1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ smells like random matrix theory :) $\endgroup$ – Ahmad Bazzi Dec 28 '18 at 13:12
0
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.