I have tried to solve the following problem but with little success.
Let $(X_l)_{l\geq1}$ be a sequence of iid random variables with the characteristic function:
$\phi(t)=\begin{cases}1-\sqrt{|t|(2-|t|)}\text{ if |t|≤1}\\0 \text{ else}\end{cases} $
Set $S_n=\sum_{k=1}^{n}X_k, n \geq 1$.
I want to show that $\frac{S_n}{n^2}$ converges in distribution as $n\to\infty$, and find the limit distribution.
Thus I start,
$\phi_{\frac{S_n}{n^2}}(t)=\mathbb{E}\Big(e^{i\big(\frac{t}{n^2}\big)S_n}\Big)=\phi_{S_n}(\frac{t}{n^2})$
Now I derive an expression for $\phi_{S_n}(t)$:
$\phi_{S_n}(t)=\mathbb{E}\Big(e^{itS_n}\Big)=\mathbb{E}\Big(e^{it\sum_{k=1}^{n}X_k}\Big)=\prod_{k=1}^{n}\mathbb{E}\Big(e^{itX_k}\Big)=\Big(\phi(t)\Big)^n$
Thus,
$\phi_{\frac{S_n}{n^2}}(t)=\Big(\phi(\frac{t}{n^2})\Big)^n=\Big(1-\sqrt{|\frac{t}{n^2}|(2-|\frac{t}{n^2}|)}\Big)^n\to e^{-\sqrt{2|t|}}$ as $n\to\infty$ (when $|t|\leq1$).
I do not recognize this as the characteristic function for any distribution. This leads me to the following two questions:
1) Is this indeed the characteristic function for some distribution? If so, which one?
or
2) Did I do the computations incorrectly? If so, where did it all go wrong?
Best Answer
$e^{-c|t|^{\alpha}}$ is a characteristic function for any $c>0$ and any $\alpha \in (0,2]$. The distribution corresponding to it is called (symmetric) stable distribution with index $\alpha$. For $\alpha =2$ it is normal but here $\alpha =\frac 1 2$.