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There are these: 1, 2, 3, but I wanna try different ways.

Let $(\Omega, \mathfrak{F}, \mathbb P)$ be such probability space with each (event) $E \in \mathfrak F$ having trivial probability. Consider a random variable $X$.


An attempt:

Show that $X$ is almost surely equal to $c = \inf\{x:F_{X}(x)=1\}$, with $F_X$ as cdf (cumulative distribution function or, well, just distribution function) of $X$.

Actually $c$ is finite. If $c=\pm \infty$, then $(x:F_{X}(x)=1) = \emptyset$ or $\mathbb R$. Then $F_X(x) < 1 \ \forall x$ or $F_X(x) = 1 \ \forall x$.

By definition of such finite c, $$P(X \le c) = F_X(c) = 1 = F_X(d) \forall \ d \ge c$$

Now either $P(X \ge c) = 1$, in which case we're done or $P(X \ge c) = 0$.

Now $P(X \ge c) = 0 \to P(X < c) = 1 \to \exists \ d < c$ s.t. $P(X \le d) = 1$

But $\forall \ d < c$,

$$P(X \le d) = F_X(d) < 1 \to P(X \le d) = 0$$


Question 1: Is this correct?


Question 2: Is there perhaps an even simpler way to go about this namely without guessing what the constant $c$ is?

Actually, what I might try to do is show the $c$ exists and then later (not sure if I have to 1st show $c$ is finite for this next part) say it's the infimum (minimum actually? idk) of when the cdf reaches 1.

I think to consider any extended real $a$ and then the event $\{X=a\}$. Well, we can't have $P(X=a)=1$ for more than 1 extended real $a$. But from this, all we can say is that $P(X=a)=1$ for at most one possible $a$.

Of course it's possible we don't have $P(X=a)=1$ for any $a$, i.e. we have $P(X=a)=0$ for all $a$ such as when $X$ is a continuous(/an absolutely continuous or whatever) random variable. (But I guess this is easy to show for every discrete random variable [even if we don't assume finite] ?)

But if $X$ were continuous then $P(X \in (g,h)) = 1$ means...yeah you could probably argue a contradiction here to show continuous random variables don't exist in such a probability space (and apparently neither do binomial distributions) but in higher probability of course not all random variables are either discrete or continuous.

Well, maybe this simpler approach works for only half the battle namely the at most 1 part, but the at least 1 part really requires more.

  • Note: Simpler need not mean easier/less machinery. Could be more difficult/more machinery. Like perhaps Lebesgue's decomposition theorem can be used? What I understand of this theorem is that every random variable basically has a 'continuous part' and a 'discrete part' (Eg this and this). Maybe I can somehow show that in such a probability space every random variable is discrete and then (immediately?) conclude every random variable is a.s. constant?
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    $\begingroup$ What is the problem with considering the events $\{X \le a\}$? They should have probability $1$ for some $a$ and $0$ for other $a$ $\endgroup$
    – Henry
    Jan 17 at 10:23
  • $\begingroup$ ah thanks! so basically that's what i'm doing in the attempt? like either they're all 1 so $X=-\infty$ a.s., they're all 0 so $X = \infty$ a.s. or they're all 1 before some $a$ and all 1 after $a$ and then this 3rd case $a$ is such $\inf$ and then we somehow rule out 1st and 2nd cases? $\endgroup$
    – BCLC
    Jan 17 at 14:57
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    $\begingroup$ Argue the contrapositive. That means assuming $X$ is a random variable that is not a.s. constant. Show this means there is a number $x$ for which $\Pr(X\le x) \gt 0$ and $\Pr(X\gt x) \gt 0.$ Derive a contradiction from this. Although of course you can use "machinery," that's not an effective question because it's personal and subjective: what constitutes "machinery" depends on what you know, how you know it, and how you think of it. $\endgroup$
    – whuber
    Jan 17 at 15:34
  • $\begingroup$ @whuber i just knew there had to be something simpler. i mean hell $c=\inf\{\cdot\}$? come on! thanks! you can post as answer Edit: wait at least for question 2. what's the answer to question 1 please? $\endgroup$
    – BCLC
    Jan 17 at 21:48
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    $\begingroup$ You did a good job with #1, because you thought of proving that $c$ has to exist (that is, is finite). $\endgroup$
    – whuber
    Jan 17 at 22:17

1 Answer 1

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Argue the contrapositive. That means assuming X is a random variable that is not a.s. constant. Show this means there is a number x for which Pr(X≤x)>0 and Pr(X>x)>0. Derive a contradiction from this. Although of course you can use "machinery," that's not an effective question because it's personal and subjective: what constitutes "machinery" depends on what you know, how you know it, and how you think of it. – whuber ♦

@whuber i just knew there had to be something simpler. i mean hell c=inf{⋅}? come on! thanks! you can post as answer Edit: wait at least for question 2. what's the answer to question 1 please? – BCLC

You did a good job with #1, because you thought of proving that c has to exist (that is, is finite). – whuber ♦

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