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Chapter 6: Problem 6

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

Short Answer

Expert verified
The MLE of \(P(X <= 2)\) is \(1 - e^{-2 / \bar{x}}\) where \(\bar{x}\) is the sample mean.

Step by step solution

01

Define the likelihood function

The likelihood function, L, for n independent observations is the product of the single observation densities: \[L(\theta; x_1, x_2, ..., x_n) = \prod_{i=1}^{n} f(x_i; \theta)\] Given \(f(x; \theta) = (1 / \theta) e^{-x / \theta}\), the likelihood function becomes \[L(\theta; x_1, x_2, ..., x_n) = \frac{1}{{\theta^n}} e^{-\frac{1}{\theta}\sum_{i=1}^{n}x_i}\]
02

Derive and solve the log-likelihood equation

Firstly, log-likelihood, \(l(\theta)\), is easier to differentiate, while yielding the same results due to the monotonicity of the logarithm function. The log-likelihood for the given likelihood function is: \[l(\theta) = -n log(\theta) - \frac{1}{\theta} \sum_{i=1}^{n}x_i\] By differentiating w.r.t. \(\theta\), equating to zero and solving for \(\theta\), it yield \(\theta_{MLE} = \frac{1}{n}\sum_{i=1}^{n}x_i = \bar{x}\)
03

Calculate P(X

Using the cumulative distribution function (CDF) for the exponential distribution \(1 - e^{-x / \theta}\) and substituting \(\theta_{MLE}\) and \(x = 2\), you will find \[P(X <= 2) = 1 - e^{-2 / \bar{x}}\]
04

Result interpretation

The MLE for \(P(X <= 2)\), denoted as \(\hat{P}(X <= 2)\), is estimated as \[\hat{P}(X <= 2) = 1 - e^{-2 / \bar{x}}\]. This equation provides the maximum likelihood estimate of the probability that a random variable X is less than or equal to 2, given the data observed.

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Most popular questions from this chapter

Prove that \(\bar{X}\), the mean of a random sample of size \(n\) from a distribution that is \(N\left(\theta, \sigma^{2}\right),-\infty<\theta<\infty\), is, for every known \(\sigma^{2}>0\), an efficient estimator of \(\theta\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) ba random sample from a distribution with one of two pdfs. If \(\theta=1\), then \(f(x ; \theta=1)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2},-\infty

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the beta distribution with \(\alpha=\beta=\theta\) and \(\Omega=\\{\theta: \theta=1,2\\} .\) Show that the likelihood ratio test statistic \(\Lambda\) for testing \(H_{0}: \theta=1\) versus \(H_{1}: \theta=2\) is a function of the statistic \(W=\) \(\sum_{i=1}^{n} \log X_{i}+\sum_{i=1}^{n} \log \left(1-X_{i}\right)\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) be independent random samples from the two normal distributions \(N\left(0, \theta_{1}\right)\) and \(N\left(0, \theta_{2}\right)\). (a) Find the likelihood ratio \(\Lambda\) for testing the composite hypothesis \(H_{0}: \theta_{1}=\theta_{2}\) against the composite alternative \(H_{1}: \theta_{1} \neq \theta_{2}\). (b) This \(\Lambda\) is a function of what \(F\) -statistic that would actually be used in this test?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) represent a random sample from each of the distributions having the following pdfs or pmfs: (a) \(f(x ; \theta)=\theta^{x} e^{-\theta} / x !, x=0,1,2, \ldots, 0 \leq \theta<\infty\), zero elsewhere, where \(f(0 ; 0)=1\) (b) \(f(x ; \theta)=\theta x^{\theta-1}, 0
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