Transformation of probability distribution function with step function Given a pdf  $f(x) = \lambda e^{- \lambda x} $, what is the pdf, $g(y)$?
Where
$$
y(x) = 
\begin{cases}
0,  & \text{if $x<i$} \\
x-i, & \text{if $x>i$}
\end{cases}
$$ 
I tried writing the inverse function $x(y)$, however, $x$ cannot seem to be precisely defined when $y=0$.

We assume that $i$ is a positive constant. Let $Y$ be defined by $Y=g(X)$ where $g(x)=0$ if $x\le i$ and by $x-i$ if $x\gt i$. 
We find the cumulative distribution function $F_Y(y)$ of $Y$. For $y\lt 0$, it is clear that $F_Y(y)=0$. 
For $y=0$ we have $F_Y(y)=\Pr(X\le i)=1-e^{-\lambda i}$. 
For $y\gt 0$, we have $F_Y(y)=\Pr(Y\le y)=\Pr(X\le y+i)=1-e^{-\lambda(y+i)}$.
Note that $Y$ has neither discrete nor continuous distribution.
A similar analysis will deal with the case $i\le 0$. The details are a little different.
