limit of gaussian process

Question

If I have a sequence of gaussian random process $X_{t}^{n}$ which converge in $L^2$ norm to a process $X_t$ for every $t$. can I say that $X_t$ is also gaussian process?

thank you

Answer 1

Let $X^n_t \sim N(a_n, b_n)$. It would be good to see that if $X^n_t \xrightarrow[]{L^2} X_t $ then $a_n \to a$, $b_n \to b$ and $X_t \sim N(a,b)$.

$L^2$ convergence implies convergence in probability and in distribution.

Therefore $$\limsup\Bbb{P}[|X^n_t|\geq K] \leq \Bbb{P}[|X_t|\geq K]$$ So for every $\frac{1}{2}>\epsilon>0$ there is a $K(\epsilon)$ such that $$\sup_n\Bbb{P}[|X^n_t|\geq K(\epsilon)] < \epsilon$$

Claim: there is a subsequence $(a_{n_k},b{n_k})\to (a,b)$.

Indeed, if there were no convergent subsequence for $a_n$ then $|a_n| \to \infty$. In this case take $a_n $ such that $|a_n|>K(\epsilon)$.

$$\Bbb{P}[|X^n_t|\geq K(\epsilon)] \geq \Bbb{P}[|X^n_t|\geq a_n] \geq \frac{1}{2} $$ A contradiction.

We must have that $b_{n_k}$ is bounded, for if it were not the case there would be $b_{n'_{k}}$ such that $|b_{n'_k}| \to \infty$. Therefore

$$\Bbb{P}[|X^n_t|\geq K(\epsilon)] = 1 - \int_{-K(\epsilon)}^{-K(\epsilon)}\frac{1}{\sqrt{2\pi b_{n'_k}}} \exp\bigg\{-\frac{|x-a_{n'_k}|^2}{2b_{n'_k}}\bigg\} \, dx\xrightarrow[b_{n'_k} \to \infty]{B.C.T.} 1 $$

Again a contradiction. (B.C. T. is the bounded convergence theorem).

So we have $(a_{n_k},b_{n_k}) \to (a,b)$. This implies that $X^n_t \xrightarrow[]{}N(a,b) $ (in distribution)

As the limit is unique we conclude that $X_t \sim N(a,b)$.

To obtain the result of normal distribution for every increment $X_{t_{1}} - X_{t_0}, X_{t_{2}} - X_{t_1}, \ldots ,X_{t_{k}} - X_{t_{k-1}}$. We just need to apply the same reasoning and note that

$X^n_i \to X_i$ in probability for $i = 1, \ldots , k$ implies that $(X^n_1,\ldots, X^n_k) \to (X_1,\ldots, X_k)$ in probability.

Hope this helps