Expected value in term of characteristic function

Question

Let $X$ be a random variable. It is known that for a function $f:\mathbb{R}^+ \rightarrow \mathbb{R}^+,$ of class $C^1,$ increasing, $f(0)=0,$ $$E[f(|X|)]=\int_0^{+\infty}f'(x)P(|X|>x)dx=\int_0^{+\infty}f'(x)(1-F_{|X|}(x))dx \ \ \ \ \ (1)$$ In particular, if $X \in L^1,$ then $E[X]=\int_0^{+\infty}(1-F_X(x)-F_X(-x))dx.$

Also $E[|X|]=\frac{2}{\pi}\int_{]0;+\infty[}\frac{1-\Re(\varphi_X(x))}{x^2}dx.$

In $(1), E[f(X)]$ is expressed in term of the distribution function $F_{|X|},$ is there an expression for $E[f(X)]$ in term of $\varphi_X$ ? In particular, if $X \in L^1,$ how can we write $E[X]$ in term of $\varphi_X$? To do so, is it possible to find $\varphi_{X^+}$ and $\varphi_{X^-}$ in term $\varphi_X$?

Answer 1

Well, let $\varphi = \varphi_X$ for convenience, and then consider

$$ \begin{align} \varphi' (0) & = \lim_{h \to 0} \frac{ \varphi(h) - \varphi(0) }{h} \\ & = \lim_{h \to 0} \int_\mathbb{R} \frac{e^{ihx} -1}{h} dP(x) \\ & = \int_\mathbb{R} \lim_{h \to 0} \frac{e^{ihx} -1}{h} dP(x) \\ & = \int_\mathbb{R} ix dP(x) \\ & = i E(X) \end{align} $$

so $E(X) = -i\varphi'(0)$. You need $|X|$ to be integrable to move the limit inside the integral using dominated convergence. I'm not sure what is true in general.

Whatever $f$ is, there's some new distribution $P_f$ and you have $-i \varphi_{f(X)}'(0) = E(f(X))$, with the same $L^1$ caveat.

Hope that helps.

Answer 2

(Not a complete answer, but too long for a comment.)

Sometimes you can write $$\mathrm E[f(X)] = \frac{1}{2\pi} \int_{-\infty}^\infty \hat f(t) \varphi_X(t) dt,$$ where $f$ is the Fourier transform of $f$, thanks to Parseval's identity. Even more often, you can write it in the distributional sense: $$\mathrm E[f(X)] = \frac{1}{2\pi} \langle \hat f , \varphi_X\rangle.$$ For example, the Fourier transform of $x^2$ is $-2\pi\delta''_0(t)$, so $$\mathrm E[X^2] = -\langle \delta_0'',\varphi_X\rangle = -\varphi''_X (0),$$ which we know does hold provided that $\mathrm E[X^2]<\infty$.