1

I have revisited an old homework problem and I can't quite wrap my head around finding the correct sets. I understand the solution of my homework and the one here.

Let $L$ be the random variable $\min(X_1,\ldots,X_n)$

and $M$ be the random variable $\max(X_1,\ldots,X_n)$

I have trouble getting to:

$$P[(M \leq x) \cap (L \leq x)] = P(M \leq x) - P[(M \leq x) \cap (L \geq x)]$$

reasoning through I thought that if $M \leq x$ then $L$ must automatically also be $\leq x$ and the former contains the latter. But this would give me $$P((M \leq x) \cap (L \leq x)) = P(M \leq x)$$ which is wrong.

then I wrote it out, if I have three uniform random variables $X_1, X_2, X_3$

$$M \leq x = [(X_1 \leq x) \cap (X_2 \leq x) \cap (X_3 \leq x)]$$ and $$L \leq x = [(X_1 \leq x) \cup (X_2 \leq x) \cup (X_3 \leq x)]$$

which still makes me arrive at the wrong solution above.

Thank you for helping me reason through this, it's driving me a bit insane that I can't see it. I tried drawing it, which often helps and saves me, but this time just couldn't find a way to visualize it.

Asaf Karagila
  • 393,674
oliver
  • 675
  • 1
    It would be more usual to look at an expression with two values rather than one, such as $P[(M \leq x) \cap (L \leq y)] = P(M \leq x) - P[(M \leq x) \cap (L \gt y)]$. Your link says this as $F_{U,V}(u,v) = P(U\leq u,V\leq v)= P(V\leq v)-P(U>u,V\leq v)$ – Henry Jul 22 '21 at 09:13
  • Hi @Henry thank you. I can see a small difference, though I tried just now I still couldn't reason through the set formulation, as y would still be less than x by default? – oliver Jul 22 '21 at 09:48
  • 1
    If $y > x$ then $P[(M \leq x) \cap (L \leq y)] = P(M \leq x)-0 = P(M \leq x)$. Since this does not vary with $y$, the density will then be $0$ if $y > x$ – Henry Jul 22 '21 at 09:50
  • @Henry Oh, thanks! I think I see a way to arrive at the desired result now. Let me try again. – oliver Jul 22 '21 at 09:52
  • @Henry apologies, I think I almost have it. I'm trying to draw a venn diagram of $M$ and $L$. The Union makes up the whole space and in the intersection we should have $y=x, y >x$ and $y <x$ on the outside, i.e in $L - M \cap L, $and$ M - M \cap L$. Which almost gives me the desired result. By these two shouldn't be disjoint. How would I draw it? Thank you for your time, I apologize for being sloow. I fear I'm not seing something obvious. – oliver Jul 22 '21 at 10:10
  • I mean this is me trying to get to the general formula $A \cap B^c = A - A \cap B$ or am I completely trying the wrong thing here? – oliver Jul 22 '21 at 10:18
  • 1
    The support region for $(M,L)$ (i.e. the closure of where the density is positive) is a triangle: half a square bounded by the extremes of the original uniform distribution, and in particular the half where $M \ge L$. Your $A \cap B^c = A - A \cap B$ is correct in general, but $A \cap B$ can be $\emptyset$ if they do not overlap and in that case $A \cap B^c = A$ – Henry Jul 22 '21 at 10:18
  • Thanks Henry! Yes, I got that triangle from your initial hint above, I'm trying to get from the triangle to a Venn diagram so I may get the formula $A \cap B = A - A \cap B^c$. I think I can do it, I probably just have to ponder for a while more, also don't want to take up too much more of your time. Thanks for the additional hint as well! – oliver Jul 22 '21 at 10:24

0 Answers0