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Find the PDF of $Z = X + Y$, where $X$ and $Y$ are independent exponential random variables with common parameter $\lambda$.

My approach

I do not want to use the convolution formula.

The joint probability density, $ f_{X,Y}(x,y) = \lambda e^{-\lambda x} \lambda e^{-\lambda y}$.

The CDF of $Z$, $F_{Z}(z) = \mathbf{P}(Z \leq z) = \mathbf{P}(X + Y \leq z) = \int\limits_{0}^{z}\int\limits_{0}^{z-x}f_{X,Y}(x,y)dydx = \int\limits_{0}^{z} \lambda e^{-\lambda x}\int\limits_{0}^{z-x}\lambda e^{-\lambda y}dydx =\int\limits_{0}^{z} \lambda e^{-\lambda x}(1 - e^{-\lambda(z-x)})dx$.

Clearly, this doesn't tally with the convolution formula. Where is my mistake?

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  • $\begingroup$ There is no mistake yet. Just go on. You cannot expect the integrand in the last expression to be the PDF yet. This because it contains (next to variable $x$) variable $z$. $\endgroup$ – drhab Mar 6 '16 at 13:41
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You are very close. Persevere in working out:

$$\int_{0}^{z}\lambda e^{-\lambda x}(1-e^{-\lambda(z-x)})dx=\int_{0}^{z}\lambda e^{-\lambda x}dx-\int_{0}^{z}\lambda e^{-\lambda z}dx=\int_{0}^{z}\lambda e^{-\lambda x}dx-z\lambda e^{-\lambda z}$$

To find the PDF we take the derivative w.r.t $z$ and arrive at:

$$\lambda e^{-\lambda z}-(\lambda e^{-\lambda z}-z\lambda^{2}e^{-\lambda z})=z\lambda^{2}e^{-\lambda z}$$

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  • $\begingroup$ yes, the mistake turned out to be stopping 1 step early. $\endgroup$ – buzaku Mar 7 '16 at 3:05
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Just use the transformation $X=U-V$ and $Y=V$ and find the joint pdf of $U$ and $V$ $$f_{U,V}(u,v)=\begin{cases}\lambda e^{-\lambda(u-v)}.\lambda e^{-\lambda v}.1&\text{if $0<u-v<\infty\,\,\,\,$ $0<v<\infty\;\;$ $0<u<\infty$}\\ 0, & \text{otherwise} \end{cases} $$ just integrate with respect to $v$ t get the pdf of $U$ i.e $$f_U(u)=\int_{0}^u\lambda^2e^{-\lambda u}dv\,\,\,\,\,\,\,,0<u<\infty$$ Thus u get $$f_U(u)=\lambda^2ue^{-\lambda u}\,\,\,\,\,\,\,\,0<u<\infty$$

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$$ F_Z(z)=\int_0^z dz'f_Z(z')=\lambda^2\int_0^z dz'\int_0^\infty dx \int_0^\infty dy e^{-\lambda (x+y)}\delta(z'-(x+y)) $$ $$ \lambda^2\int_0^z dz'\int_0^\infty dy e^{-\lambda z'}\Theta(z'-y)=\lambda^2 \int_0^z dz' e^{-\lambda z'}\int_0^{z'}dy=\lambda^2\int_0^z dz' e^{-\lambda z'}z'=1-e^{-\lambda z} (\lambda z+1)\ , $$ where $\Theta(x)$ is the Heaviside step function, and $z\geq 0$. This perfectly agrees with your final formula. I don't see any mistake.

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