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I was looking at Determine density of $\min(X,Y)$ and $\max(X,Y)$ for independently uniform distributed variables

There's a very detailed answer, but while I was doing the same exercise by myself I was thinking in a similar but slightly different way, which lead me to a different answer. Here I'll share with you my work so I can understand the why of the difference.

Also, I am making focus on the value of the CDF, and not yet working on the joint pdf.

Given $(Z,W)=(min(X,Y), max(X,Y))$ for $X,Y$ iid Uniform(0,1), find the CDF.

I thought of : \begin{equation} P(Z\leq a,W\leq b)=\\ P(min(X,Y)\leq a,max(X,Y)\leq b)=\\ P( \{X\leq a, a \leq Y\leq b\} \cup \{a \leq X \leq b, Y\leq a\} )=(1)\\ P(\{X\leq a, a \leq Y\leq b\}) + P(\{a \leq X \leq b, Y\leq a\})= (2)\\ P(X\leq a) P( a \leq Y\leq b) + P(a \leq X \leq b) P(Y\leq a)= (3)\\ F_X(a)( F_Y(b)-F_Y(a)) + ( F_X(b)-F_X(a)) F_Y(a)=\\ F(a)(F(b)-F(a))+ (F(b)-F(a)) F(a)= (4)\\ 2F(a)(F(b)-F(a)) =\\ 2F(a)F(b) - 2F(a)^2 \end{equation} Where $a$ would be the minimum, $b$ the maximum and I use:

  1. {X is the minimum, Y the maximum} union {Y is the minimum, X the maximum}
  2. Probability of the union of disjoint set
  3. Independence of $X$ and $Y$
  4. The fact that $F_X$ = $F_Y$ = $F$

This result is different to $F_{X,Y}(a,b) + F_{X,Y}(b,a) - F_{X,Y}(a,a)$, as given in the original question (there are differences in variable naming, but the setting is essentially the same)

I notice that if I use the idea (4) about equality of CDFs, it would be really close: with the answer to the original question being $- F(a)^2$ compared to $- 2F(a)^2$ in the way I am doing it.

I did the drawings of the regions considered and I can clearly see the missing "infinite square" $(X<a, Y<a)$, but my argument is that if $X$ is the minimum, then $Y$ should be integrated over a region where $Y$ would be the maximum (and vice-versa), hence the difference in the proposed sets considered.

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  • $\begingroup$ In (1), what about the case $X\leq a, Y\leq a$? $\endgroup$
    – gunes
    Aug 13 at 10:52
  • $\begingroup$ Since I derive the $X$,$Y$ limits from $Z$,$W$ I would say that if $Z< \leq a$ and $W \leq b$ it is a necessary condition that $a \leq b$ by definition of $max$ and $min$, therefore that case should not occur. I think that the problem in my argument is the following: Taking $min(5,50)$, $max(5,50)$ I can say that $5 \leq 100$ and that does not imply that $50 \geq 100$. In general: saying that the $min$ of two numbers is less than another number, doesn't mean that the $max$ of the two is greater than that number. And that is the flaw of my previous reasoning. $\endgroup$
    – nico_so
    Aug 13 at 12:40
  • $\begingroup$ I don't understand, if $X\leq a$ and $Y\leq a$, then automatically, $\min(X,Y)\leq a$ and $\max(X,Y)\leq a\leq b$. But, this case is not covered. $\endgroup$
    – gunes
    Aug 13 at 12:42
  • $\begingroup$ Exactly, that is what I was trying to say in my comment. You put it more clearly in mathematical terms, but you're right. That was the flaw of my argument, since I was saying that $min(X,Y)\leq a$ implies $a \leq max(X,Y)$, which now I see it is not necessary the case. $\endgroup$
    – nico_so
    Aug 13 at 12:47
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Your

$$P( \{X\leq a, a \leq Y\leq b\} \cup \{a \leq X \leq b, Y\leq a\} )$$

Is excluding the cases $X<Y<a$ and $X<Y<a$, when both $X$ and $Y$ are smaller than $a$.

This is because your expression with either $a \leq Y\leq b$ or $a \leq X\leq b$ you make the restriction that at least one of the variables must be above $a$.

When you incorporate these cases with both $X\leq a, Y \leq a$, then your end result will be increased by $P(X \leq a,Y \leq a) = F(a)^2$. You will get a result of $2F(a)F(b)-F(a)^2$ instead of $2F(a)F(b)-2F(a)^2$

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