# Prove that $E[g(X)] = \int_{-\infty}^{\infty}G(t)\phi(t) dt$

Let $X$ denote a real-valued random variable with characteristic function $\phi$. Suppose that $g$ is a real-valued function on $\mathbb{R}$ that has the representation

$\hspace{25mm}g(x) = \int_{-\infty}^{\infty}G(t)\exp(itx)\,dt$

for some $G$ satisfying

$\hspace{25mm}\int_{-\infty}^{\infty}|G(t)|dt < \infty$

Prove that

$\hspace{15mm}$$E[g(X)] = \int_{-\infty}^{\infty}G(t)\phi(t) dt Attempt: \hspace{15mm} E[g(X)] = \int_{-\infty}^{\infty}G(t)\exp(itX)\cdot p(x) dx Since \int_{-\infty}^{\infty}\exp(itX)\cdot p(x) dx = E[\exp(itX)] = \phi(t), we then have \hspace{15mm}$$E[g(X)] = \int_{-\infty}^{\infty}G(t)\phi(t) dt$

• Actually, (assuming that the sample space for $X$ is $\mathbb{R}$) we have $E(g(X))=\int_{-\infty}^\infty\int_{-\infty}^\infty G(t)exp(itX)p(x)dtdx$. You have the right idea about how to show this, but you need to think about whether or not you can change the order of integration. Commented Jun 27, 2014 at 10:26
• See en.wikipedia.org/wiki/Plancherel_theorem for an indication of where this result comes from.
– whuber
Commented Nov 27, 2017 at 22:48

Simply rewrite$$Eg(X) = \int \left[\int G(t) e^ {itx}dt\right]p(x) dx = \int \left[\int p(x) e^ {itx}dx\right]G(t) dt = \int G(t)\phi(t) dt$$
You can use Fubini theorem because $$\int \left[\int \left|G(t) e^ {itx}\right|dt\right]p(x) dx = \int |G(t)|dt\int p(x) dx = \int |G(t)|dt<\infty$$