I am trying to understand the following:

Let $X$ and $Y$ be independent and exponential random variables with parameter $1$, so $f_X(x) =e^{-x} $ and $f_Y(y) =e^{-y} $. Let $Z=X+Y, X<Y$ and $Z=X-Y, X\geq Y$. Determine the pdf of $Z$.

The solution to the exercise first calculates the cdf $F_Z(z)=P(Z\leq z, X<Y)+P(Z\leq z, X\geq Y)$ and then takes its derivative to obtain $f_Z(z)=\frac{e^{-z}}{2}(1+z)$.

This is fine but it is a bit long so I would like to do the exercise by computing the convolution to obtain $f_Z(z)$ directly, so I did: $f_Z(z)=\int_{0}^{z}e^{-x}e^{-(z-x)}dx=ze^{-z}$ for the $X<Y$ part but when doing the $X\geq Y$ part I get $f_Z(z)=\int_{0}^{x}e^{-z-y}e^{-y}dy=\frac{1}{2}(e^{-z}-e^{-z-2x})$ which is wrong.

Could someone please explain to me what I am doing wrong?



1 Answer 1


Note: $X$ and $Y$ are non-negative random variables.

We have $Z=X+Y$ (ie $Y=Z-X$), exactly when $Z>X$, which is only supported when also $X<Z-X$, ie $2X<Z$.

We have $Z=X-Y$ (ie $Y=X-Z$), exactly when $Z\leqslant X$, which isn't further restricted by the support.

Thus we have the support joint pdf for $X$ and $Z$, and this gives the bounds for your integration.

$$\begin{align}f_{X,Z}(x,z) &= f_{X,Y}(x,z-x)\cdot\mathbf 1_{0\leqslant 2x<z}+f_{X,Y}(x,x-z)\cdot\mathbf 1_{0\leqslant z\leqslant x}\\[1ex] &=\mathrm e^{-z}\cdot\mathbf 1_{0\leqslant 2x<z}+\mathrm e^{-2x+z}\cdot\mathbf 1_{0\leqslant z\leqslant x}\\[2ex] f_{Z}(z)&=\left( \int_0^{z/2}\mathrm e^{-z}\mathrm d x+\int_z^\infty \mathrm e^{-2x+z}\mathrm d x\right)\cdot \mathbf 1_{0\leqslant z}\\[1ex]&=\tfrac {z+1}2\mathrm e^{-z}\cdot\mathbf 1_{0\leqslant z}\end{align}$$

  • $\begingroup$ It is a bit confusing to write $f_{X,Z}(x,z)$ in stead of $f_{Z|X}(z|x)f_{X}(x)$. $\endgroup$
    – Tan
    Jan 29, 2021 at 1:45
  • $\begingroup$ Uh, ... Why? @Tan $\endgroup$ Jan 29, 2021 at 1:51
  • $\begingroup$ Because $f_{Z|X}(z|x)=f_{Y+x|X}(Y+x=z|x)=f_{Y|X}(z-x|x)=f_{Y}(z-x)$, where independence between $X$ and $Y$ is demonstrated clearly. However, $f_{X,Z}(x,z)$ becomes $f_{X,Y}...$ is a bit confusing. $\endgroup$
    – Tan
    Jan 29, 2021 at 2:06
  • $\begingroup$ When $Z=X+Y$ then clearly $f_{\small X,Z}(x,z) = f_{\small X,X+Y}(x,z)= f_{\small X,Y}(x,z-x)$; whether or not we have any independence. $\tiny\text{where, of course, such pdf exist}$ The fact of independence is only required to say $f_{\small X,Y}(x,z-x)=f_{\small X}(x)\,f_{\small Y}(z-x)$ . $\endgroup$ Jan 29, 2021 at 2:50
  • $\begingroup$ @Graham Kemp great answer, thank you. $\endgroup$
    – lorenzo
    Jan 29, 2021 at 9:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .