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Chapter 6: Problem 12

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the Poisson distribution with \(0<\theta \leq 2\). Show that the mle of \(\theta\) is \(\widehat{\theta}=\min \\{\bar{X}, 2\\}\).

Short Answer

Expert verified
The maximum likelihood estimate (MLE) of the Poisson distribution parameter \(\theta\) given that \(0<\theta \leq 2\) from a random sample is \(\widehat{\theta}=min\{\bar{X}, 2\}\).

Step by step solution

01

Formulate the Likelihood Function

The likelihood function is derived from the joint probability density function (pdf) of the observed sample. For the Poisson distribution, the pdf is given by \(P(X=k)=\frac{e^{-\theta}\theta^k}{k!}\). Therefore, the likelihood function for the sample is given by \(L(\theta; X)=e^{-n\theta}\theta^\Sigma x_i/\Pi x_i!\), where \(x_i\) are the observed values of the random sample.
02

Compute the Log-Likelihood Function

The logarithm of the likelihood function (known as the log-likelihood function) is often used because it simplifies the maximization problem. Continuing from the previous step, the log-likelihood function is \(l(\theta; X)=-n\theta+\Sigma x_i \log \theta - \Sigma \log x_i!\).
03

Differentiate the Log-Likelihood Function and Set it Equal to Zero

To find the maximum point(s) of a function, the derivative of the function is set equal to zero. Therefore, differentiating the log-likelihood function with respect to \(\theta\) and setting it equal to zero gives \(\frac{\partial l}{\partial \theta}=-n+\frac{1}{\theta}\Sigma x_i =0\). This implies \(\Sigma x_i=n\theta\). Thus, the MLE of \(\theta\) without constraint is \(\widehat{\theta} = \bar{X}\), with \(\bar{X}\) being the sample mean.
04

Apply the Bound Constraint on the MLE

The problem states that \(0<\theta \leq 2\), which implies the MLE of \(\theta\) cannot be larger than 2. Therefore, given this constraint, the MLE becomes \(\widehat{\theta} = min\{\bar{X}, 2\}\), since if \(\bar{X}\) is less than 2, it is the MLE, otherwise the MLE cannot exceed 2, so it would be equal to 2.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be iid, each with the distribution having pdf \(f\left(x ; \theta_{1}, \theta_{2}\right)=\) \(\left(1 / \theta_{2}\right) e^{-\left(x-\theta_{1}\right) / \theta_{2}}, \theta_{1} \leq x<\infty,-\infty<\theta_{2}<\infty\), zero elsewhere. Find the maximum likelihood estimators of \(\theta_{1}\) and \(\theta_{2}\).

Let \(X_{1}, X_{2}, X_{3}, X_{4}, X_{5}\) be a random sample from a Cauchy distribution with median \(\theta\), that is, with pdf. $$f(x ; \theta)=\frac{1}{\pi} \frac{1}{1+(x-\theta)^{2}}, \quad-\infty

Recall that \(\widehat{\theta}=-n / \sum_{i=1}^{n} \log X_{i}\) is the mle of \(\theta\) for a \(\operatorname{Beta}(\theta, 1)\) distribution. Also, \(W=-\sum_{i=1}^{n} \log X_{i}\) has the gamma distribution \(\Gamma(n, 1 / \theta)\). (a) Show that \(2 \theta W\) has a \(\chi^{2}(2 n)\) distribution. (b) Using Part (a), find \(c_{1}\) and \(c_{2}\) so that $$P\left(c_{1}<\frac{2 \theta n}{\hat{\theta}}

Given \(f(x ; \theta)=1 / \theta, 00\), formally compute the reciprocal of $$n E\left\\{\left[\frac{\partial \ln f(X: \theta)}{\partial \theta}\right]^{2}\right\\}$$ Compare this with the variance of \((n+1) Y_{n} / n\), where \(Y_{n}\) is the largest observation of a random sample of size \(n\) from this distribution. Comment.

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