Variance of Wishart distribution?

Question

$X\sim W_p(I_p,1)$, i.e. $X$ is a $p\times p$ Wishart matrix. For a vector $p\times 1$ $\beta$ such that $|\beta|=1$, $E(X\beta)=E(X)\beta=\beta$. How do I get ${\rm var}(X\beta)$? Thank you for your help in advance.

Answer 1

I'd start with the special case $\beta = [1,\,\underbrace{0,0,\ldots,0}_\text{all 0s}\,]^T$. Then $X\beta$ is the first column of $X$. The first entry in $X$ has a $\chi^2_1$ distribution, so its variance is $2$. The other entries are of the form $Z_1 Z_k$, where $Z_1,\ldots,Z_p\sim \text{i.i.d. } N(0,1)$ and $2\le k\le p.$ For those we have \begin{align} \operatorname{var}(Z_1 Z_k) & = \operatorname{var}(\operatorname{E}(Z_1 Z_k \mid Z_k)) + \operatorname{E}(\operatorname{var}(Z_1 Z_k \mid Z_k)) = 0 + \operatorname{E}(Z_k^2) = 1, \\[10pt] \operatorname{cov}(Z_1 Z_k, Z_1 Z_j) & = \operatorname{cov}(\operatorname{E}(Z_1 Z_k\mid Z_1), \operatorname{E}(Z_1Z_j\mid Z_1)) + \operatorname{E}(\operatorname{cov}(Z_1 Z_k,Z_1 Z_j \mid Z_1)) \\ &= 0 + {}0 = 0, \\[10pt] \operatorname{cov}(Z_1^2, Z_1Z_k) & = \operatorname{cov}(\operatorname{E}(Z_1^2 \mid Z_1), \operatorname{E}(Z_1 Z_k\mid Z_1)) + \operatorname{E}(\operatorname{cov}(Z_1^2,Z_1 Z_k \mid Z_1)) \\ & = 0 + {}0 = 0. \end{align} So the variance in that case is $$ \begin{bmatrix} 2 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{bmatrix} = I_p + \beta \beta^T. $$ Cleary this applies equally if $\beta$ is any of the other members of the standard basis. If $\beta$ is a linear combination of those, then use what you know about variances and covariances of linear combinations.