Skip to content

Question

Math 6040 • Probability

Spring 2023 • Question 4

Recall the inversion formula: if \(\mu \) is a probability measure and \(\widehat {\mu }\) is its characteristic function, then for all \(a < b\):

\[ \lim _{T \to \infty } \frac {1}{2\pi }\int _{-T}^{T}\int _{a}^{b} e^{-ity}\widehat {\mu }(t)\,dy\,dt = \mu (a,b) + \frac {1}{2}\mu (\{a,b\}). \]
  1. 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

    \[ f(y) = \frac {1}{2\pi }\int _{-\infty }^{\infty } e^{-ity}\widehat {\mu }(t)\,dt. \]
  2. 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

    \[ f(y) = \frac {1}{\pi }\int _{0}^{\infty } \left (\frac {\sin t}{t}\right )^{n}\cos (ty)\,dt. \]

Raw LaTeX

Recall the inversion formula: if \(\mu\) is a probability measure and \(\widehat{\mu}\) is its characteristic
function, then for all \(a < b\):

\[
\lim_{T \to \infty} \frac{1}{2\pi}\int_{-T}^{T}\int_{a}^{b}
e^{-ity}\widehat{\mu}(t)\,dy\,dt
= \mu(a,b) + \frac{1}{2}\mu(\{a,b\}).
\]

\begin{enumerate}

\item
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

\[
f(y) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-ity}\widehat{\mu}(t)\,dt.
\]



\item
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

\[
f(y) = \frac{1}{\pi}\int_{0}^{\infty} \left(\frac{\sin t}{t}\right)^{n}\cos(ty)\,dt.
\]


\end{enumerate}