Question

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

Short answer

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

Step-by-step solution

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 \[L(\theta; x_1, x_2, ..., x_n) = \frac{1}{{\theta^n}} e^{-\frac{1}{\theta}\sum_{i=1}^{n}x_i}\]

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 \(\theta_{MLE} = \frac{1}{n}\sum_{i=1}^{n}x_i = \bar{x}\)

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 \[P(X <= 2) = 1 - e^{-2 / \bar{x}}\]

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.