0

Seems like I'm rather stuck on this issue, which should be pretty simple.
Say I have n i.i.d variables that distribute uniformly. Denote $X$ as their maximum, $Y$ as their minimum.
I know that $_{,}(,)=(−1)(−)^{−2}$, and I need to find $P(X+Y\leq z)$. It should be done by a double integral over the required domain, but I am afraid that even finding the domain gives me a hard time.
We require $0\leq y\leq x\leq 1 \wedge x+y\leq z$ and so it should be: $$ \int\limits_{0}^{1}\int\limits_{y}^{z-y}(−1)(−)^{−2}dxdy=\frac{z^n}{2} $$ or $$ \int\limits_{0}^{z/2}\int\limits_{0}^{x}(−1)(−)^{−2}dydx + \int\limits_{z/2}^{z}\int\limits_{0}^{z-x}(−1)(−)^{−2}dydx = 1-\frac{1}{2}(2-z)^n $$ Neither one of this is the right answer by simply checking for the condition that $0\leq P(X+Y\leq z) \leq 1$. Yet I believe that I have minor issue in finding the right domain. I'd be thankful for any kind of help.

Thanks!

EDIT: U1…Un are taken from uniform distribution.

Lior Pollak
  • 133
  • What is the distribution of your variables? You wrote that they distribute normally (so I assume you meant normal distribution), but the link you provided refers to uniform distribution. – mowzorn Jan 03 '23 at 08:43
  • Correct, thank! They are uniform. – Lior Pollak Jan 03 '23 at 09:57

1 Answers1

1

We already know that if $X$ is the maximum and $Y$ is the minimum, then the joint PDF is $$f_{X,Y}(x,y) = n(n-1)(x-y)^{n-2}\cdot\textbf{1}_{0\leq y\leq x\leq 1}(x,y),$$ where $\textbf{1}_A(x)$ is the indicator function. We are interested in finding the CDF of $X+Y$ (in other words $P(X+Y\leq z)$). Notice that $0\leq X+Y\leq 2$. Let's consider 2 cases:

  • $z\in [0,1)$: In this case our domain is $\{0\leq x\leq \frac{z}{2},0\leq y\leq x\}\cup\{\frac{z}{2}\leq x\leq z,0\leq y\leq z-x\}$. So $$\begin{split}P(X+Y\leq z)&=\int_0^\frac{z}{2}\int_0^xn(n-1)(x-y)^{n-2}dy\ dx+\\ &+\int_\frac{z}{2}^z\int_0^{z-x}n(n-1)(x-y)^{n-2}dy\ dx,\end{split}$$ which evaluates to $\frac{z^n}{2}$.
  • $z\in[1,2]$: Now our domain is diffirent, because $x\leq 1$. So we need to integrate $f_{X,Y}(x,y)$ over $\{0\leq x\leq \frac{z}{2},0\leq y\leq x\}\cup\{\frac{z}{2}\leq x\leq 1,0\leq y\leq z-x\}$. So $$\begin{split}P(X+Y\leq z)&=\int_0^\frac{z}{2}\int_0^xn(n-1)(x-y)^{n-2}dy\ dx+\\ &+\int_\frac{z}{2}^1\int_0^{z-x}n(n-1)(x-y)^{n-2}dy\ dx,\end{split}$$ which evaluates to $1-\frac{(2-z)^n}{2}$ In the end we get $$P(X+Y\leq z)=\left\{\begin{array}{ll} 0, & z < 0\\ \frac{z^n}{2},& z\in[0,1)\\ 1-\frac{(2-z)^n}{2},&z\in[1,2)\\ 1,& z\geq 2. \end{array}\right.$$
mowzorn
  • 472
  • Thanks for the informative answer! Do you think there’s a way to deduce this from basic probability (i.e. without integrals)? – Lior Pollak Jan 03 '23 at 10:44
  • 1
    Maybe there is, but I think that this is the easiest way to go about it. I'd be impressed if there's an easier way to prove this without integrals. – mowzorn Jan 03 '23 at 10:47
  • Could you perhaps elaborate on how did you find the domain of integration please? – Lior Pollak Jan 16 '23 at 20:01
  • I mean, I drew the integration zone and it made sense. But is there a more analytic way to find this out? And to verify that the domain is indeed correct? – Lior Pollak Jan 16 '23 at 20:17
  • I think that drawing the area is your best bet. You simply need to 'swap' the axes and go from there. – mowzorn Jan 16 '23 at 21:10