Question

Suppose \(\mathbf{X}\) is distributed \(N_{n}(\boldsymbol{\mu}, \mathbf{\Sigma}) .\) Let \(\bar{X}=n^{-1} \sum_{i=1}^{n} X_{i}\). (a) Write \(\bar{X}\) as aX for an appropriate vector a and apply Theorem \(3.5 .2\) to find the distribution of \(\bar{X}\). (b) Determine the distribution of \(\bar{X}\) if all of its component random variables \(X_{i}\) have the same mean \(\mu\).

Short answer

The distribution of \(\bar{X}\), when \(\mathbf{X}\) is distributed as $N_{n}(\boldsymbol{\mu}, \mathbf{\Sigma})$, is given by $N_{1}(a\boldsymbol{\mu}, a\mathbf{\Sigma}a')$, where 'a' is a vector with 1/n for all entries. When all component random variables \(X_{i}\) have the same mean \(\mu\), the distribution of \(\bar{X}\) is still Gaussian, with mean \(\mu\) and variance being the variance of each \(X_{i}\) divided by n.

Step-by-step solution

Express \(\bar{X}\) as aX

Express \(\bar{X}\) as aX for an appropriate vector 'a'. In this case a=\[1/n, 1/n, ..., 1/n\], the vector of length n with all entries equal to 1/n. This leads to \(\bar{X}\) = a\(\mathbf{X}\), where \(\mathbf{X}\) is the random variables vector of length n.

Apply Theorem 3.5.2 to find the distribution of \(\bar{X}\)

This step would require applying Theorem 3.5.2, which should provide a formula for determining the distribution of a linear combination of random variables. The exact content will vary based on the exact wording of the theorem, but generally this will give the mean and the variance of the new distribution. For \(\bar{X}\) = a\(\mathbf{X}\) the mean of \(\bar{X}\) will be a\(\boldsymbol{\mu}\), the geometric mean of vector \(\mathbf{X}\) and the variance will be given by a\(\mathbf{\Sigma}a'\), where \(\mathbf{\Sigma}\) is the covariance matrix of the random variable \(\mathbf{X}\) and a' is the transpose of the vector a.

Determine the distribution of \(\bar{X}\) under specified conditions

In this step, we are to determine the distribution of \(\bar{X}\) given that all its component random variables \(X_{i}\) have the same mean \(\mu\). Given all the \(X_{i}\) have the same mean \(\mu\), The mean of \(\bar{X}\)= a\(\boldsymbol{\mu}\) = \(1/n * n\mu = \mu\), regardless of the number of variables. For the variance, as all \(X_{i}\) have the same mean, the off-diagonal elements of the covariance matrix will be zero. Hence, the variance will be 1/n^2 times the sum of all variances of the \(X_{i}\), which reduces to the variance of each \(X_{i}\) divided by n if all \(X_{i}\) have the same variance.