Does anyone have an idea how to prove the following? It is a step in the proof of some theorem in a book about gaussian processes.


Let $f\_n$ be an orthonormal sequence of gaussian variables. Consider $\sum\_{n \geq 1} a\_n f\_n$ and assume that it is convergent in $L^2$. Show that it converges also almost everywhere.


The hint is that every orthonormal family of gaussian variables is automatically independent, but I have no idea how to proceed.



Convergence of a series of orthonormal gaussian variables
You can assume $f\_n$ have mean zero, independence implies


$$E[ \sum\_{n=1}^{\infty} f\_n| \mathcal{A}m ] = \sum\_{n=1}^m f\_n $$ 


where $\mathcal{A}\_m$ is the sigma field generated by $f\_1,..,f\_m$
This means that $Z\_m=\sum\_{n=1}^mf\_n$ is a martingal with respect to $\mathcal{A}\_m$. 


You then have to apply doob martingal convergence theorem
<http://en.wikipedia.org/wiki/Doob%27s_martingale_convergence_theorems>


This is brutal force as the result you ask for is originally due to Kolmogorov, 
Hence if you want to reproduce the direct proof of kolmogorov you have to show the following lemma (called Kolmogorov inequality):


$$ P(\max\_{k=1,..,m} |Z\_m|\geq \epsilon )\leq E[Z\_m^2]/\epsilon^2$$


You will get it by using the partition 


$$( \max\_{k=1,..,m} |Z\_m|\geq \epsilon ) =\cup\_{n=1}^m B\_n$$ 
where $B\_n$ is the event:


$$B\_n =(|Z\_1|<\epsilon,.. |Z\_{n-1}|<\epsilon, |Z\_n|\geq \epsilon )$$


