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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$

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  • $\begingroup$ 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. $\endgroup$
    – MånsT
    Jun 27, 2014 at 10:26
  • $\begingroup$ See en.wikipedia.org/wiki/Plancherel_theorem for an indication of where this result comes from. $\endgroup$
    – whuber
    Nov 27, 2017 at 22:48

1 Answer 1

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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 $$ \int \left[\int \left|G(t) e^ {itx}\right|dt\right]p(x) dx = \int |G(t)|dt\int p(x) dx = \int |G(t)|dt<\infty$$

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