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Let $X\sim \text{Exponential}(\frac{1}{\lambda})$ and let $Y \sim \text{Exponential}(\frac{1}{\mu})$. Let $Z = X+Y$. I want to find the pdf of $Z$. I start by using the convolution $$f_Z(z) = \int_{-\infty}^{\infty} f_X(z-t)f_Y(t)dt$$

$$f_Z(z) = \int_{0}^{\infty} \frac{1}{\lambda}e^{-\frac{z-t}{\lambda}}\cdot \frac{1}{\mu}e^{-\frac{t}{\mu}}dt = \frac{e^{-\frac{z}{\lambda}}}{\lambda\mu}\int_{0}^{\infty}e^{t(\frac{1}{\lambda}-\frac{1}{\mu})}dt$$

But the final integral I have is undefined. I assume I have made an error somewhere. Any help is appreciated.

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  • $\begingroup$ Asked here many times before. Here are some relevant threads: math.stackexchange.com/questions/655302/…, math.stackexchange.com/questions/635443/…, math.stackexchange.com/questions/2018282/… $\endgroup$ – StubbornAtom Oct 6 '18 at 9:55
  • $\begingroup$ I had already seen the first two links you gave. The first one the parameters are the same. In the second one, there is some notation that I am unfamiliar with, and the question is several years old, so I didn't think I'd get a response posting on it. I hadn't seen the third link. From reading that, it seems my limit should be up to $z$ rather than $\infty$, is that correct? $\endgroup$ – mrnovice Oct 6 '18 at 10:06
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    $\begingroup$ Consider the support of the densities. $$z-t>0,t>0\implies 0<t<z$$ $\endgroup$ – StubbornAtom Oct 6 '18 at 10:09
  • $\begingroup$ Ah thanks, that solved my problem. Feel free to submit it as an answer if you want. $\endgroup$ – mrnovice Oct 6 '18 at 10:11

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