# Thread: characteristic function of triangular distribution

1. ## characteristic function of triangular distribution

Suppose I have:

$\displaystyle f(x)=(1-|x|)1_{(-1,1)}(x)$

I want to show that its characteristic function is:

$\displaystyle \varphi (u)= 2(1-cosu)/u^2$

I let

$\displaystyle E(e^{iux})=\int_{-1}^1e^{iux}(1-|x|).dx$
$\displaystyle =\int_{-1}^0e^{iux}(1+x).dx+\int_{0}^1e^{iux}(1-x).dx$
...
$\displaystyle =\frac{1}{(iu)^2}(e^{-iu}+e^{iu})$
$\displaystyle =2(\cos u)/u^2$

May I know what went wrong?

2. Originally Posted by noob mathematician
Suppose I have:

$\displaystyle f(x)=(1-|x|)1_{(-1,1)}(x)$

I want to show that its characteristic function is:

$\displaystyle \varphi (u)= 2(1-cosu)/u^2$

I let

$\displaystyle E(e^{iux})=\int_{-1}^1e^{iux}(1-|x|).dx$
$\displaystyle =\int_{-1}^0e^{iux}(1+x).dx+\int_{0}^1e^{iux}(1-x).dx$
...
$\displaystyle =\frac{1}{(iu)^2}(e^{-iu}+e^{iu})$
$\displaystyle =2(\cos u)/u^2$

May I know what went wrong?
The error lies in your ellipsis, show us what you have in that gap and then we may have a chance of seeing what went wrong.

CB

3. Hey CB.. Thanks for the pointer.

The error is indeed found in the (...)