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I am solving the next exercise about a property of a Wishart Distribution:

$$M_1\sim W_p(\Sigma,n_1)$$ $$M_2\sim W_p(\Sigma,n_2)$$ are independent, then $M_1+M_2\sim W_p(\Sigma,n_1+n_2)$

I have the next:

Since $M_1,M_2$ are wishart distributed I can write $M_1=\underline{Y}_1^T\underline{Y}_1$ and $M_2=\underline{Y}_2^T\underline{Y}_2$ where $\underline{Y}_k=(Y_1,...,Y_{n_k})\sim \mathcal{N}_{n_k\times p}(0,\Sigma)$

Using the previous notations I have that $$M_1+M_2=\underline{Y}_1^T\underline{Y}_1+\underline{Y}_2^T\underline{Y}_2$$ that is $$=\displaystyle\sum_{i=1}^{n_1}\underline{Y}_{1_i}^T\underline{Y}_{1_i}+\displaystyle\sum_{j=1}^{n_2}\underline{Y}_{2_j}^T\underline{Y}_{2_j}$$

I wrote the last equation as:

$$\displaystyle\sum_{i=1}^{n_1+n_2}(1_{\{i> n_1\}}\underline{Y}_{1_i}^T\underline{Y}_{1_i}+\underline{Y}_{2_i}^T\underline{Y}_{2_i})$$ w.l.o.g $n1>n2$ The last expression is equal to : $$\underline{Y}^T\underline{Y}$$, where $\underline{Y}$ follows a $\mathcal{N}_{(n_1+n_2)\times p}(0,\Sigma)$

Is ok ok my approach, I am not very sure of my solution.

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  • $\begingroup$ Is the $\Sigma$ a covariance matrix in your parametrization? How would $n_1$ and $n_2$ be of different length but have the same $n \times n$ covariance matrix? $\endgroup$ – AdamO Jan 25 '18 at 17:56
  • $\begingroup$ Is not included in the parametrization since $\Sigma\in\mathbb{R}^{p\times p}$ $\endgroup$ – Boris Jan 25 '18 at 18:06

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