Transformation on a random variable Can someone please help me with formatting this question?
$Y$ is an exponential random variable with parameter $1$. Let $Z=-Y$, what is the pdf of $Z$?

Attempt:
$$\Pr(-Y< y)=\Pr(Y>-y) ,$$ 
but 
$$
\begin{align}
f_Y(y)&=\frac{d}{dy}[ \Pr(Y>-y)]\\
&=\frac{d}{dy}[1-\Pr(Y< y]\\
&=\frac{d}{dy}[\exp(y)]\\
&=\exp(y)
\end{align}
$$
 A: We want $f_Z(z)$, the density function of $z$. We will first look for the cdf of $Z$.
When $z\gt 0$, we have $\Pr(Z\le z)=\Pr(Y\ge -z)=1$. So for $z\gt 0$, we have $F_z(z)=1$, and therefore $f_Z(z)=0$.
Now let $z\lt 0$. Then 
$$F_Z(z)=\Pr(Z\le z)=\Pr(Y\ge -z)=1-F_Y(-z).$$
Differentiate with respect to $z$, using the Chain Rule, and the fact that the derivative of $F_Y(y)$ with respect to $y$ is $e^{-y}$. We get
$$f_Z(z)=-(-1)e^{-(-z)}=e^z.$$
Remark: It may be a little clearer to note that $F_Y(y)=1-e^{-y}$, so if $z\lt 0$ then
$$F_Z(z)=1-F_Y(-z)=1-(1-e^{-(-z)})=e^z,$$
 and therefore $f_Z(z)=e^z$. 
A: Using theorem of transformation variables:
$$
g_Z(z)=f_Y(y)\ |J|=f_Y(y)\ \left|\frac{dy}{dz}\right|,
$$
where $J$ is Jacobian. We have
$$
f_Y(y)=\lambda e^{-\lambda y}=e^{- y}\quad;\text{ since $\lambda=1$ and for }y\ge0
$$
and $y=-z$. Therefore
$$
g_Z(z)=e^{- y} \left|\frac{dy}{dz}\right|=e^{-(-z)} \left|\frac{d}{dz}(-z)\right|=e^z\ |-1|=\Large\color{blue}{e^z}\quad;\text{ for }z\le0.
$$
It can also be done as follows
$$
\begin{align}
\Pr[Z\le z]&=\Pr[-Y\le z]\\
&=\Pr[Y> -z]\\
&=1-\Pr[Y\le -z]\\
&=1-\left(1-e^{-(-z)}\right)\\
\Pr[Z\le z]&=e^{z}\\
\end{align}
$$
and
$$
\begin{align}
f_Z(z)&=
\frac{d}{dz}\Pr[Z\le z]\\
&=\frac{d}{dz}e^{z}\\
&=\Large\color{blue}{e^z}
\end{align}
$$
