Is it probability density function too?

Question

Is it true that if $f(x)$ is probability density function then $f(-x)$ is probability density function too?

I don't even know how to start solving this problem. I tried to use a property of $\int_{-\infty}^{\infty} f(x) = 1$, but this way of proof didn't look right for me as I don't make any progress. Thank You!

Answer 1

Another proof : if a random variable $X$ has $f(x)$ as its pdf, then $-X$ has $f(-x)$ for its pdf, proving that $f(-x)$ is a pdf "per se".

Answer 2

Suppose $f(x)$ is a probability density function.

Since $f(x)\ge 0$ for all $x\in\mathbb{R}$, the same is true of $f(-x)$.

To verify that $f(-x)$ is a probability density function, simply note that \begin{align*} &\int_{-\infty}^{\infty} f(-x)\,dx\\[4pt] =\;&\int_{\infty}^{-\infty} f(u)\,(-1)\,du&&\text{[using the substitution $x=-u$]}\\[4pt] =\;&-\int_{\infty}^{-\infty} f(u)\,du\\[4pt] =\;&\int_{-\infty}^{\infty} f(u)\,du\\[4pt] =\;&\,1\\[4pt] \end{align*}