# Sufficient condition for convergence of sequence of random variables in $L^2(\mathbb{P})$

convergence-divergencerandom variables

Let $$W_n$$ is sequence of random variables. Is following condition $$\mathbb{E}\;W_n^2 \to c$$
sufficient for $$W_n$$ to converge in $$L^2$$-norm i.e. $$\exists W\in L^2(\mathbb{P})$$ such that $$\; \mathbb{E}(W_n-W)^2\to 0,;n\to \infty$$.

I was reading book about Brownian motion from Schilling. In chapter about consturction of Brownian motion, he creates sequence
$$W_N(t)= \sum_{n=1}^{N-1} G_n \int_{0}^t \phi_n(s)ds$$
where $$G_n$$ are iid $$\mathcal{N}(0,1)$$ for every $$t\in [0,1]$$.
He wants to prove that $$W_n(t)$$ converges in $$L^2$$ for every $$t$$. He only prove $$\mathbb{E} W_N^2(t)$$ converges to $$0$$ and state that $$L^2 – \lim_{N\to \infty} W_N(t)$$ exists. Can someone explain me how did he get it?

In general, condition $$\mathbb E[W_n^2] \to c$$ for some $$c \in \mathbb R$$ isn't sufficient to state that $$W_n \to W$$ in $$L_2$$ for some random variable $$W$$. Indeed, take $$(\Omega,\mathcal F,\mathbb P)= ([0,1],\mathcal B([0,1]),\mathcal L)$$ where $$\mathcal L$$ is Lebesgue measure on $$[0,1]$$. Define $$W_n = \sqrt{n}1_{[0,\frac{1}{n}]}$$. Then $$\mathbb E[W_n^2] = n\mathcal L([0,\frac{1}{n}]) = 1 \xrightarrow[n \to \infty]{} 1$$. Note that $$W_n \to 0$$ almost surely, hence in probability, too. If there is some $$W$$ such that $$W_n \to W$$ in $$L_2$$, then $$W_n \to W$$ in probability, too. Since limit in probability is unique almost surely, we would get that $$W=0$$ almost surely. But $$\mathbb E[(W_n-0)^2] = \mathbb E[W_n^2] = 1$$ which doesn't converge to $$0$$.
For the second example, if we have $$W_n$$ in the special form, that is $$W_n(t) = \sum_{k=1}^n G_k f_k(t)$$ for some deterministic functions $$f_k$$ and family $$\{G_k\}_k$$ of i.i.d standard normal, and we know that $$\mathbb E[W_n^2(t)] = \sum_{k=1}^n f_k^2(t) \to \sum_{k=1}^\infty f_k^2(t) < \infty$$ then it is enough to claim that $$W_n$$ converges in $$L_2$$ to some limit $$W$$. Indeed, note that $$W_n$$ is Cauchy in $$L_2$$, because for $$m > n$$ we have $$\mathbb E[|W_m(t)-W_n(t)|^2] = \sum_{k=n+1}^m f_k^2(t) \le \sum_{k=n+1}^\infty f_k^2(t) \xrightarrow[n \to \infty]{} 0$$ since the latter is a tail of convergent series. So that $$(W_n(t))_n$$ is Cauchy in $$L_2$$. Since $$L_2$$ is a complete space, then $$(W_n(t))_n$$ is convergent, which means that there is some $$W(t) \in L_2$$ such that $$W_n(t) \to W(t)$$ in $$L_2$$
In your example $$f_k(t) = \int_0^t \phi_k(s)ds$$, which by Parseval theorem and the fact that $$\{\phi_n\}_n$$ forms an orthonormal basis, gives $$\sum_{k=1}^\infty f_k^2(t) = t < \infty$$.