Probability density function of the negative of a random variable (Exercise 4.1.2 from Grimmett and Stirzaker) Find the density function of $Y = a X$, where $a > 0$, in terms of the density function of $X$. Show that the continuous random variables $X$ and $-X$ have the same distribution function if and only if $f_X(x) = f_X(-x)$ for all $x \in \mathbb{R}$.
I managed to prove the first part but is the second part correct? (I'm not a 100% sure about the forward direction.)
($\rightarrow$ direction)
Suppose that $X$ and $-X$ have the same distribution function. Then this implies that for any $x \in \mathbb{R}$
\begin{align}
P(X \leq x) &= P(-X \leq x), \\
\implies P(X \leq x) &= P(X \geq -x), \\
\implies \int_{-\infty}^x f_X(u)\,du &= \int_{-x}^{\infty} f_X(u)\,du.
\end{align}
Now let $u=-v$. Then
\begin{align}
\int_{-\infty}^{x} f_X(u)\,du &= \int_{x}^{-\infty}-f_X(-v)\,dv, \\
\implies \int_{-\infty}^{x} f_X(u)\,du &= \int_{-\infty}^{x}f_X(-v)\,dv, \\
\implies f_X(x) &= f_X(-x).
\end{align}
($\leftarrow$ direction)
Suppose that $f_X(x) = f_X(-x)$. Reversing the argument above gives for any $x \in \mathbb{R}$
\begin{align}
f_X(x) &= f_X(-x), \\
\implies \int_{-\infty}^{x}f_X(u)\,du &= \int_{-\infty}^{x}f_X(-u)\,du.
\end{align}
Now let $u = -v$. Then,
\begin{align}
\int_{-\infty}^{x}f_X(u)\,du &= \int_{\infty}^{-x}-f_X(v)\,dv,\\
\implies \int_{-\infty}^{x}f_X(u)\,du &= \int_{-x}^{\infty}f_X(v)\,dv,\\
\implies P(X \leq x) &= P(X \geq -x), \\
\implies P(X \leq x) &= P(-X \leq x). \hspace{2cm} \square
\end{align}
 A: Although I think your answer is correct, I find it easier to think of it in the following way:
Once you've figured out the density function of $Y=aX$, which should by $f_{Y}(y)=|\frac{1}{a}| f_{X}(x=\frac{y}{a})$. You could use the result.
($\rightarrow$ direction)
Assume X and -X has the same distribution function, $Y_1=X$ and $Y_2=-X$ would have the same distribution. According to the pdf we derived above, the pdf for $Y_1$ is $f_{Y_1}(y)=f_{X}(x=y)$, and the pdf for $Y_2$ is $f_{Y_2}(y)=f_{X}(x=-y)$. Since $Y_1$ and $Y_2$ have the same distribution, their pdf should be the same.
So we have $f_{Y_1}(y)=f_{Y_2}(y)$, which means $f_{X}(x=y)=f_{X}(x=-y)$. We can rewrite this as $f_{X}(x)=f_{X}(-x)$.
($\leftarrow$ direction)
Assume $f_{X}(x)=f_{X}(-x)$. Let $a=-1$, $Y_1=X$, and $Y_2=-X$.
The pdf of $Y_1$ is $f_{Y_1}(y)=f_{X}(x=y)=f_{X}(x)$, and the pdf for $Y_2$ is $f_{Y_2}(y)=f_{X}(x=-y)=f_{X}(-x)$. Since $f_{X}(x)=f_{X}(-x)$, $f_{Y_1}(y)=f_{Y_2}(y)$. According to the uniqueness of pdf, $Y_1$ and $Y_2$ have the same distribution, which means $X$ and $-X$ have the same distribution.
A: $$F_{-X}(x) = P(-X \leq x) = P(X \geq -x) = 1 - P(X \leq -x)$$
Hence $f_{-X}(x) = f_{X}(-x)$. 
If $X$ and $-X$ have the same distribution function then 
$f_{-X}(x) = f_{X}(x)$, whence the claim follows.
Conversely, if $f_{X}(-x) = f_{X}(x)$ for all $x$, then
by substituting $u = -x$,
$$
\begin{align}
P(-X \leq y) & = P(X \geq -y) \\
 &  \\ 
 & = \int_{-y}^{\infty} f_X(x) \: \text{dx} \\
 &  \\ 
 & = \int_{-\infty}^{y} f_X(-u) \: \text{du} \\
 &  \\ 
 & = \int_{-\infty}^{y} f_X(u) \: \text{du} \\
 &  \\ 
 & = P(X \leq y) \\
\end{align}
$$
whence $X$ and $-X$ have the same distribution function.
