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If $X$ and $Y$ are independent and identically distributed RVs with pdfs $f_X(x)=e^{-x}U(x)$ and $f_Y(y)=e^{-y}U(y)$, find the PDF of $Z$ if:

a. $Z=X-Y$

b. $Z=\min(X,Y)/\max(X,Y)$

$U$ represents the unit step function.

For the first part, I tried looking at the range of values that $y$ could be, and I got $y\geq x-z$ since $x-y \leq z$ for $F_Z(z)$. I set up my CDF integral to be

$$\int_0^\infty\int_0^{z+y}e^{-x-y} dx dy$$

and got $F_Z(z)= 1-e^{-z}/2$. I differentiated this to get my PDF, which is $e^{-z}/2$. This answer is incorrect, and I'm not sure what I did wrong.

For the second part, I again tried looking at the range of values that $y$ could be, and I got $y \leq zx$ where $0 \leq z \leq 1$. I set up my CDF integral to be $$\int_0^\infty \int_0^{y/z}e^{-x-y} dx dy$$

and got $F_Z(z)= 1-\frac {z} {z+1}$. I differentiated this to get $\frac {-1} {(z+1)^2}$. This answer is also incorrect. I'm not sure what to do on either of these two problems at this point.

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2 Answers 2

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Just expanding drhab's answer.

For the first case you should consider when, for any given $z\in \Bbb R$, when $Y > X-z$. When $z>0$ you have the situation depicted below.

enter image description here

Thus, for $z>0$, \begin{eqnarray} F_Z(z) &=& \int_0^{+\infty}\int_0^{z+y} e^{-x-y} dx dy=\\ &=&1-\frac12 e^{-z}. \end{eqnarray}

For $z<0$ you can look at the figure below.

enter image description here

Then you can compute, in such interval, e.g. as \begin{eqnarray} F_Z(z) &=& \int_{0}^{+\infty}\int_{x-z}^{+\infty} e^{-x-y} dy dx=\\ &=&\frac12 e^{z}. \end{eqnarray}

Differentiation yields $$\boxed{f_Z(z) = \frac12 e^{-|z|}}.$$


As for the second question, consider that, if $X>Y$ you need to consider the event $Y<zX$; whereas, if $X<Y$, your event becomes $Y>\frac1{z}X$.

It is thus immediate to conclude that, for $z>1$, $$F_Z(z) = 1.$$

For $0<z<1$, consider then the shaded area in the figure below.

enter image description here

Therefore, for $0<z<1$, \begin{eqnarray} F_Z(z) &=&1- \int_0^{+\infty} \int_{zx}^{\frac{x}{z}} e^{-x-y}dy dx=\\ &=&\frac{2z}{1+z}. \end{eqnarray}

And, finally, by differentiation $$\boxed{f_Z(z) = \frac2{(1+z)^2}}, \ \ 0<z<1.$$

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    $\begingroup$ Ah, so I was just was missing half of the integral on both steps. These answers make sense. Thanks! $\endgroup$
    – SD'Anc
    Commented Oct 7, 2019 at 12:49
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a) Your CDF integral is almost correct.

It should be:$$P\left(X-Y\leq z\right)=\int_{0}^{\infty}\int_{0}^{\infty}\left[x\leq y+z\right]e^{-x-y}dxdy=\int_{0}^{\infty}\int_{0}^{\max\left(0,y+z\right)}e^{-x-y}dxdy$$For calculation you must discern the cases $z\geq0$ and $z<0$.

I haven't looked at b) but there is a good chance that you made a similar mistake there.

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  • $\begingroup$ I understand that I need to discern the two cases, but I'm not sure I understand the new integral bound min(0,y+z). If the min was 0 on the upper bound, wouldn't you just be integrating 0? $\endgroup$
    – SD'Anc
    Commented Oct 7, 2019 at 12:29
  • $\begingroup$ @SD'Anc You are wright. There was a mistake in my answer (sorry). $\min$ must be $\max$. I repaired. $\endgroup$
    – drhab
    Commented Oct 7, 2019 at 13:42

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