Density of $X_1 + \cdots+X_n$ when $X_i$'s are independent $U(-1,1)$ variables

Question

How can we show for independent random variables uniformly distributed over $(1,-1)$ that $X_1 + \cdots+X_n$ has density

$$\pi^{-1}\int^\infty_0 \left(\frac{\sin t}{t}\right)^n \cos tx \; dt \textrm{ for }n \geq 2\text{?} $$

This is problem 26.6 from Billingsley "Chapter : CONVERGENCE OF DISTRIBUTIONS" (3rd edition).

I would really appreciate it, if you show it analytically .

Answer 1

Take $Y_n:=\sum_{k=1}^nX_k$

Compute the characteristic function of $Y_n$ :

we know that $\varphi_{X_1}(x)=\frac{\sin(x)}{x}$ if $x \neq 0,$ and $\varphi_{X_1}(0)=1,$ using independence we obtain that $\varphi_{Y_n}(x)=\frac{\sin^n(x)}{x^n}$ if $x \neq 0$ and $\varphi_{Y_n}(0)=1.$

Notice that $\varphi_{Y_n}$ is an odd function

Show it is an integrable function:

notice that $n \geq2,$ we have $$\int_{\mathbb{R}}\frac{|\sin^n(x)|}{|x^n|}dx=\int_{|x| \leq 1}\frac{|\sin^n(x)|}{|x^n|}dx+\int_{|x|>1}\frac{|\sin^n(x)|}{|x^n|}dx\leq \int_{|x| \leq 1}1dx+\int_{|x|>1}\frac{1}{|x|^n}dx<\infty.$$

Conclude using Fourier inversion formula:

Then $Y_n$ has a density such that $$f_{Y_n}(x)=\frac{1}{2\pi}\int_{\mathbb{R}}e^{-ixy}\frac{\sin^n(y)}{y^n}dy=\frac{1}{\pi}\int_{0}^{+\infty}\frac{\sin^n(y)}{y^n}\cos(xy)dy.$$

Answer 2

Hint

If $X,Y$ are independent and continuous, then $X+Y$ has density function $$f_X*f_Y(t)=\int_{\mathbb R}f_X(t-u)f_Y(u)\,\mathrm d u.$$