For independent random variables X ∼ Exp(1) and Y ∼ Exp(2), find the density of the random variable Z = X + Y .

My work: For any exponential distribution with parameter $\lambda$ the function is $f(x) = \lambda e^{-\lambda x}$

$f_x(x) = e^{-x}$

$f_y(x) = 2e^{-2y}$

Therefore the joint density function is: $f_{xy}(x) = 2e^{-x-2y} , x,y\ge 0$ and $0$ otherwise

After this I do not know how to $f_{X+Y}$


You can use convolution and get $$f_Z(z) = \int_0^z f_X(x)f_{Y}(-x+z)|1|\,dx.$$

I suspect that you might rather compute the cdf of $Z$ $$P(Z\leq z) = P(X+Y\leq z).$$

  • $\begingroup$ I ended up getting $f_Z(z) =2ze^{-z} for X+Y \le z$ is this correct? $\endgroup$ – FutureUIUXDeveloper Mar 9 '16 at 3:33
  • $\begingroup$ I don't think so. Note $f_z$ is the pdf, $F_Z$ is the cdf. $\endgroup$ – Em. Mar 9 '16 at 3:44
  • $\begingroup$ By the way, the answer is $-2[e^{-2z}-e^{-z}]$. $\endgroup$ – Em. Mar 9 '16 at 5:25
  • $\begingroup$ I integrated the pdf to get the cdf.I got the answer was that you just posted. Thank you! $\endgroup$ – FutureUIUXDeveloper Mar 9 '16 at 5:27
  • $\begingroup$ @Future Great. Consider up voting and giving a check mark. Good luck! $\endgroup$ – Em. Mar 9 '16 at 5:31

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