The probability density function $f(x)$ of a random variable $X$ is symmetric about $0$. Then we have:

Question

The pdf $f(x)$ of a random variable $X$ is symmetric about $0$. Then $\int_{-2}^{2}\int_{-\infty}^{x}f(u)dudx= $?

My input:

$\int_{-2}^{2}\int_{-\infty}^{x}f(u)dudx$

$\int_{-2}^{2}F(x)dx\ $

This tells that $F(x)\sim U(0,1)$

pdf=$f(F(x))=\dfrac{1}{1-0}$

Therefor

$\int_{-2}^{2}F(x)dx\ =4$

But this is wrong answer can someone help me here?

Edit : my friend solved this question as

CDF of Uniform distribution is $\frac{x-a}{b-a}$

$\int_{-2}^{2}F(x)dx\ =$ $\int_{-2}^{2} \frac{x+2}{4}\implies 2 $ (I don't know the reasoning that's how just he solved I am not in contact with him so if anyone can relate?)

Answer 1

Since $f(x)$ is an even function, $F(0)=\frac12$ and $F(x)-\frac12$ is an odd function. Then $$\int_{-2}^2F(x)\,dx=\int_{-2}^2(F(x)-1/2)\,dx+\int_{-2}^2\frac12\,dx$$ Since the integral of an odd function from $-a$ to $a$ is zero: $$=0+\int_{-2}^2\frac12\,dx=\frac12\cdot4=2$$ and the answer is 2.