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Find minimum variance unbiased estimators of \(\lambda^{2}, e^{\lambda}\), and \(e^{-n \lambda}\) based on a random sample \(Y_{1}, \ldots, Y_{n}\) from a Poisson density with mean \(\lambda\). Show that no unbiased estimator of \(\log \lambda\) exists.

Short Answer

Expert verified
The unbiased estimators are: \( \hat{\lambda}^2 = \left(\frac{T}{n}\right)^2 - \frac{\hat{\lambda}}{n} \), \( e^{\lambda} = \left(1 + \frac{1}{n}\right)^T \), \( e^{-n \lambda} = \left(1 - \frac{1}{n}\right)^T \); no unbiased estimator for \( \log \lambda \).

Step by step solution

01

Identify the Sufficient Statistic

The sufficient statistic for a random sample from a Poisson distribution with parameter \( \lambda \) is the sample sum \( T = \sum_{i=1}^{n} Y_i \). This is because the Poisson distribution is in the exponential family, and the sum \( T \) summarizes all the information about \( \lambda \).
02

Find the Unbiased Estimator for \( \lambda \)

The expected value of \( T \) is \( n \lambda \). Therefore, an unbiased estimator for \( \lambda \) is \( \hat{\lambda} = \frac{T}{n} \).
03

Derive the Unbiased Estimator for \( \lambda^2 \)

We use the Rao-Blackwell theorem to find the unbiased estimator for \( \lambda^2 \). The unbiased estimator of \( \lambda^2 \) using \( \hat{\lambda}^2 \) is derived by noting that the second moment of \( T \) is given by \( E[T^2] = n^2\lambda^2 + n\lambda \). Hence, the unbiased estimator is \( \hat{\lambda}^2 = \left( \frac{T}{n} \right)^2 - \frac{\hat{\lambda}}{n} \).
04

Derive the Unbiased Estimator for \( e^{\lambda} \)

Since \( E(e^{T}) \) can be expressed as a function of \( \lambda \), the unbiased estimator for \( e^{\lambda} \) is \( \left(1 + \frac{1}{n}\right)^T \), as \( E\left(\left(1+\frac{1}{n}\right)^T\right) = e^\lambda \).
05

Derive the Unbiased Estimator for \( e^{-n \lambda} \)

For \( e^{-n \lambda} \), consider \( E\left( \left(1 - \frac{1}{n}\right)^T \right) = e^{-n \lambda} \). Thus, the unbiased estimator is \( \left(1 - \frac{1}{n} \right)^T \).
06

Show No Unbiased Estimator for \( \log \lambda \) Exists

Using Lehmann-Scheffé theorem, a function solely of the sufficient statistic \( T \) must exist to find the unbiased estimator of \( \log \lambda \). However, due to the nature of \( \log \lambda \), which involves division by zero as \( \lambda \to 0 \), no such unbiased estimator exists for \( \log \lambda \).

