Let $X \sim U([0,1])$ and $Y \sim \operatorname{Exp}(\lambda)$.
$X$ and $Y$ are independent random variables. Find the distribution of the convolution $Z=X+Y$.
I proceed like this:
$$f_z(z) = \int_{-\infty}^\infty f_x(z – y) \cdot f_Y(y) \, dy= \int_0^\infty \lambda e^{-\lambda y} \cdot I_{[0,1]} (z-y) \, dy$$
the professor told me to do the change of variable and call $t=z-x$
and do the integral from $-\infty$ to $y$
But I don't understand how to proceed … can anybody help me please??
Best Answer
First I would recommend you to move the indicator function into the integral by splitting the integral into seperate cases: $$\int_0^\infty \lambda e^{-\lambda y} \cdot \underbrace{\mathbf{1}_{[0,1]} (z-y)}_{= \mathbf{1}_{[z, z-1]}(y)} \, dy = \mathbf{1}_{[0, 1]}(z) \cdot \int_0^z \lambda e^{-\lambda y} \, dy \; + \mathbf{1}_{(1, \infty)}(z) \int_{z-1}^z \lambda e^{-\lambda y} \, dy$$ Now continue by integrating the terms.
Edit (Explanation)
Basically we are splitting up the integral in three different cases as our indicatorfunction $\mathbf{1}_{[z-1, z]}(y)$ behaves different for different $z$. So technically $$\int_0^\infty \lambda e^{-\lambda y} \cdot \mathbf{1}_{[z, z-1]}(y) \, dy = \underbrace{\mathbf{1}_{(-\infty, 0]}(z) \cdot \int_0^\infty \lambda e^{-\lambda y} \cdot \mathbf{1}_{[z, z-1]}(y) \, dy}_{= 0} \\ + \mathbf{1}_{(0, 1)}(z) \cdot \int_0^\infty \lambda e^{-\lambda y} \cdot \mathbf{1}_{[z, z-1]}(y) \, dy +\mathbf{1}_{[1, \infty)}(z) \int_0^\infty \lambda e^{-\lambda y} \cdot \mathbf{1}_{[z, z-1]}(y) \, dy$$ The first integral is $0$ because $y$ has to be $\ge 0$ and $\in [z-1, z]$ but $z < 0$.
The choice of the splitting borders is useful because for $z \in [0, 1]$ the lower bound of the integral has to respect both the $0$ from $\mathbf{1}_{[0, \infty)}(y)$ but also the $z-1$ from $\mathbf{1}_{[z-1, z]}(y)$ so we have to take the maximum of both, which is exactly $0$ if $z \in [0,1]$. For $z \in [1, \infty)$ follows $z-1 > 0$ so the maximum of the lower bound is $z-1$ not $0$.