# How to calculate the expectation of the Wishart distribution?

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?

## 2 Answers

This can be shown with the characteristic function.

If $X$ is a random matrix, its characteristic function is $Z \mapsto E\Bigl[\exp\bigl(\textrm{tr}(iZX)\bigr)\Bigr]$.

The application $Z \mapsto \exp\bigl(\textrm{tr}(iZX)\bigr)$ is differentiable and its differential at $Z_0$ is $\textsf{D}_{Z_0}\phi(Z) = \exp\bigl(\text{tr}(iZ_0X)\bigr)\text{tr}(iZX)$. For $Z_0=0$ (the null matrix), $\textsf{D}_{0}\phi(Z) = \text{tr}(iZX)$, and if $X$ is integrable then $\phi$ is differentiable at $0$ and $\textsf{D}_0\phi(Z) = \text{tr}\bigl(iZE[X]\bigr)$. Conversely, if $\phi$ is differentiable at $0$ then $X$ is integrable and $\textsf{D}_0\phi(Z) = \text{tr}\bigl(iZE[X]\bigr)$

Therefore, if you prove that $\phi$ is differentiable at $0$ and $\textsf{D}_0\phi(Z) = \text{tr}\bigl(iZA\bigr)$ then $E[X]=A$.

The characteristic function of the Wishart distribution $W_p(\nu, \Sigma)$ is $\phi(Z) = \det(I - 2iZ\Sigma)^{-\frac{\nu}{2}}$.

Let's set $V(Z) = I - 2iZ\Sigma$ and $U(Z) = \det\bigl(V(Z)\bigr)$. Then $V$ is differentiable and $\textsf{D}_{Z_0}V(Z) = -2iZ\Sigma$. By Jacobi's formula, $U$ is differentiable and $\textsf{D}_{Z_0}U(Z) = \text{tr}\bigl({V(Z_0)}^\# \textsf{D}_{Z_0}V(Z)\bigr)$ where $M^\#$ denotes the adjugate of a square matrix $M$. For $Z_0=0$, this gives $\textsf{D}_{0}U(Z) = \text{tr}(-2iZ\Sigma)$. Now, $\phi$ is differentiable and $$\textsf{D}_{Z_0}\phi(Z) = -\frac{\nu}{2}{\det(I - 2iZ_0\Sigma)}^{-\frac{\nu}{2}-1} \textsf{D}_{Z_0}U(Z)$$ and for $Z_0=0$, $$\textsf{D}_{Z_0}\phi(Z) = -\frac{\nu}{2} \text{tr}(-2iZ\Sigma) = \text{tr}\bigl(iZ(\nu\Sigma)\bigr).$$ That shows that the expectation is $\nu\Sigma$.

$$X$$ follows a wishart distribution, $$X \sim W_{p}(\Sigma,v)$$, if $$$$X = \sum_{i=1}^{v}Z_{i}Z_{i}^{T}$$$$ where $$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 $$$$\mathbb{E}(X) = v\Sigma.$$$$

• Thanks! I should have been more explicit in saying that I was interested in the general way to do it for matrix valued distributions. Your answer only works for distributions that can be represented as sums of outer products of random vectors and only in the specific case where $\nu$ is a positive integer. Nov 15, 2020 at 19:31