Recall the inversion formula: if \(\mu \) is a probability measure and \(\widehat {\mu }\) is its characteristic function, then for all \(a < b\):
\[ \lim _{T \to \infty } \frac {1}{2\pi }\int _{-T}^{T}\int _{a}^{b} e^{-ity}\widehat {\mu }(t)\,dy\,dt = \mu (a,b) + \frac {1}{2}\mu (\{a,b\}). \]
-
Prove that if \(\widehat {\mu } \in L^{1}\), then \(\mu (\{a\}) = 0\) for all \(a \in \mathbb {R}\) and \(\mu \) has a probability density function given by
\[ f(y) = \frac {1}{2\pi }\int _{-\infty }^{\infty } e^{-ity}\widehat {\mu }(t)\,dt. \] -
Show that if \(X_{1}, \ldots , X_{n}\) are independent and uniformly distributed on \((-1, 1)\), then for \(n \ge 2\), \(X_{1} + \cdots + X_{n}\) has density
\[ f(y) = \frac {1}{\pi }\int _{0}^{\infty } \left (\frac {\sin t}{t}\right )^{n}\cos (ty)\,dt. \]