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?



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 at 1:45
  • $\begingroup$ Uh, ... Why? @Tan $\endgroup$ Jan 29 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 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 at 2:50
  • $\begingroup$ @Graham Kemp great answer, thank you. $\endgroup$
    – lorenzo
    Jan 29 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.