Show that $S_n = X_1+ \cdots X_n $ the probability density function $f_{s{_n}}$

Question

We consider $(\Omega, \mathcal{F}, \mathcal{P})$ a probability space and $(X_{n})_{n \in \mathbb{N}}$ a sequence of independent and identically distributed random variables such that $X_1 \sim \operatorname{Unif}(-1,1)$. Show that $S_n = X_1+ \cdots X_n $ for $n \ge 2$ has the probability density function $$f_{s{_n}} = \frac{1}{n} \int_{0}^{\infty} \Bigg(\frac{\sin(t)}{t} \bigg)^n \cos(xt)dt $$

I have found out that I could use that $|\frac{\sin(t)}{t}|\le1$ for $\forall t \in \mathbb{R}$ and I have also calculated that $$\int_{0}^{\infty} \Bigg(\frac{\sin(t)}{t} \bigg)^2 dt = \frac{\pi}{2}$$ At this point I have no clue how should I continue it I would really appreciate any kind of help since I'm learning the topics on my own.

Answer 1

The characteristic function of $S_n$ is $$\phi_{S_n}(t) = E[e^{itS_n}] = \prod_{k=1}^n E[e^{itX_k}] = \left(\frac{e^{it} - e^{-it}}{2it}\right)^n = \left(\frac{\sin t}{t}\right)^n.$$

The inversion theorem shows how you can compute the density $f_{S_n}$ if you know the characteristic function $\phi_{S_n}$.

\begin{align} f_{S_n}(x) &= \frac{1}{2\pi} \int_{-\infty}^\infty e^{-itx} \phi_{S_n}(t) \, dt \\ &= \frac{1}{2\pi} \int_0^\infty e^{-itx} \phi_{S_n}(t) \, dt + \frac{1}{2\pi} \int_0^\infty e^{iux} \phi_{S_n}(-u) \, du & u = -t \\ &= \frac{1}{2\pi} \int_0^\infty (e^{itx} + e^{-itx}) \phi_{S_n}(t) \, dt & \phi_{S_n}(-t) = \phi_{S_n}(t) \\ &= \frac{1}{\pi} \int_0^\infty \cos(tx) \phi_{S_n}(t) \, dt. \end{align}

Answer 2

Characteristic function will work. Ley $f(x)$ be density function for $X_k$. $\phi(t)=\int_R e^{itx}f(x)dx$. Then density function for $S$ is given by $f_S(x)=\frac{1}{2\pi}\int_R e^{-itx}\phi^n(t)dt$.