Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(N\left(\theta_{1}, \theta_{2}\right)\) distribution. (a) Show that \(E\left[\left(X_{1}-\theta_{1}\right)^{4}\right]=3 \theta_{2}^{2}\). (b) Find the MVUE of \(3 \theta_{2}^{2}\).

Short answer

For a normal distribution, the fourth central moment is calculated as \( E\left[\left(X_{1}-\theta_{1}\right)^{4}\right] = 3 \theta_{2}^{4} \). The MVUE (minimum variance unbiased estimator) of \( 3 \theta_{2}^{2} \) can be evaluated as \( 3s^2 \), where \( s^2 \) is sample variance.

Step-by-step solution

Identify the Moments of Normal Distribution

First, recognize that for a normally distributed random variable \(X \sim \mathcal{N}(\mu, \sigma^2)\), the fourth central moment, i.e., \(E\left[(X-\mu)^{4}\right] = 3 \sigma^{4}\).

Prove Given Equation

Substitute \( \mu = \theta_1 \) and \( \sigma = \theta_{2} \) into the equation we have, given that X is distributed normally with parameters \(\theta_{1}\) and \(\theta_{2}\) : \( E\left[\left(X_{1}-\theta_{1}\right)^{4}\right] = 3 \theta_{2}^{4} \).

Compute the MVUE of \(3 \theta_{2}^{2}\)

For point b), you should know that the minimum variance unbiased estimator (MVUE) of \( \sigma^2 = \theta_{2}^{2} \) in normal distribution is \( s^2 = \frac{1}{n-1} \sum_{i=1} ^{n} (X_i - \bar{X})^2 \), where \( \bar{X} = \frac{1}{n} \sum_{i=1} ^{n} X_i \). Following from the previous equation, for \( 3 \theta_{2}^{2} \), the MVUE would be \( 3s^2 \).