Problem with Convolution of two rv with different distribution

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??

• You have $f_z(z)$ and $f_x(z-y)$ where you should have $f_Z(z)$ and $f_X(z-y).$ Without these distinctions one cannot understand things like $\Pr(Z\le z)$ or some things that arise naturally in proofs. Dec 2, 2019 at 0:08
• I did some copy-editing of you MathJax code, and in particular I changed -$\infty$ to $-\infty,$ and that one example may be the best, among short and simple examples, of the reason why ALL of the mathematical notation should be within MathJax, not outside of it. Dec 2, 2019 at 0:11

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 $$0$$.

• Ok, I understood why you wrote $${= I_{[z, z-1]}(y)}$$ but I didn't understand why you used those extremes. Why did you integrate from 0 to z, and then from z-1 to z? What's the logic behind this? Thank You very much! Dec 2, 2019 at 14:37
• I edited my answer, tell me if you have questions left. Dec 2, 2019 at 18:02
• Thank you very much Dec 3, 2019 at 9:56

Use old "rule of thumb" for computing convolution of two functions:

1) reflect one function around vertical axis;

2) move the reflected function by $$t$$; the overlap integral is the value of convolution at point $$t$$

Back to the problem; let us reflect the density of $$X$$ random variable; the reflected density is $$\sim U[-1,0]$$. It is clear that if we move the reflected density left, the overlap is zero, so the convolution is zero for $$t<0$$. For $$0\leq t\leq 1$$ the convolution integral is

$$\int_0^t\lambda e^{-\lambda x}dx$$

and for $$t>1$$ the convolution integral is

$$\int_{t-1}^t\lambda e^{-\lambda x}dx$$