0
$\begingroup$

Let $x_i$ be samples from gaussian distribution with mean $0$ and variance $\sigma^2$ and $s_n=\sum_{i=0}^n2^ix_i$. What can one say about the distribution of $s_n$ at $n\rightarrow\infty$?

Sum of weighted Gaussians is a Gaussian.

$\mu(s_n)=\sum_i2^i\mu_i=0$

$\sigma^2(s_n)=\sum_i2^{2i}\sigma_i^2=\sigma^2\sum_i2^{2i}=\sigma^2\frac{2^{2n+1}-1}{2^2-1}$

So, $\lim_{n\rightarrow\infty}\sigma^2(s_n)=\infty$

Does the distribution become an almost $\epsilon$ function in distribution since mass of the total distribution has to be $1$?

$\endgroup$
2
$\begingroup$

First note that $2^iX_i\sim N(0,2^{2i}\sigma^2)$. If $f_i$ is the pdf corresponding to $N(0,2^{2i}\sigma^2)$, then the distribution of $s_n$ is the convolution of $f_1,\cdots,f _n$. Now to find the asymptotic distribution of $s_n$, you need some kind of Central Limit Theorem for non-i.i.d. random variables. I think Lyapunov Condition will suffice.

$\endgroup$
1
$\begingroup$

If the samples are independent then $s_n$ has a gaussian distribution with mean $Es_n=0$ and variance $Es_n^2=\Sigma_{i=0}^n2^{2i}E[x_i^2]=\sigma^2\Sigma_{i=0}^n2^{2i}=\sigma^2\frac{4^{n+1}-1}{4-1}$. The result remains valid as n tends to infinity.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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