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Say $X,Y \sim U(0,1)$ be two independent uniform random variables, and $T=|X-Y|$. I would like to find the CDF of $M=\max(X,Y)$, $L=\min(X,Y)$, and $T$. I find the CDF of $M$ and $L$ as $$P(M\leq t ) = P(X\leq t)P(Y \leq t) =t^2$$ $$P(L \leq t) = 1-P(X\geq t ) P(Y\geq t) = 1-(1-t)^2$$ To find the CDF of $T$, I draw a rectangle with unit length and width, and compute the area within the region $|X-Y| \leq t$, which turns out to be $1-(1-t)^2$.

My question is how come the CDF of $L$ and $T$ are the same when one is the minimum of two uniform r.vs and the other is the absolute difference of two uniform r.vs? Is there something wrong with my computation? Thanks.

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Your calculations are correct! Just for the sake of completeness, you have to write

$$F_M(t)=t^2\mathbb{1}_{(0;1)}(t)+\mathbb{1}_{[1;\infty)}(t)$$

...and similar notation for the other CDF's. This because the CDF is defined over all $\mathbb{R}$

Why the law of L and T are the same? As you did to find $F_T$ do the same drawing to find $F_L$ and you will realize that the integration area is the same

enter image description here

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    $\begingroup$ Cool explanation. Thanks a ton. $\endgroup$ – ForumWhiner Sep 19 '20 at 10:20
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Imagine a line segment $OA$ of unit length.

On this line segment imagine two points $X,Y$ each at a distance of $x, y$ from point $O$ respectively.

These two points divide $OA$ into three smaller segments, whose length is equal to $min(x,y)$; $|x-y|$ and $1 - max (x,y)$.

From symmetry it can be seen that the three values should have same Expectation values.

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  • $\begingroup$ $max(X;Y)$ has not the same expectation of the other rv's. This is self evident if you look at the drawing in my answer. $\endgroup$ – tommik Sep 19 '20 at 8:24
  • $\begingroup$ @tommik yes, it's a typo. I'll correct it. $\endgroup$ – xax Sep 19 '20 at 8:45
  • $\begingroup$ Sorry I don't get why all three segments should have same expectations. I find $E[M]=2/3$ and $E[L]=1/3$. Am I correct? $\endgroup$ – ForumWhiner Sep 19 '20 at 10:19
  • $\begingroup$ Symmetry. The only rule you have is the sum of lengths of three segments should be one. And this rule is same to all three segments, which means their Expectation should be equal and be equal to one. $\endgroup$ – xax Sep 19 '20 at 19:02
  • $\begingroup$ @ForumWhiner Yes, E[L] = E[T] = E[1 - M] = 1/3 $\endgroup$ – xax Sep 19 '20 at 19:03

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