The distribution of the product of a Bernoulli & an exponential random variable Let $X$ be an exponential random variable $f(x) = c e^{-c x} \text{ if }x >  0; 0 \text{ otherwise.}$ Let $Z$ be a Bernoulli RV with $Pr(Z=1)=0.45$ and $Pr(Z=0)=0.55$.
$X$ and $Z$ are independent.
Let $Y$ be the product of $X$ and $Z$: $Y=XZ$. What's the distribution of $Y$?
Here's my attempt:
$Pr(Y\leqslant y)=Pr(XZ\leqslant y)=Pr(XZ\leqslant y \cap Z=1)+Pr(XZ\leqslant y \cap Z=0)$
Due to independence, the above equation becomes
$$
\begin{align}
&Pr(XZ\leqslant y)Pr(Z=1)+ Pr(XZ\leqslant y)Pr(Z=0)\\
=&Pr(X\leqslant y)Pr(Z=1)+ Pr(0\leqslant y)Pr(Z=0)\\
=&0.45Pr(X\leqslant y)+0.55Pr(y\geqslant 0)
\end{align}
$$
Plug in $Pr(X\leqslant y)=F_X(y)$ and $Pr(y\geqslant 0)=1$:
$Pr(Y\leqslant y)=0.45(1-e^{-cy})+0.55=1-0.45e^{-cy}$
But this can't be right: it's not a valid cdf. What did I do wrong here? Thanks in advance!
Edit: it is a correct cdf. I had a brain freeze for a moment.
 A: I don't know why you call this an invalid CDF. 
You have a mixed random variable $Y$: 
$$Y=XZ=\begin{cases}0&,\text{ if }Z=0\\X&,\text{ if }Z=1\end{cases}$$
So the distribution function of $Y$ must be
\begin{align}
P(Y\le y)&=P(XZ\le y)
\\\\&=P(XZ\le y\mid Z=0)P(Z=0)+P(XZ\le y\mid Z=1)P(Z=1)
\\\\&=\begin{cases}P(Z=0)+P(X\le y)P(Z=1)&,\text{ if }y\ge0 \\0&,\text{ if }y<0\end{cases}
\\\\&=\begin{cases}0.55+0.45(1-e^{-cy})&,\text{ if }y\ge0\\ 0&,\text{ if }y<0\end{cases}
\end{align}
That is, 
$$F(y)=P(Y\le y)=\begin{cases}1-0.45e^{-cy}&,\text{ if }y\ge0\\ 0&,\text{ if }y<0\end{cases}$$
Taking $c=1$, the plot of $F(y)$ looks like

If you check the conditions of a valid CDF, you will see that $F$ satisfies all those conditions.
A: The first three central moments of $Y$ are as follows:
The expectation of $Y$ can be written as:
$$\displaystyle \mu_{{1}}\,  = \, E(Y) \,  = \,  0 \,  (1-p) \, + \, \int_{0}^{\infty }\!\, p \, y \, c \,{{\rm e}^{- c \,y}}\,{\rm d}y  \,= \, \frac {p}{ c }$$
The variance of $Y$ can be writt
en as:
$$\displaystyle \mu_{{2}}\,  =  \, Var(Y) \,  = \, (1-p) \,  \left(0-\frac{p}{ c }\right) ^2 +\int_{0}^{\infty }\!p \, \left( y-{\frac {p}{ c }} \right) ^{2} c \,{{\rm e}^{- c \,y}}\,{\rm d}y \,  =  \,-{\frac {p \left( p-2 \right) }{{ c }^{2}}}$$
The third central moment can be written as:
$$\displaystyle \mu_{{3}}\,  =  \, (1-p) \,  \left(0-\frac{p}{ c }\right) ^3 +\int_{0}^{\infty }\!p \, \left( y-{\frac {p}{ c }} \right) ^{3} c \,{{\rm e}^{- c \,y}}\,{\rm d}y \,  =  \,2\,{\frac {p \left( {p}^{2}-3\,p+3 \right) }{{ c }^{3}}}$$
with $p = Pr(Z=1)$ and $(1-p) = Pr(Z=0)$.
