Let $X_i \sim \operatorname{Ber}(0.5)$ and $X_i$'s independent. Let $Y$ be a random
variable with the same distribution as $\sum_{i=1}^n iX_i$.
Prove that $\frac{Y-E(Y)}{\sqrt{\operatorname{Var}(Y)}}$ converges in distribution to $Z\sim N(0,1)$ as $n\to \infty$.
In the previous question I calculated
$$E(Y) = \frac{n(n+1)}{4}$$ and $$\operatorname{Var}(Y) = \frac{n(n+1)(2n+1)}{24}$$
I've been staring at this for a while now. I've tried showing it converges in probability, which would imply it converges in distribution. I also tried to do it with the MGF, which is $M_Y(t) = \prod_{j=1}^n 0.5(1+e^{jt})$.
But neither approaches worked.
I would like to solve this myself so I would appreciate a hint on which technique to use over a full answer. Thank you in advance!
Best Answer
You're on the right track, but as it's easier to work with a sum than a product. We'll repeatedly use $\sum_{i=1}^ni^p\in O(n^{p+1})$ for $p\ne-1$, and abbreviate the operator $\sum_{i=1}^n$ as $\sum_i$.
Since $X_1$ has cgf $\ln\cos t=-\tfrac12t^2-O(t^4)$ (we won't need the $t^4$ coefficient), and $Y$ has zero mean and zero skew, $Y$ has cgf$$H_Y(t):=\sum_i\ln\cos it=-\tfrac12\sigma^2\cdot t^2-O(n^5t^4).$$You want the asymptotic distribution of $Z:=Y/\sigma$, which has cgf $H_Z(t)=-\frac12t^2-O(n^5t^4/\sigma^4)$. Since $\sigma^2\in O(n^3)$, the last term is $O(t^4/n)$, which disappears in the $n\to\infty$ limit as required.