1

Let $X\perp Y\sim Exp(\lambda)$.

- Find the distribution of $Z=\frac{X}{X+Y}$

- Find the distribution of $W=U-V$ where $U=max(X,Y)$ and $V=min(X,Y)$.

- Is it possible determine if $U\perp V$ are indipendent without making calculations? If yes, why?

For the first point we have $F_Z(z)=\mathbb{P}(Z\leq z)=\mathbb{P}(Y\leq \frac{1-Z}{Z}X)=\int_{0}^{+\infty}[\int_{\frac{1-Z}{Z}X}^{+\infty}… dy]dx=z$ $\Rightarrow Z\sim U(0,1)$.

For the second point we have:

if… $\left.\begin{matrix} Y<X\rightarrow U=max(X,Y)=X;V=min(X,Y)=Y\Rightarrow U-V=X-Y \\ Y>X\rightarrow U=max(X,Y)=Y;V=min(X,Y)=X\Rightarrow U-V=Y-X \end{matrix}\right\}$ $\Leftrightarrow max(X,Y)-min(X,Y)=|X-Y|$, …so:

$F_W(w)=\mathbb{P}(W\leq w)=\mathbb{P}(max(X,Y)-min(X,Y)\leq w)=\mathbb{P}(|X-Y|\leq w)=\mathbb{P}(Y\leq X+w,Y<X)+\mathbb{P}(Y\geq X-w,Y>X)$

Now i've trouble to fix the extremes of integration. I've tried with…

1) $\int_{0}^{+\infty}[\int_{0}^{x+w}… dy]dx+\int_{0}^{+\infty}[\int_{x-w}^{+\infty}… dy]dx=1-\frac{1}{2}e^{-\lambda w}+\frac{1}{2}e^{\lambda w}$;

2) $\mathbb{P}(|X-Y|\leq w)=\mathbb{P}(-w\leq X-Y \leq w)=2[\mathbb{P}(X-Y\leq w)-\mathbb{P}(X-Y\leq 0)]=2[\mathbb{P}(Y\geq X-w)-\mathbb{P}(Y\geq X)]=2[\int_{0}^{+\infty}[\int_{x-w}^{+\infty}… dy]dx-\int_{0}^{+\infty}[\int_{x}^{+\infty}… dy]dx]=e^{\lambda w}-1$.

…but the correct result is $1-e^{-\lambda w}\Rightarrow W\sim Exp(\lambda)$.

Every time I'm in front of a sum or a difference in module, I don't understand how proceed. Could you please explain me, in the specific case but (more important) in general, how do you calculate the sum/difference in module?

Thanks in advance for any answer!

Marco Pittella
  • 565
  • $U$ and $V$ are not independent, rather $U-V$ and $V$ are independent. See https://math.stackexchange.com/questions/2240822/let-u-operatornamemin-x-y-and-v-operatornamemax-x-y-show-that, https://math.stackexchange.com/questions/3394576/properties-of-exponential-random-variables-memoryless-property-and-sums-differe. – StubbornAtom Nov 02 '19 at 11:58
  • Since $\max(X,Y) \ge \min(X,Y)$ you have $U \ge V$. Meanwhile $V$ is unbounded above, so $U $ and $V$ cannot be independent – Henry Nov 02 '19 at 13:28
  • What does $X\perp Y$ denote? – Math1000 Nov 02 '19 at 15:04
  • @StubbornAtom Thanks for your answer. I've watched the links that you sent me but they don't explain how to set the extremes of integration with a sum or a difference in module. This is my problem. – Marco Pittella Nov 02 '19 at 16:25
  • @Henry Thanks for your help. Would you care to clarify? Both $U$ and $V$ are defined in $\mathbb{R}^+$... – Marco Pittella Nov 02 '19 at 16:28
  • @Math1000 $X\perp Y$ means that $X$ and $Y$ are independent. – callculus42 Nov 02 '19 at 17:15
  • So why would you ask "determine if $U\perp V$ are independent"? Writing $U\perp V$ is already implying that they are dependent (which they are not, as has been pointed out earlier. – Math1000 Nov 02 '19 at 17:35
  • 1
    @MarcoPittella the support of $U$ is $[0,\infty)$, while conditionally given $V=v$ the support of $U$ is $[v,\infty)$ since $U \ge V$. So $U$ and $V$ are not independent – Henry Nov 02 '19 at 17:59
  • @Henry Thanks again, now it's clear! Could you please help me with the second point? Which is the reasoning to do for setting the correct extremes of integration starting from $\mathbb{P}(Y\leq X+w,Y<X)+\mathbb{P}(Y\geq X-w,Y>X)$? – Marco Pittella Nov 03 '19 at 09:52

0 Answers0