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

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.