Expected value, from characteristic function

Question

I have a random variable, $Z$ with a its characteristic function given as $\varphi_Z(x)=e^{-|x|}, x\in \mathbb{R}$. How do I calculate $\mathbb{E}[|Z|]$?

• I do not have the assumption that $\mathbb{E}[|Z|]<\infty$, so I cannot use that $\mathbb{E}[Z]=\frac{1}{i}\varphi_Z(0)$, thus I am stuck.

Answer 1

The pdf of a Cauchy distributed random variable with parameters $\gamma=1,x_0=0$ is $$\frac1{\pi(1+x^2)}.$$

Then the pdf of $|X|$ is $$\frac2{\pi(1+x^2)}\text{ if }x\geq 0.$$

The corresponding expectation (times $\frac{\pi}2$) is

$$\int_0^{\infty}\frac x{1+x^2}\ dx=\int_0^1\frac x{1+x^2}\ dx+\int_1^{\infty}\frac x{1+x^2}\ dx>$$$$>\int_1^{\infty}\frac x{x^2+x^2}\ dx=\frac12\int_1^{\infty}\frac1x\ dx=\infty.$$

Answer 2

If you don't know what zoli's answer requires (that $\varphi_Z$ is the characteristic function of a standard cauchy distribution):

If for $n \in \Bbb N$ it holds $$E[|Z|^n]< \infty$$ then $\varphi_Z$ is $n$-times continuously differentiable in $0$.

Because your given characteristic function is not differentiable in 0 it cannot hold $$E[|Z|]< \infty$$ hence $$E[|Z|^n]= \infty$$