# Proving this sequence converges in $L^2(\mathbb{P})$

We have some IID sequence, $$\left\{ {{X_n}} \right\}_{n = 1}^\infty$$, of standard normal random variable on the probability space $$\left( {\Omega ,\mathcal{F},\mathbb{P}} \right)$$. Also $$\left\{ {{\xi _n}} \right\}_{n = 1}^\infty$$ is an orthonormal basis for $${L^2}\left( {[0,\infty ),\mathcal{B}\left( {[0,\infty )} \right),\mu } \right)$$. Mu refers to the Lebesgue measure. I have been trying to prove that the following sequence converges in $$L^2(\mathbb{P})$$ without much success. I have been thinking of proving that it is a Cauchy sequence and exploiting the completeness of the space but that hasn't paned out. Any ideas?

$$Y_t^{(k)} = \sum\limits_{n = 1}^k {{X_n}\int\limits_0^\infty {{\xi _n}(u){1_{[0,t]}}(u)d\mu } } (u)$$

• Do you want to prove that you have a Cauchy sequence for each fixed $t$? – Davide Giraudo May 23 at 17:16
• Yes, I am trying to prove that it converges for every $t \in [0,\infty )$. – Anonymous May 23 at 17:19

1. The series $$\sum_{n\geqslant 1}c_nX_n$$ is convergent in $$\mathbb L^2$$ if $$\sum_{n\geqslant 1}c_n^2$$ converges (show that the sequence is Cauchy in $$\mathbb L^2$$ using the fact that $$\sum_{n=N}^Mc_nX_n$$ has a centered normal distribution with variance $$\sum_{n=N}^Mc_n^2$$.
2. For a fixed $$t$$, define $$c_n:=\int\limits_0^\infty \xi _n(u) 1_{[0,t]} (u)d\mu (u)$$. Then $$c_n=\langle \xi_n,\mathbf 1_{[0,t]}\rangle$$ and square summability of $$\left(c_n\right)_{n\geqslant 1}$$ follows from Parseval's inequality.