# Limit of a sequence of multivariate normal vectors

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? en.wikipedia.org/wiki/Convergence_of_random_variables – leonbloy Apr 22 '12 at 14:30

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 $B = \lim_{k \to \infty} B_k$.
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
@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