Convergence in distribution and asymptotic distribution

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?

• $E[U_N] = 0$. Do you know what $Var(U_N)$ is or at least what it converges to? – Henry Aug 13 '18 at 9:37
• Var(U_N) should be 1/Nsum(sin(npi)/N), which also converges to zero for N->\infty – Jonathan Kiersch Aug 13 '18 at 9:45
• I suspect $Var(U_N) \to \int\limits_0^1 \sin^2(\pi x)\, dx = \frac12$ – Henry Aug 13 '18 at 11:10
• How did you find that? – Jonathan Kiersch Aug 13 '18 at 11:57
• A combination of thinking that without the $\sin(\frac{n\pi}{N})$ term the variance would be $1$ as in the central limit theorem, plus simulation and basic calculus. – Henry Aug 13 '18 at 12:01

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.
• $s_N=\sum_{i=1}^N\operatorname{Var}\left(X_{N,i}\right)\sim c\sqrt N$.
• $$\mathbb E\left[X_{N,i}^2\mathbf 1\left\{\left\lvert X_{N,i}\right\vert\gt\varepsilon s_N \right\}\right]=\mathbb E\left[X_1^2\sin^2\left(i\pi/N\right)\mathbf 1\left\{\left\lvert \sin\left(i\pi/N\right) X_{1}\right\vert\gt\varepsilon s_N\right\}\right]\leqslant\mathbb E\left[X_1^2\sin^2\left(i\pi/N\right)\mathbf 1\left\{\left\lvert X_{1}\right\vert\gt\varepsilon s_N \right\}\right].$$
• @DavideGiraudo - the limiting variance is $\frac12$ (it would have been $1$ without the sine term) – Henry Aug 13 '18 at 14:04