Convergence in distribution and asymptotic distribution

Question

I am having a problem with proving convergence in distribution (or by law).

Consider that the sequence $X_n$ of random variables are IID and that $E[X_n]=0$ and $V[X_n]=1$.

Now define the variable $U_N$ as:

$$U_N= \frac{1}{\sqrt{N}}\sum_{n=1}^N X_n\cdot \sin\left(\frac{n\pi}{N}\right).$$

For $N\rightarrow \infty$, I want to show that $U_N$ converges by distribution, furthermore, I also want to determine the asymptotic distribution of $U_N$.

For the first part, I have tried to show convergence in probability, because this implies conv. in distribution (by law), but this was not possible since I dont have the asymptotic distribution of the $X_n$.

For the second part, I tried with the delta-method, but this did not work because of the summation of in the expression for $U_N$.

Does someone have an idea to this?

Answer 1

One can try to check Lindeberg's condition with $X_{N,i}:= X_i\sin\left(i\pi/N\right)$, using the following three facts:

• the limits $\lim_{N\to +\infty}N^{-1}\sum_{n=1}^N\sin^2\left(n\pi/N\right)$ exists and is computable, as a limit of Riemann sums.
• $s_N=\sum_{i=1}^N\operatorname{Var}\left(X_{N,i}\right)\sim c\sqrt N$.
• $$\mathbb E\left[X_{N,i}^2\mathbf 1\left\{\left\lvert X_{N,i}\right\vert\gt\varepsilon s_N \right\}\right]=\mathbb E\left[X_1^2\sin^2\left(i\pi/N\right)\mathbf 1\left\{\left\lvert \sin\left(i\pi/N\right) X_{1}\right\vert\gt\varepsilon s_N\right\}\right]\leqslant\mathbb E\left[X_1^2\sin^2\left(i\pi/N\right)\mathbf 1\left\{\left\lvert X_{1}\right\vert\gt\varepsilon s_N \right\}\right].$$