Find the density of the random variable with characteristic function $\varphi(t)=(1-|t|)^+$.

Question

Suppose a $X$ is a random variable, I am asked to find the density of the random variable with characteristic function $\varphi(t)=(1-|t|)^+$.

I am trying to use the inversion formula for the charactersic functions

$$P(a<X<b)+\frac{1}{2}(P(X=a)+P(X=b))=\frac{1}{2\pi}\lim_{T\rightarrow \infty}\int_{-T}^T\frac{e^{-iat}-e^{-ibt}}{it}\varphi(t)dt$$

We end up needing to compute the following integral,

$$\int_{-1}^1\frac{e^{-iat}-e^{-ibt}}{it}(1-|t|)^+dt $$

Which I am finding very difficult becuase of the $1/t$ factor. Any other ideas?

Answer 1

Hint:

Use the following inversion formula:

$$p(x) = \int_{\mathbb{R}} e^{\imath \, t x} \phi(t) \, dt;$$

here $p$ denotes the density of $X$. Using the definition of $\phi$ and the fact that $2\cos(tx) = e^{\imath \, tx} + e^{-\imath \, tx}$, show that

$$p(x) = 2 \int_0^1 \cos(xt) (1-t) \, dt.$$

Solution: $$p(x) = 2 \frac{1-\cos(x)}{x^2}.$$