# Identifying limit distribution of sum of iid random variables

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?

$$e^{-c|t|^{\alpha}}$$ 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 $$\alpha =\frac 1 2$$.