TITLE: How to calculate the expectation of the Wishart distribution?
QUESTION [5 upvotes]: Let $X \sim \mathcal{W}_p(V,\nu)$ follow a central Wishart distribution with scale matrix $V$ and $\nu$ degrees of freedom.
Its p.d.f. is given by:
$$
\frac{|\mathbf{X}|^{(\nu-p-1)/2} e^{-\operatorname{tr}(\mathbf{V}^{-1}\mathbf{X})/2}}{2^\frac{\nu p}{2}|{\mathbf V}|^{\nu/2}\Gamma_p(\frac \nu 2)} 
$$
Its expectation is given by:
$$
E[X]=\nu V
$$
How do we actually calculate this expected value? What is the general procedure for matrix valued distributions?

REPLY [3 votes]: $X$ follows a wishart distribution, $X \sim W_{p}(\Sigma,v)$, if
\begin{equation}
 X = \sum_{i=1}^{v}Z_{i}Z_{i}^{T}
\end{equation}
where $Z_i$ are independent identically p-dimensional normally distributed such that $Z_{i} \sim N_{p}(0,\Sigma)$, and $\Sigma$ is the covariance matrix of $Z_{i}$.
The expectation of a random matrix A is defined as
$\mathbb{E}(A)_{i,j} = \mathbb{E}(A_{i,j})$, i.e it's the expectation of each of its elements. This definition is compatible with random vectors.
Now
\begin{align*}
 \mathbb{E}(X) &= \mathbb{E} \left( \sum_{i=1}^{v}Z_{i}Z_{i}^{T} \right)
\\&= \sum_{i=1}^{v} \mathbb{E}(Z_{i}Z_{i}^{T})
\\ &= \sum_{i=1}^{v} \mathbb{E}(Z_{1}Z_{1}^{T})
\\ &= v\mathbb{E}(Z_{1}Z_{1}^{T})
\end{align*}
Where the equality follows from the linearity of the expectation and that $Z_{i}$ are identically distributed.
But for any random vector $Y$,
$Cov(Y) = \mathbb{E}(YY^{T}) - \mathbb{E}(Y)\mathbb{E}(Y^{T})$. The mean of $Z_{1}$ is zero and hence
$\mathbb{E}(Z_{1}Z_{1}^{T}) = Cov(Z_{1}) = \Sigma$.
Therefore
\begin{equation}
 \mathbb{E}(X) = v\Sigma.
\end{equation}