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I have a question regarding sequences of multivariate normal vectors:

Let $ (X_k)_{k \geq 1} $ be a sequence of random vectors of fixed length $ n \geq 1 $ with multivariate normal distribution on a probability space $ (\Omega, \mathcal{F}, P) $ , such that $ X_k \rightarrow X, \ P-a.s.,$ as $ k \rightarrow + \infty $.

Is is true then that $ X $ has again multivariate normal distribution? And would you happen to know where I can find a proof of this?

Thanks a lot for your help & have a nice week!

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"...such that $X_k \to X$" in what sense? – leonbloy Apr 22 '12 at 14:30
up vote 1 down vote accepted

It is true that if $X_k$ is multivariate normal random vector and $X_k$ converge in distribution to $X$, then $X$ is also a multivariate normal random vector.

According to Levy continuity theorem convergence in distribution implies point-wise convergence of characteristic functions. Characteristic function $\phi_{X_k}(t) = \exp(- t.A_k.t + B_k.t)$, being characteristic function of a Gaussian random vector. Then

$$ \lim_{k \to \infty} \phi_{X_k}(t) = \lim_{k \to \infty} \exp(- t.A_k.t + B_k.t) = \exp(- t.A.t + B.t) $$ where $A = \lim_{k\to \infty} A_k$ and $B = \lim_{k \to \infty} B_k$.

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I'm sorry, but how do I know that $ \lim_{k \rightarrow \infty} A_k $ and $ \lim_{k \rightarrow \infty} B_k $ must exist at all? I see it if I have convergence w.r.t. $ L^2 $, but what if I only have convergence $ P-a.s.,$ or even only in distribution? Thanks for your help! – Mad Si Apr 22 '12 at 14:48
@MadSi You know that limits exist, because you said that $X_k$ converges in distribution, and Levy theorem tells that $\lim_{k \to \infty} \phi_{X_k}(t) = \phi_{X}(t)$ for every $t$, i.e. the limit exists. The functional form of $\phi_{X_k}(t)$ does not change with $k$, hence the limiting ch.f. will have the same functional form, and it follows that the limit of $A_k$ and $B_k$ will exist. – Sasha Apr 22 '12 at 14:59
Ok, thanks! I think I got it. – Mad Si Apr 22 '12 at 21:58
@MadSi: Like you, I don't understand why the limits $\lim_{k \rightarrow \infty} A_k$ and $\lim_{k \rightarrow \infty} B_k$ must exist. Have you found a satisfactory explanation by now? – Evan Aad Feb 28 '14 at 21:44

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