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.

  • 1
    $\begingroup$ The hint is not actually correct as you state it. An orthonormal family of jointly Gaussian variables is independent, but it's easy to construct a nonindependent orthonormal sequence of random variables, each of which is Gaussian. $\endgroup$ – Mark Meckes Jul 22 '10 at 14:38

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:

$$B_n =(|Z_1|<\epsilon,.. |Z_{n-1}|<\epsilon, |Z_n|\geq \epsilon )$$


See Theorem 2.5.3 in R. Durrett's Probability: Theory and Examples. It's available online at http://www.math.cornell.edu/~durrett/PTE/PTE4_Jan2010.pdf.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.