Absolute Value of a normally distributed random variable. Below is a problem I did. My answer matches the back of the book, but some how, I do not have confidence in my answer. I am hoping somebody here can confirm that my solution is right.
Problem:
Let $X$ have the normal density $N(0,\sigma^2)$. Find the density of $Y =|X|$.
Answer:
Clearly the density function for $Y$ is $0$ when $Y < 0$ because Y cannot be negative. Here is the density function for $X$.
$$ f_x = \frac{e^ {-\frac{x^2}{2 \sigma^2} } }{\sigma \sqrt{2 \pi}} $$
Now, we are interested in $P(Y <= y_0)$ where $Y = |X|$. Call the density function we are look for $f_y(x)$
\begin{align*}
P(Y <= y_0) &= \int_{-y_0}^{y_0} f_x(x) \, dx \\
P(Y <= y_0) &= \int_{-y_0}^{y_0} \frac{e^ {-\frac{x^2}{2\sigma^2} } }{\sigma \sqrt{2 \pi}} \, dx \\
P(Y <= y_0) &= \int_{0}^{y_0} \frac{2e^ {-\frac{x^2}{2\sigma^2} } }{\sigma \sqrt{2 \pi}} \, dx \\
\end{align*}
Now to find $f_y(x)$ we differentiate the last equation. This gives us:
$$ f_y(x) =  \frac{2e^ {-\frac{x^2}{2\sigma^2} } }{\sigma \sqrt{2 \pi}} $$

As has already been stated in the comments, your solution is correct.
