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??

  • 1
    $\begingroup$ 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. $\endgroup$ – Michael Hardy Dec 2 '19 at 0:08
  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Dec 2 '19 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 $0$.

  • $\begingroup$ 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! $\endgroup$ – LukePower Dec 2 '19 at 14:37
  • $\begingroup$ I edited my answer, tell me if you have questions left. $\endgroup$ – claimes Dec 2 '19 at 18:02
  • $\begingroup$ Thank you very much $\endgroup$ – LukePower Dec 3 '19 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

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


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