Let $Y$ be a random variable formed from $X$ as $Y=X\mathsf 1_{\{X>0\}} + 2X^2\mathsf 1_{\{X\leqslant 0\}}$.
(a) Compute the cdf of $Y$, $F_Y(y)$, in terms of the cdf of $X$, $F_X(x)$.
(b) Find the pdf of $Y$, $f_Y(y)$, when $f_X(x) = \frac1{\sqrt{2\pi}} e^{-x^2/2}$.
I did the first part in the following way:
For $X>0$:
$$F_Y(y) = \mathbb P(Y\leqslant y) = \mathbb P(X\leqslant Y) = F_X(y).$$
For $X\leqslant 0$:
$$
F_Y(y) = \mathbb P(Y\leqslant y) = \mathbb P(2X^2\leqslant y) = F_X\left(\sqrt{y/2})\right).
$$
Therefore,
$$
F_Y(y) = F_X(y)\mathsf 1_{(0,\infty)}(x) + F_X\left(\sqrt{y/2}\right)\mathsf 1_{(-\infty)}(y).
$$
For the second part I wasn't able to solve as I don't know whether the first part is correct or wrong.
Best Answer
Fairly close. You have noticed that for positive $y$, the event of $\{Y\leqslant y\}$ is the event of $\{-\sqrt{y/2~}\leqslant X\leqslant 0\}$ or $\{0< X \leqslant y\}$, so $$F_Y(y) = (F_X(y)-F_X(-\surd(y/2)))~\mathbf 1_{y\geqslant 0}$$
[Note: be careful about case sensitivity. The value $y$ and random variable $Y$ should not be confused.]
Now for the second, evaluate the derivative and substitute:$$f_Y(y)=\dfrac{\mathrm d ~~}{\mathrm d y}F_Y(y)$$