I'd like to plug-in the estimate of $\Sigma$ in a likelihood calculation,
$$ p(\{x\}) = \prod _{i=1}^{n} \frac{1}{C} \exp \bigg[-\frac{ 1}{2}(x_i-\mu)^T\Sigma^{-1}(x_i-\mu) \bigg] $$
where $C = \frac{1}{(2\pi)^{k/2} \det(\Sigma)^{1/2}}$. The maximum likelihood estimate is, as we know,
$$ \hat{\Sigma} = \frac{1}{n} \sum _{i=1}^{n} (x_k-\hat{\mu}) (x_k-\hat{\mu})^T $$
And since
$$ p(\{x\}) = \frac{1}{C^n} \exp \bigg[\sum _{i=1}^{n} -\frac{ 1}{2}(x_i-\mu)^T\Sigma^{-1}(x_i-\mu) \bigg] $$
Could I plug-in the estimate (for both $\mu$ and $\Sigma$) like this?
$$ = \frac{1}{C^n} \exp \bigg[\sum _{i=1}^{n} -\frac{ 1}{2n}(x_i-\hat{\mu})^T \bigg[ \frac{1}{n} \sum _{i=1}^{n} (x_k-\hat{\mu}) (x_k-\hat{\mu})^T \bigg]^{-1} (x_i-\hat{\mu}) \bigg] $$
$$ = \frac{1}{C^n} \exp \bigg[\sum _{i=1}^{n} -\frac{ 1}{2} \bigg[ \frac{1}{n} \sum _{i=1}^{n} (x_i-\hat{\mu})^{-T} (x_k-\hat{\mu}) (x_k-\hat{\mu})^T (x_i-\hat{\mu})^{-1} \bigg]^{-1} \bigg] $$
$$ = \frac{1}{C^n} \exp \bigg[\sum _{i=1}^{n} -\frac{ 1}{2} \bigg[ \frac{1}{n} \sum _{i=1}^{n} 1 \bigg]^{-1} \bigg] $$
$$ = \frac{1}{C^n} \exp \bigg[\sum _{i=1}^{n} -\frac{ 1}{2} \bigg[ \frac{1}{n} n \bigg]^{-1} \bigg] $$
$$ = \frac{1}{C^n} \exp \bigg[\sum _{i=1}^{n} -\frac{ 1}{2} \bigg] $$
$$ = \frac{1}{C^n} \exp \bigg[-\frac{n}{2} \bigg] $$
Thanks,
I am addressing only the issue of how to rewrite the likelihood with plug in estimates.
Let $z_i=x_i-\hat{\mu}$. We assume $z_i$ are $p\times 1$ vectors. At some point we need to evaluate $$\sum_{i=1}^n z_i^T\left(\sum_{k=1}^n z_kz_k^T\right)^{-1}z_i$$ We assume the invertibility of the above matrix. We use the trace operator. It has some remarkable properties, like $\operatorname{tr}(A+B)=\operatorname{tr}(A)+\operatorname{tr}(B)$ and $\operatorname{tr}(AB)=\operatorname{tr}(BA)$. Note that $$\begin{aligned} \sum_{i=1}^n z_i^T\left(\sum_{k=1}^n z_kz_k^T\right)^{-1}z_i & = \operatorname{tr}\left[\sum_{i=1}^n z_i^T\left(\sum_{k=1}^n z_kz_k^T\right)^{-1}z_i\right] \\ & = \sum_{i=1}^n\operatorname{tr}\left[ z_i^T\left(\sum_{k=1}^n z_kz_k^T\right)^{-1}z_i\right] \\ & = \sum_{i=1}^n\operatorname{tr}\left[ \left(\sum_{k=1}^n z_kz_k^T\right)^{-1}z_iz_i^T\right] \\ & = \operatorname{tr}\left[ \left(\sum_{k=1}^n z_kz_k^T\right)^{-1}\sum_{i=1}^nz_iz_i^T\right] = \operatorname{tr}(I_p)=p \end{aligned}$$
Hence $$\begin{aligned}\exp\left[-\frac12\sum_{i=1}^n z_i^T\left(\frac1n\sum_{k=1}^n z_kz_k^T\right)^{-1}z_i\right] & = \exp\left[-\frac{n}2\sum_{i=1}^n z_i^T\left(\sum_{k=1}^n z_kz_k^T\right)^{-1}z_i\right]=\exp(-np/2)\end{aligned}$$