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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:

graph

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!

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  • $\begingroup$ $\int f_X(x)f_Y(t-x)\,\mathrm{d}x$ 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
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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!

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

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