Question

Let a random sample of size \(n\) be taken from a distribution that has the pdf \(f(x ; \theta)=(1 / \theta) \exp (-x / \theta) I_{(0, \infty)}(x)\). Find the mle and MVUE of \(P(X \leq 2)\).

Short answer

The MLE for \(P(X \leq 2)\) is \(1 - e^{-2 \sum_{i}^{n} x_i / n}\) and MVUE for \(P(X \leq 2)\) is obtained by integrating the product of \(1 - e^{-2 / y}\) with the pdf of minimum order statistic Y over y > 0.

Step-by-step solution

Calculate MLE for \(\theta\)

The likelihood function is given by \(L(\theta) = \prod_{i=1}^{n} (1/\theta) e^{(-x_i/\theta)} = (1/\theta^n) \exp(-\Sigma_{i} x_i/ \theta).\n Take the logarithm of the likelihood function to derive the log-likelihood function, \(l(\theta) = -n \log(\theta) - \Sigma_{i} x_i/ \theta.\n Differentiate it with respect to \(\theta\) and equate to zero to get the MLE, \(\hat{\theta}_{MLE} = \frac{\Sigma_{i} x_i}{n}\).

Find the distribution of X

We have X following an exponential distribution with parameter \(\theta = 1/ \lambda\) where \(\lambda\) is usually known as the rate parameter. Hence, the value of X follows a distribution of \(Exp(\lambda)\) where \(\lambda = 1 / \hat{\theta}_{MLE}\).

Calculate \(P(X \leq 2)\)

The cumulative distribution function \(F(x)\) of an exponential distribution is given by \(F(x) = 1 - e^{-\lambda x}\). Hence, using the derived \(\lambda\) in step 2, \(P(X \leq 2) = F(2) = 1 - e^{-2 \lambda}\). Substitute \(\lambda = 1 / \hat{\theta}_{MLE}\) to calculate the MLE of \(P(X \leq 2)\).

Calculate MVUE of \(P(X \leq 2)\)

To get MVUE, first look for an unbiased estimator. We use the fact that Expectation of a minimum order statistic for exponential distribution follows the distribution \(E(Y) = \frac{\theta}{n}\) where \(Y\) is the minimum order statistic. And, \(E(Y) = \hat{\theta}_{MLE}\), this implies \(Y\) is an unbiased estimator. Then apply Rao-Blackwell Theorem to get the MVUE of \(P(X \leq 2)\). As per the theorem, the MVUE is given by expectation of \(P(X \leq 2)\) conditional on \(T = Y\), where \(T\) is sufficient statistic. Therefore, MVUE is calculated as \(E[F(2)|Y] = E[1 - e^{-2 \lambda(Y)}]\). Where, \(\lambda(Y) = 1 / Y\).

Simplify MVUE

The expectation of a function g(Y) of Y is integral of g(y) times the probability density function (pdf) of Y. Therefore calculate \(E[1 - e^{-2 * 1/Y}]\) using the pdf of Y which is \(\frac{n e^{-y/( \theta/n )}}{(\theta/n)}\), on domain y > 0. This results in the MVUE.