The textbook I am using, using convolution in order to find the CDF of the $X+Y$ when $X\sim exp(\alpha)$ $Y\sim exp(\beta)$, and X and Y are are independent.

However, I have no background with convolutions at all (although my lecturer assumes I do, but that's another issue), so I am trying to figure this out without using convolutions, and I would really really appreciate if someone can point out what is wrong in my approach. of course I will also do my best to learn the convolution thing myself, but I am still very curious about my mistake here.


Since both X and Y are non negative, I am only interested in the following region:


The numbers are arbitrary of course.

So my integration of the region is:

$$ F_Y(t)=P(Z\leq t)=P(X+Y\leq t)=\int_{0}^{t} \int_{0}^{t-x} \alpha e^{-\alpha x} \beta e^{-\beta y} dy dx $$

The result of the integration is this Wolfarm

While the convolution result is this Wolfarm

I really don't know what is so wrong about my work, can someone please help? Thanks!

  • $\begingroup$ $\int f_X(x)f_Y(t-x)\,\mathrm{d}x$ gives the pdf $f_{X+Y}(t)$, and $\mathbb{P}(X+Y\leq t)=F_{X+Y}(t)$ is the cdf. $\endgroup$ – user10354138 Feb 16 at 17:18

In the convolution you have calculated the density of $Z = X + Y$ and in your result you essentially calculate the CDF of $Z = X + Y$. Differentiating your result with respect to $t$ leads indeed to the same result!

Since you are new in this field. Let me explain what happens in a discrete setting. For example, assume $X,Y\geq 0$ take on integer values and you want to know the distribution of $Z = X+Y$. Then what you generally do is write it as follows $$ P(Z=z) = P(X+Y = z) = P(X=z - Y) \\ = P((X=z,Y=0)\ \text{or}\ (X=z-1,Y=1)\ \text{or}\ \cdots\ \text{or}\ (X=0,Y=z)) \\ = \sum_{i=0}^{z}P(X=z-i,Y=i) = \sum_{i=0}^{z}P(X=z-i)P(Y=i). $$ On the left, we have the convolution of $X$ and $Y$. In a continuous setting, this would generalise to $$ f_Z(z) = \int_0^z f_X(z-y)f_Y(y)\,\mathrm{d} y. $$

I hope this makes it a bit more clear for you!

  • $\begingroup$ Gosh I am dumb. It is the density, not the CDF. Sorry you wasted your time $\endgroup$ – superuser123 Feb 16 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.