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Is is true to say that k-dimensional Normal distribution is equivalent to the multiplication of k 1-dimensional Normal distributions if variance is equal in all dimensions?

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up vote 3 down vote accepted

No. That's true only if all of the covariances are zero. One of the simplest counterexamples is this: suppose $X \sim N(0,1)$. Then $(X,X)$ has a bivariate normal distribution in which the two components are perfectly correlated. The covariance matrix in this case is singular.

Another simple case is this: Suppose $X \sim N(0,1)$ and $Y\sim N(0,1)$ and $X,Y$ are independent, so that the pair $(X,Y)$ has just the sort of distribution that your question anticipates. Then the pair $(X,X+Y)$ has a bivariate normal distribution with correlated components that is not of the kind that your question contemplates. The matrix of covariances is $\begin{bmatrix} 1 & 1 \\ 1 & 2\end{bmatrix}$. The correlation between the two components is $1/\sqrt{2}$, which is about $0.7$

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