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If I have a vector $X=(x_1,...,x_n)$ with all $x_i$ normal variables, and I have all the correct conditionals, does this imply $X$ is a multivariate normal?

For example, in two dimensions: Does the fact that $x_1$ and $x_2$ are normal, and $x_1|x_2$ and $x_2|x_1$ are normal with the correct $\mu,\sigma^2$ for a multivariate normal, imply that $X=(x_1,x_2)$ is a multivariate normal?

I tried to fiddle around with the resulting distribution but got nowhere.

Thanks.

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What do you call the correct μ,σ2 for a multivariate normam? –  Did Nov 7 '12 at 7:19
    
The conditionals that would exist if this were a multivariate normal: en.wikipedia.org/wiki/… –  ido Nov 7 '12 at 7:24
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The joint density of $X_1$ and $X_2$ is the product of the marginal density of $X_1$ and the conditional density of $X_2$ given $X_1$: $$f_{X_1,X_2}(x_1,x_2) = f_{X_1}(x_1) f_{X_2|X_1}(x_2|x_1)$$ Thus suppose $X_1$ is normal and the conditional distribution of $X_2$ given $X_1$ is normal with mean of the form $a X_1 + b$ (with $a$ and $b$ constant) and constant variance. Then $f_{X_1,X_2}(x_1,x_2)$ is a constant times the exponential of a quadratic in $x_1$ and $x_2$, therefore $X_1$ and $X_2$ are jointly normal.

More generally, what we need for $X_1, \ldots, X_n$ to be jointly normal is that $X_1$ is normal and that for each $k$ from $2$ to $n$, the conditional distribution of $X_k$ given $X_1, \ldots, X_{k-1}$ is normal with mean an affine function of $X_1, \ldots, X_{k-1}$ and constant variance.

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