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Chapter 6: Problem 10

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(N\left(\theta, \sigma^{2}\right)\) distribution, where \(\sigma^{2}\) is fixed but \(-\infty<\theta<\infty\) (a) Show that the mle of \(\theta\) is \(\bar{X}\). (b) If \(\theta\) is restricted by \(0 \leq \theta<\infty\), show that the mle of \(\theta\) is \(\widehat{\theta}=\max \\{0, \bar{X}\\}\).

Short Answer

Expert verified
The MLE of \(\theta\) is \(\bar{X}\) in the unrestricted case, and \(max\{0, \bar{X}\}\) with the restriction \(0 \leq \theta<\infty\).

Step by step solution

01

Derive the Likelihood Function

Given a random sample \(X_{1}, X_{2}, \ldots, X_{n}\) from a normalized distribution with known variance \(\sigma^{2}\) and unknown mean \(\theta\), the joint probability density function can be written as \(L(\theta)=\prod_{i=1}^{n} \frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{(X_{i}-\theta)^{2}}{2 \sigma^{2}}}\). This is the so-called likelihood function, which we want to maximize with respect to \(\theta\).
02

Apply Logarithmic Function to Simplify

To simplify derivation, take the natural logarithm (ln) forming the log-likelihood function, \(ln(L(\theta))=-\frac{n}{2} ln(2 \pi \sigma^{2})-\frac{1}{2 \sigma^{2}} \sum_{i=1}^{n}(X_{i}-\theta)^{2}\). The first term is a constant with respect to \(\theta\) and can be ignored while maximizing the likelihood.
03

Finding the Derivative to Locate Maximum

Find the derivative of the log-likelihood function with respect to \(\theta\) and equate to 0 to locate the maximum. We get \(\frac{d ln(L(\theta))}{d \theta}=\frac{1}{\sigma^{2}} \sum_{i=1}^{n}(X_{i}-\theta)=0\). Rearranging gives \(\sum_{i=1}^{n}(X_{i}-\theta)=0\). Solving for \(\theta\) gives \(\theta=\frac{1}{n}\sum_{i=1}^{n} X_{i}\), which is the sample mean, \(\bar{X}\). Hence the MLE of \(\theta\) is \( \bar{X}\) in the unrestricted case.
04

Finding Restricted Maximum Likelihood Estimate

When \(0 \leq \theta<\infty\), the likelihood function is still maximized at \(\bar{X}\). However, if \(\bar{X}\) falls below 0, the maximum likelihood within the restricted range is at 0 because the normal distribution is symmetric about \(\theta\). Therefore, the maximum likelihood estimate for \(\theta\), given the restrictions is \(max\{0, \bar{X}\}\).

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Most popular questions from this chapter

Recall that \(\widehat{\theta}=-n / \sum_{i=1}^{n} \log X_{i}\) is the mle of \(\theta\) for a 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}}

Suppose \(X_{1}, X_{2}, \ldots, X_{n}\) are iid with pdf \(f(x ; \theta)=(1 / \theta) e^{-x / \theta}, 0

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(\Gamma(\alpha, \beta)\) distribution where \(\alpha\) is known and \(\beta>0\). Determine the likelihood ratio test for \(H_{0}: \beta=\beta_{0}\) against \(H_{1}: \beta \neq \beta_{0}\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(N(0, \theta)\) distribution. We want to estimate the standard deviation \(\sqrt{\theta}\). Find the constant \(c\) so that \(Y=\) \(c \sum_{i=1}^{n}\left|X_{i}\right|\) is an unbiased estimator of \(\sqrt{\theta}\) and determine its efficiency.

Let \(X_{1}, \ldots, X_{n}\) and \(Y_{1}, \ldots, Y_{m}\) be independent random samples from the distributions \(N\left(\theta_{1}, \theta_{3}\right)\) and \(N\left(\theta_{2}, \theta_{4}\right)\), respectively. (a) Show that the likelihood ratio for testing \(H_{0}: \theta_{1}=\theta_{2}, \theta_{3}=\theta_{4}\) against all alternatives is given by $$\frac{\left[\sum_{1}^{n}\left(x_{i}-\bar{x}\right)^{2} / n\right]^{n / 2}\left[\sum_{1}^{m}\left(y_{i}-\bar{y}\right)^{2} / m\right]^{m / 2}}{\left\\{\left[\sum_{1}^{n}\left(x_{i}-u\right)^{2}+\sum_{1}^{m}\left(y_{i}-u\right)^{2}\right] /(m+n)\right\\}^{(n+m) / 2}},$$ where \(u=(n \bar{x}+m \bar{y}) /(n+m)\). (b) Show that the likelihood ratio test for testing \(H_{0}: \theta_{3}=\theta_{4}, \theta_{1}\) and \(\theta_{2}\) unspecified, against \(H_{1}: \theta_{3} \neq \theta_{4}, \theta_{1}\) and \(\theta_{2}\) unspecified, can be based on the random variable $$F=\frac{\sum_{1}^{n}\left(X_{i}-\bar{X}\right)^{2} /(n-1)}{\sum_{1}^{m}\left(Y_{i}-\bar{Y}\right)^{2} /(m-1)}$$
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