Let $X$ be an exponentially distributed random variable with parameter $\beta>0$ and let $Y$ be a random variable with $P(Y=0)=1-p$ and $P(Y=1)=p$, where $0<p<1$. Assume that $X$ and $Y$ are independent. Find the density of $Z=X+Y$.
In the answer teacher provided the following density function for $Z$: $$ f_Z(z) = \begin{cases} \beta \cdot e^{-\beta \cdot z}(1-p) + \beta \cdot e^{-\beta \cdot (z-1)}(p) &\text{for } z \geq 1, \\ \beta \cdot e^{-\beta \cdot z}(1-p) &\text{for } 0 < z < 1, \\ 0 &\text{else}. \end{cases} $$
I don't understand why my teacher specified a different density function for $0<z<1$ and didn't use the same as the density for $z \geq 1$. Did he just make a mistake, or am I ignoring something?
Thanks in advance.
edit:
my calculations were: $$f_Z(z)=\int f_X(z-y)dP^Y(y)=f_X(z-0)P(Y=0)+f_X(z-1)P(Y=1)$$ After writing this down it finally makes sense, i think it is because $f_X(z-1)$ which is density for exponential distribution is $0$ for $z<1$. thanks.
If you observe $Z = z$ for some $0 < z < 1$, that means you could not have observed $Y = 1$, because $Z = X+Y < 1$ and $Y$ can only take on values $0$ or $1$. So you automatically know that $Y = 0$ if $Z < 1$, hence $Z = X$ if $Z < 1$. That is why the density has two cases.
On the other hand, if you observed $Z > 1$, then it could be because $X > 1$ and $Y = 0$, or it could be because $X > 0$ and $Y = 1$.
With this in mind, can you prove the teacher's answer? How would you go about such a proof?