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Suppose we have three independent normally distributed random variables $$ X_0 \sim \mathcal{N}(\mu_0, \sigma_0^2), $$ $$ X_1 \sim \mathcal{N}(\mu_1, \sigma_1^2), $$ $$ X_2 \sim \mathcal{N}(\mu_2, \sigma_2^2).$$

Now, define two new random variables $Y_0 = X_0+X_1$ and $Y_1 = X_1+X_2$.

Let $\vec{Y} = [Y_0 \;\;\; Y_1]^T$

What can we say about the distribution of $\vec{Y}$? Obviously, $Y_0$ and $Y_1$ are not independent. If they were, then $\vec{Y}$ would have been a multivariate normal variable. Any ideas?

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Yes, $Y$ is a multivariate random vector. This is a duplicate question addressed repeatedly on both math.SE and stats.SE. But, the Wikipedia article linked to should suffice. –  cardinal Jan 4 '12 at 1:20
    
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@cardinal I think you meant to say "Yes, $\vec{Y}$ is a multivariate normal random variable". That $\vec{Y}$ is a multivariate random vector is a tautology. But this is an issue that comes up repeatedly on math.SE as well as stats.Se where many assume that marginally normal automatically means jointly normal as well and many assume (as eakbas has done) that joint normality requires independence. The overlap between the two groups of people may be large too. –  Dilip Sarwate Jan 4 '12 at 2:36
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@eakbas: You supplied the additional assumption that $X_0, X_1,$ and $X_2$ were independent. This is sufficient to guarantee the multivariate normality of $Y$. (Read the second point under definition on the Wiki page.) –  cardinal Jan 4 '12 at 2:55
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Thanks @cardinal, I missed the "second point under definition" part in your previous answer. Now it's clear. –  emrea Jan 4 '12 at 3:15

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