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I have two multivariate normal populations:

$\boldsymbol{A_1}$~$N_r(\boldsymbol{\mu_1},\Sigma)$ with $n_1$ observations, and

$\boldsymbol{A_2}$~$N_r(\boldsymbol{\mu_2},\Sigma)$ with $n_2$ observations.

I am wondering if there is some way to express the likelihood of $\boldsymbol{A_1}-\boldsymbol{A_2}$. Separately, I know the likelihoods of $\boldsymbol{A_1}$ and $\boldsymbol{A_2}$ are:

$L(\boldsymbol{A_1})=L(\boldsymbol{\mu_1},\Sigma)=(2\pi)^{-\frac{1}{2}pn_1}|\Sigma|^{-\frac{1}{2}n_1}exp{[-\frac{1}{2}\sum_{\alpha=1}^{n_1}(\boldsymbol{A_{1,\alpha}-\mu_1})^T\Sigma^{-1}(\boldsymbol{A_{1,\alpha}-\mu_1})]}$

and

$L(\boldsymbol{A_2})=L(\boldsymbol{\mu_2},\Sigma)=(2\pi)^{-\frac{1}{2}pn_2}|\Sigma|^{-\frac{1}{2}n_2}exp{[-\frac{1}{2}\sum_{\alpha=1}^{n_2}(\boldsymbol{A_{2,\alpha}-\mu_2})^T\Sigma^{-1}(\boldsymbol{A_{2,\alpha}-\mu_2})]}$,

respectively.

The end goal is to find the MLEs of $\boldsymbol{A_1}-\boldsymbol{A_2}$ (both the mean vector and covariance matrix). If I cannot express the Likelihood of $\boldsymbol{A_1}-\boldsymbol{A_2}$, I am wondering if it is appropriate to find the MLEs $\boldsymbol{A_1}$ and $\boldsymbol{A_2}$ separately?

For the mean vector, I believe the MLE will be $\boldsymbol{\bar{A_1}-\bar{A_2}}$. The covariance matrix is a bit more tricky, but I believe the MLE will be

$\frac{1}{n_1+n_2}\{\frac{n_1}{n_1+n_2}\sum_{n=1}^{n_1}(\boldsymbol{A_{1,n}-\bar{A_1}})(\boldsymbol{A_{1,n}-\bar{A_1}})^T+\frac{n_2}{n_1+n_2}*\sum_{n=1}^{n_1}(\boldsymbol{A_{2,n}-\bar{A_2}})(\boldsymbol{A_{2,n}-\bar{A_2}})^T\}$

In summary, 1) How can I express the likelihood of $\boldsymbol{A_1}-\boldsymbol{A_2}$, 2) Is it possible to find the MLEs of $\boldsymbol{A_1}-\boldsymbol{A_2}$, and 3) Are these the correct MLEs?

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  • $\begingroup$ What's the relationship between $A_1$ and $A_2$? Are they independent? $\endgroup$ – user795305 Oct 24 '17 at 1:38
  • $\begingroup$ Yes, they are independent. $\endgroup$ – user178237 Oct 24 '17 at 2:43
  • $\begingroup$ Ah, okay, so it turns out that they're jointly normal so that their difference is normal. You just have to determine what the mean and covariance is, which is a lot less messy. (Do you see how you could do that?) $\endgroup$ – user795305 Oct 24 '17 at 12:12
  • $\begingroup$ Yes, I believe they would have mean vector \mu_1-\mu_2 and covariance matrix 2*\Sigma. However, I am not sure how I would determine the MLE for these because the original samples sizes were different. $\endgroup$ – user178237 Oct 24 '17 at 12:23
  • $\begingroup$ I assume that since they are independent, I can just multiply their likelihoods together and proceed this way. Thanks! $\endgroup$ – user178237 Oct 24 '17 at 12:56

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