Variance of a sum of weighed random variables with standard normal distribution

Question

I have to find the pdf of the following random variable:

$$ Z = \lim_{n \rightarrow \infty} \sum_{k=1}^{n}(\frac{3}{4})^{n-k}X_{k}$$

Where $X_{k} \sim N(0,1)$, and they are independent. I think the central limit theorem applies, so the resulting distribution should be normal with median equal to zero:

$$ \mathbb{E}[Z] = \lim_{n \rightarrow \infty} \sum_{k=1}^{n}(\frac{3}{4})^{n-k} \mathbb{E}[X_{k}] = 0$$

But I'm not sure about how I can obtain the variance analytically. The CLT I know uses non-weighed terms. By running a simulation and making a histogram, I think it the standard deviation should be $\sigma \approx 2$.

Thanks.

Answer 1

The CLT does not apply here, because the summed variables are not identically distributed, they have different (and "very different") variances.

What is thue, however is that the sum of normal variables is also normal. Specifically, if $Z = Y_1 + Y_2 + \cdots Y_n$ where $Y_i \sim N(0, \sigma_i^2)$ then $Z \sim N(0,\sigma^2_Z)$ where $\sigma^2_Z = \sum \sigma_i^2$.

Can you go on from here?