{cases} $$ (e) Show that \(\hat{\zeta}(j) \rightarrow \hat{\zeta}\) and \(\hat{\sigma}_{(j)}^{2} \rightarrow \hat{\sigma}^{2}\), the ML estimates of \(\zeta\) and \(\sigma^{2}\).

The property of asymptotic relative efficiency was defined (Definition \(6.6\) ) for estimators that converged to normality at rate \(\sqrt{n} .\) This definition, and Theorem \(6.7\), can be generalized to include other distributions and rates of convergence. 9 As has been pointed out by Stigler (1973) such models for heavy-tailed distributions had already been proposed much earlier in a forgotten work by Newcomb (1882, 1886). Theorem 9.1 Let \(\left\\{\delta_{i n}\right\\}\) be two sequences of estimators of \(g(\theta)\) such that $$ n^{\alpha}\left[\delta_{i n}-g(\theta)\right] \stackrel{\mathcal{L}}{\rightarrow} \tau_{i} T, \alpha>0, \tau_{i}>0, i=1,2 $$ where the distribution \(H\) of \(T\) has support on an interval \(-\infty \leq A
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