# Trying to find the CDF of $X+Y$ when $X\sim exp(\alpha)$ and $Y \sim exp(\beta)$ (independent) without convolution, but it doesn't seem to work

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.

So:

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!

• $\int f_X(x)f_Y(t-x)\,\mathrm{d}x$ gives the pdf $f_{X+Y}(t)$, and $\mathbb{P}(X+Y\leq t)=F_{X+Y}(t)$ is the cdf. – 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 $$f_Z(z) = \int_0^z f_X(z-y)f_Y(y)\,\mathrm{d} y.$$