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Key Concepts

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

Poisson Distribution
The Poisson Distribution is a discrete probability distribution, widely known for modeling the number of events occurring within a fixed interval, provided these events happen with a constant mean rate and independently of each other. In statistical terms, it sees a random variable as distributed according to a Poisson process, described by the parameter \( \lambda \), the rate or average number of occurrences.
To give a concrete example, it often applies to scenarios like the number of phone calls received by a call center in an hour or the number of buses arriving at a given stop. In this exercise, you're looking at random samples \( Y_1, \ldots, Y_n \) from a Poisson distribution with mean \( \lambda \). This foundational concept is key to understanding the estimations we explore further, such as those for \( \lambda^2 \), \( e^{\lambda} \), and \( e^{-n\lambda} \).
Sufficient Statistics
Sufficient statistics are essential tools in statistical inference. They help summarize all necessary information from a sample data set about a parameter of a probability distribution.
In this exercise, the sum \( T = \sum_{i=1}^{n} Y_i \) serves as the sufficient statistic for the Poisson distribution with parameter \( \lambda \). This means that "T" condenses all the data's information about this parameter.
This principle makes estimation tasks easier and less cumbersome as only this reduced form needs to be considered to make inferences about \( \lambda \). This approach is what paves the way to various estimator definitions, including unbiased estimators employed later on.
Rao-Blackwell Theorem
The Rao-Blackwell Theorem is a fundamental concept in statistics that provides a method to improve an estimator by conditioning on a sufficient statistic.
In this scenario, it indicates that any unbiased estimator can be improved given a sufficient statistic, thereby leading to the minimum variance unbiased estimator (MVUE).
It was used in this exercise to refine the unbiased estimator \( \lambda^2 \). Initially based on raw calculations, it employed \( \hat{\lambda}^2 \) as an estimator by understanding the second moment of the sufficient statistic \( T \).
Thus, the Rao-Blackwell Theorem ensures we have the most accurate, stable estimator form, reflecting both efficiency and unbiased nature.
Unbiased Estimators
Unbiased estimators are the cornerstone for making accurate inferences about population parameters. An estimator is unbiased if, on average, it equals the true parameter value of a population from which a random sample is drawn.
Throughout the provided solution, various unbiased estimators are derived for parameters related to \( \lambda \) itself, including \( \lambda^2 \), \( e^{\lambda} \), and \( e^{-n\lambda} \).
For instance, the unbiased estimator for \( \lambda \) is expressed as \( \hat{\lambda} = \frac{T}{n} \). Such expressions ensure that when accounting for the sample data, you have an accurate center point around the actual parameter value, offering a reliable framework for statistical analysis.
Lehmann-Scheffé Theorem
The Lehmann-Scheffé Theorem is a guiding principle in the search for a minimum variance unbiased estimator (MVUE). This theorem states that if there is a complete sufficient statistic, the function of this statistic that is unbiased towards estimating a parameter is indeed the unique MVUE.
However, the application of this theorem sometimes shows the limitation of finding unbiased estimators in certain scenarios. In the case of trying to find an unbiased estimator for \( \log \lambda \), the Lehmann-Scheffé Theorem indicates no such estimator is possible. This is due to the problematic nature of \( \log \lambda \) becoming undefined as \( \lambda \approaches 0 \), leaving no unbiased estimation pathway due to boundary constraints.

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Most popular questions from this chapter

Let \(\bar{Y}\) be the average of a random sample from the uniform density on \((0, \theta)\). Show that \(2 \bar{Y}\) is unbiased for \(\theta\). Find a sufficient statistic for \(\theta\), and obtain an estimator based on it which has smaller variance. Compare their mean squared errors.

Show that when the data are normal, the efficiency of the Huber estimating function \(g_{c}(y ; \theta)\) compared to the optimal function \(g_{\infty}(y ; \theta)\) is $$ \frac{\\{1-2 \Phi(-c)\\}^{2}}{1+2\left\\{c^{2} \Phi(-c)-\Phi(-c)-c \phi(c)\right\\}} $$ Hence verify that the efficiency is \(0.95\) when \(c=1.345\).

Consider testing the hypothesis that a binomial random variable has probability \(\pi=1 / 2\) against the alternative that \(\pi>1 / 2\). For what values of \(\alpha\) does a uniformly most powerful test exist when the denominator is \(m=5\) ?

Independent random samples \(Y_{i 1}, \ldots, Y_{i n_{i}}\), where \(n_{i} \geq 2\), are drawn from each of \(k\) normal distributions with means \(\mu_{1}, \ldots, \mu_{k}\) and common unknown variance \(\sigma^{2}\). Derive the likelihood ratio statistic \(W_{\mathrm{p}}\) for the null hypothesis that the \(\mu_{i}\) all equal an unknown \(\mu\), and show that it is a monotone function of $$ R=\frac{\sum_{i=1}^{k} n_{i}\left(\bar{Y}_{i \cdot}-\bar{Y}_{. .}\right)^{2}}{\sum_{i=1}^{k} \sum_{j=1}^{n_{i}}\left(Y_{i j}-\bar{Y}_{i}\right)^{2}} $$ where \(\bar{Y}_{i}=n_{i}^{-1} \sum_{j} Y_{i j}\) and \(\bar{Y}_{. .}=\left(\sum n_{i}\right)^{-1} \sum_{i, j} Y_{i j}\). What is the null distribution of \(R ?\)

Let \(Y_{1}, \ldots, Y_{n}\) be a random sample from an unknown density \(f\). Let \(I_{j}\) indicate whether or not \(Y_{j}\) lies in the interval ( \(\left.a-\frac{1}{2} h, a+\frac{1}{2} h\right]\), and consider \(R=\sum I_{j}\). Show that \(R\) has a binomial distribution with denominator \(n\) and probability $$ \int_{a-\frac{1}{2} h}^{a+\frac{1}{2} h} f(y) d y $$ Hence show that \(R /(n h)\) has approximate mean and variance \(f(a)+\frac{1}{2} h^{2} f^{\prime \prime}(a)\) and \(f(a) / n h\), where \(f^{\prime \prime}\) is the second derivative of \(f\). What implications have these results for using the histogram to estimate \(f(a)\) ?

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