Give that $X_1$ and $X_2$ are random variables and the joint of $$f_{x_1x_2}(x_1,x_2) = e^{-x_1},\quad 0\le x_2\le x_1\le \infty$$

Given that $Y=X_1-X_2$ Find the pdf for Y. So far I have. $$P(Y \le y)=\iint_D e^{-x_1}\,dx_1\,dx_2.$$ $$P(Y \le y)=\int_{0} ^{\infty}\int_{0} ^{y+x_2} e^{-x_1}\,dx_1\,dx_2.$$ After I evaluate the integral I get $$e^{-y-x_2}+x_2$$ which if I try to evaluate from 0 to infinity it diverges and that doesn't seem right.

  • $\begingroup$ There is a typo in the problem. $\endgroup$ Commented Apr 10, 2014 at 1:06
  • $\begingroup$ Yeah. I typed the $x_1$ and $x_2$ in the wrong order in the inequality. But that doesn't change that it diverges which doesn't make sense since it's asking for the expectation in the next part. $\endgroup$ Commented Apr 10, 2014 at 1:37

1 Answer 1


We will use $X$ instead of $X_1$, and $Y$ instead of $X_2$. So we need a new letter for what you called $Y$. Let $W=X-Y$.

Take $w\ge 0$. We have $W\le w$ if $X\le Y+w$. Thus $$\Pr(W\le w)=\iint_D e^{-x}\,dx\,dy,$$ where $D$ is the part of the first quadrant between the lines $y=x-w$ and $y=x$. This strip is slightly unpleasant to integrate over. It is easier to note that $$\Pr(W\le w)=1-\Pr(X\gt Y+w)=1-\iint_E e^{-x}\,dx\,dy,$$ where $E$ is the part of the first quadrant below the line $y=x-w$. We have $$\iint_E e^{-x}\,dx\,dy=\int_{x=w}^\infty \left(\int_{y=0}^{x-w} 1\,dy\right)e^{-x}\,dx.$$ Integrate. The inner integral is $x-w$, so we end up with $\int_w^\infty (x-w)e^{-x}\,dx$, which is $e^{-w}$.

It follows that (for $w\gt 0$), we have $F_W(w)=1-e^{-w}$.

Added: There is a poor choice of order of integration above, which resulted in an integration by parts for the outer integral. It is better to note that $$\iint_E e^{-x}\,dx\,dy=\int_{y=0}^\infty \left(\int_{x=y+w}^{\infty} e^{-x}\,dx\right)\,dy.$$


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