# Deriving the sampling distribution of MLE for Normal distribution

Let $$X_1,\ldots,X_n$$ be an observed random sample from $$N_p(\mu, \Sigma)$$.

I know that the MLE of $$\Sigma$$ is $$\frac{1}{n} \sum_i^n(X_i -\bar X)(X_i -\bar X)^T$$, which is biased.

We define $$S = \frac{1}{n-1}\sum_i^n(X_i-\bar X)(X_i-\bar X)^T$$ which is unbiased.

I am proving that $$(n-1)S \sim W_p(\Sigma, n-1)$$ where $$W$$ represents the Wishart distribution.

I let $$X$$ be the $$n\times p$$ matrix with $$X_i^T$$ as its i-th row. If $$H = (I-\textbf{11}^T/n)$$, then $$HX$$ returns the column-centered data matrix, and hence $$S=X^THX/(n-1)$$. Furthermore, $$H=H^T$$ and $$\operatorname{Tr}(H)=n-1$$, and hence $$H$$ is a $$n\times n$$ orthogonal projection matrix with rank $$n-1$$. It therefore has the spectral decomposition $$H=\sum_{j=1}^{n-1} u_j u_j^T$$, where the $$u_j$$s are orthonormal. Using this spectral decomposition, I get $$(n-1)S=\sum_{j=1}^{n-1}(X^Tu_j)(X^Tu_j)^T$$.

If I can prove that the vectors $$(X^Tu_j)_j$$ are jointly MVN, and are $$\text{i.i.d. } N_p(0,\Sigma)$$, then the Wishart distribution is shown. Showing the expectation is 0 and variance is $$\Sigma$$ however is what I am stuck on. I wonder whether it has something to do with the column-centering matrix which subtracts the mean?

• smells like random matrix theory :) – Ahmad Bazzi Dec 28 '18 at 13:12

## 1 Answer

Disclaimer: This is not my proof, but rather an understanding of one of Lukács theorems.

First transform $$X_k$$ using orthogonal matrix $$V$$ so that it contains entries equal to $$\frac{1}{ \sqrt{n}}$$ in the last row. This means that the sum of entries in first $$n-1$$ rows are all zeros. Let's call the transformed vectors $$Y_k$$, i.e. $$\begin{equation} Y_i = \sum_{k=1}^n v_{ik} X_k \end{equation}$$ for all $$i = 1 \ldots n$$, where $$v_{ij}$$ is the $$(i,j)^{th}$$ entry of $$V$$. Equivalently, we can write $$\begin{equation} Y = XV^T \end{equation}$$ We can write $$S$$ as follows $$\begin{equation} S = \sum_{k=1}^n (X_k - \bar{X})(X_k - \bar{X})^T = \sum_{k=1}^n X_kX_k^T - n \bar{X}\bar{X}^T \end{equation}$$ Using some straightforward math, the above could be shown to be $$\begin{equation} S =\sum_{k=1}^n Y_kY_k^T - n (\frac{Y_n}{\sqrt{n}}) (\frac{Y_n}{\sqrt{n}})^T = \sum_{k=1}^{n-1} Y_kY_k^T \end{equation}$$ i.e. a sum of $$n-1$$ outer-product. Now, we can see that $$S$$ is the dyadic sum of $$n-1$$ independent normal vectors of mean $$0$$ and variance $$\Sigma$$. Therefore, by using the Wishart-distribution definition, we have that $$\begin{equation} S \sim \mathcal{W}_p(n-1, \Sigma) \end{equation}$$