Sufficient statistic for bivariate or multivariate normal Here on page 7, example 2.7. The claim is that sufficient statistics for $d$ dimensional multivariate normal $\mathbf{x}_i \sim N(\vec{\mu}, \Sigma)$  is $$\left(n^{-1}\sum_{i=1}^n \mathbf{x}_i, \hat{\Sigma} \right)\,,$$ where $\hat{\Sigma} := \sum_{i=1}^n\mathbf{x}_i\mathbf{x}^T_i - \left(n^{-1}\sum_{i=1}^n \mathbf{x}_i\right)\left(n^{-1}\sum_{i=1}^n \mathbf{x}_i\right)^T$. 
Where can I find derivation of this result? Thanks!
 A: As W. Huber tried to lead you to conclude, the sufficiency is a simple consequence of looking at the likelihood:
\begin{align*}
f(\mathbf{x}_1,\ldots,\mathbf{x}_n|\boldsymbol{\mu},\boldsymbol{\Sigma}) &\propto |\boldsymbol{\Sigma}|^{-n/2}\,\exp\frac{-1}{2}\left\{\sum_{i=1}^n (\mathbf{x}_i-\boldsymbol{\mu})^\text{T}\boldsymbol{\Sigma}^{-1}(\mathbf{x}_i-\boldsymbol{\mu})\right\}\\
&=|\boldsymbol{\Sigma}|^{-n/2}\,\exp\frac{-1}{2}\left\{\sum_{i=1}^n (\bar{\mathbf{x}}-\boldsymbol{\mu})^\text{T}\boldsymbol{\Sigma}^{-1}(\bar{\mathbf{x}}-\boldsymbol{\mu})\right\}\\
&\quad \times\exp\frac{-1}{2}\left\{\sum_{i=1}^n (\mathbf{x}_i-\bar{\mathbf{x}})^\text{T}\boldsymbol{\Sigma}^{-1}(\mathbf{x}_i-\bar{\mathbf{x}})\right\}\\
&=|\boldsymbol{\Sigma}|^{-n/2}\,\exp\frac{-n}{2}(\bar{\mathbf{x}}-\boldsymbol{\mu})^\text{T}\boldsymbol{\Sigma}^{-1}(\bar{\mathbf{x}}-\boldsymbol{\mu})\\
&\quad \times\exp\frac{-1}{2}\left\{\sum_{i=1}^n (\mathbf{x}_i-\bar{\mathbf{x}})^\text{T}\boldsymbol{\Sigma}^{-1}(\mathbf{x}_i-\bar{\mathbf{x}})\right\}\\
&=|\boldsymbol{\Sigma}|^{-n/2}\,\exp\frac{-n}{2}(\bar{\mathbf{x}}-\boldsymbol{\mu})^\text{T}\boldsymbol{\Sigma}^{-1}(\bar{\mathbf{x}}-\boldsymbol{\mu})\\
&\quad \times\exp\frac{-1}{2}\text{tr}\left\{\sum_{i=1}^n (\mathbf{x}_i-\bar{\mathbf{x}})^\text{T}\boldsymbol{\Sigma}^{-1}(\mathbf{x}_i-\bar{\mathbf{x}})\right\}\\
&=|\boldsymbol{\Sigma}|^{-n/2}\,\exp\frac{-n}{2}(\bar{\mathbf{x}}-\boldsymbol{\mu})^\text{T}\boldsymbol{\Sigma}^{-1}(\bar{\mathbf{x}}-\boldsymbol{\mu})\\
&\quad \times\exp\frac{-1}{2}\text{tr}\left\{\sum_{i=1}^n \boldsymbol{\Sigma}^{-1}(\mathbf{x}_i-\bar{\mathbf{x}})(\mathbf{x}_i-\bar{\mathbf{x}})^\text{T}\right\}\\
&=|\boldsymbol{\Sigma}|^{-n/2}\,\exp\frac{-n}{2}(\bar{\mathbf{x}}-\boldsymbol{\mu})^\text{T}\boldsymbol{\Sigma}^{-1}(\bar{\mathbf{x}}-\boldsymbol{\mu})\\
&\quad \times\exp\frac{-1}{2}\text{tr}\left\{ \boldsymbol{\Sigma}^{-1}\widehat{\boldsymbol{\Sigma}}\right\}\\
\end{align*}
This likelihood thus only depends on two functions of the sample, $\bar{\mathbf{x}}$ and $\widehat{\boldsymbol{\Sigma}}$. $\qquad{ }$Q.E.D.

Note: the author of the attached document should be identified. The more when he is the stellar statistician Jon Wellner!

