Question

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

The maximum likelihood estimator \(\widehat{\theta}\) for the Poisson distribution with the given condition is \(\min \{\bar{X}, 2\}\).

Step-by-step solution

Define the log likelihood function

From the definition of the Poisson probability mass function (PMF), the likelihood function can be written as \( L(\theta) = \prod_{i=1}^{n} \frac{\theta^{x_i}e^{-\theta}}{x_i!}\). The log-likelihood function is then \(l(\theta)= \log(L(\theta)) = \sum_{i=1}^{n} [x_i \log(\theta) - \theta - \log(x_i!)]\).

Compute the derivative of the log likelihood function

To find the maximum likelihood estimate (MLE), we take the derivative of the log-likelihood with respect to \(\theta\) and set it equal to 0. The derivative of \(l(\theta)\) is \(\frac{d}{d\theta}l(\theta) = \sum_{i=1}^{n} [x_i/\theta - 1]\).\n

Solve for theta

Setting the derivative equal to zero gives \(0 = \sum_{i=1}^{n} x_i/\theta - n\). Solving for \(\theta\) we obtain \(\theta = \bar{X}\).

Incorporate given condition

However, from given condition, \(0 < \theta \leq 2\), so the maximum likelihood estimate of \(\theta\) under this condition is \(\widehat{\theta} = \min \{\bar{X}, 2\}\).