To calculate the density of $X+Y$, if $X \sim Poisson(\lambda)$ and $Y \sim U(0;1)$ are independent, my professor wrote:
$f_Z(z)=\sum_{x=0}^{\infty}1_{(0,1)}((z-x)).e^{-\lambda}.\frac{\lambda^x}{x!}=\frac{\lambda^{\lfloor z\rfloor}.e^{-\lambda}}{\lfloor z\rfloor}$ for all $z\ge 0$ and $f_Z(z)=0$ if $z<0$.
What's the explanation for $\sum_{x=0}^{\infty}1_{(0,1)}((z-x)).e^{-\lambda}.\frac{\lambda^x}{x!}=\frac{\lambda^{\lfloor z\rfloor}.e^{-\lambda}}{\lfloor z\rfloor}$?
Best Answer
Hint: $1_{(0,1)}(z-x)=1_{\{0<z-x < 1\}}$ so
$$1_{(0,1)}(z-x)=1$$
$$\Leftrightarrow$$ $$0<z-x < 1$$ $$\Leftrightarrow$$ $$\lfloor z-x \rfloor =0$$ since x is a integer
$$\Leftrightarrow$$
$$\lfloor z \rfloor -x=0$$
$$\Leftrightarrow$$
$$\lfloor z \rfloor =x$$
so in $$\sum_{x=0}^{\infty}\left(1_{(0,1)}(z-x)\right) e^{-\lambda}\frac{\lambda^x}{x!}$$ for all $x$ , $1_{(0,1)}(z-x)=0$ except $x=\lfloor z \rfloor$