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Let $X_1$ and $X_2$ be two vectors mutually independent such that $X_i\sim N_p(\mu_i,\Sigma)$ with $\mu_i$ a vector ${p\times 1}$ and the covariance matrix $\Sigma$ $(p\times p)$. Let $Y_i=(\mu_1-\mu_2)^T\Sigma^{-1}X_i$. Which is the joint distribution of the vector $(Y_1,Y_2)^T$? Are the components independent?

First I get that each $Y_i$ is univariate normal where $$Y_1\sim N((\mu_1-\mu_2)^T\Sigma^{-1}\mu_1,(\mu_1-\mu_2)^T\Sigma^{-1}(\mu_1-\mu_2))$$ $$Y_2\sim N((\mu_1-\mu_2)^T\Sigma^{-1}\mu_2,(\mu_1-\mu_2)^T\Sigma^{-1}(\mu_1-\mu_2))$$

Then \begin{align} \begin{bmatrix} Y_1 \\ Y_2 \end{bmatrix} \sim \mathcal{N}_2 \left( \begin{bmatrix} (\mu_1-\mu_2)^T\Sigma^{-1}\mu_1 \\ (\mu_1-\mu_2)^T\Sigma^{-1}\mu_2 \end{bmatrix}, \begin{bmatrix} (\mu_1-\mu_2)^T\Sigma^{-1}(\mu_1-\mu_2)) & 0\\ 0 & (\mu_1-\mu_2)^T\Sigma^{-1}(\mu_1-\mu_2)) \end{bmatrix} \right). \end{align}

Since $X_1$ and $X_2$ are mutually independent, then $Y_1$ and $Y_2$ are also independent because they are linear combinations of each of the vectors, so $Cov(Y_1,Y_2)=0$. However the components of $(Y_1,Y_2)^T$ are not independent because they have the same covariance matrix and $Cov=0$ only implies in independence under multivariate normality.

Is it right?

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You are correct that $Y_1$ and $Y_2$ are independent since $Y_i$ is a function of $X_i$ and the $X_i$ are independent. Your calculations of the distributions look right too.

However when you say the components of $(Y_1,Y_2)^T$ are not independent this does not make sense. $Y_1$ and $Y_2$ are the components of this vector, and as you said, they are independent. Not sure if this is what's causing the confusion here, but note that $Y_1$ and $Y_2$ are scalars. $(\mu_1-\mu_2)^T\Sigma^{-1}(\mu_1-\mu_2)$ is a number... the variance of $Y_1$ (and also of $Y_2$).

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