Let $Z \sim Z(0, 1)$ be a standard normal random variable. Find
the PDF of $|Z|$.
My approach:
I let $Y = |Z|$, found the CDF and then took the first derivative to get the PDF.
For $0 \leq y \leq 1$, the CDF is given by:
$$F_Y(y) = P(Y \leq y)$$
$$= P(-y \leq Z \leq y)$$
$$= F_Z(y) – F_Z(-y) \tag{$*$}$$
Hence the PDF is given by,
$$f_Y(y) = \frac{d}{dy}(F_Y(y))$$
$$= \frac{d}{dy}(F_Z(y) – F_Z(-y)) \; \text{by }(*)$$
$$= \frac{d}{dy}(F_Z(y)) – (\frac{d}{dy}F_Z(-y))$$
$$=1\cdot f_Z(y)-(-1\cdot f_Z(-y))$$
$$=f_Z(y)+f_Z(-y))$$
The PDF of a standard normal distribution is given by $f_X(x) = \frac{1}{\sqrt{2\pi}} e^\frac{-x^2}{2}$ so we have,
$$=\frac{1}{\sqrt{2\pi}} e^\frac{-y^2}{2}+\frac{1}{\sqrt{2\pi}} e^\frac{y^2}{2}$$
I am not too sure if this is the correct answer and approach to this question so any help would be really appreciated.
Best Answer
You almost had the right answer, but you need to be careful when evaluating $f_Z(-y)$.
Since $(-y)^2 = y^2$, we have $f_Z(-y) = \dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{(-y)^2}{2}} = \dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{y^2}{2}}$.
Thus, your final answer ends up being $f_Y(y) = \dfrac{2}{\sqrt{2\pi}}e^{-\dfrac{y^2}{2}}$ for $y \ge 0$.