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Let $(\Omega, \mathfrak{F}, P)$ be a probability space such that $\forall F \in \mathfrak{F}, P(F) = 0 \ or \ 1$. Show that for all random variables X on $(\Omega, \mathfrak{F}, P)$, $\exists \ c \in \mathbb{R}$ such that P(X=c)=1.

Hint: Let $c = \inf(x:F_{X}(x)=1)$.

My attempt:

$P(X=c)$

$=P(X \in {c})$

$=P(\omega \in X^{-1}(c))$

$=0, 1 \because X^{-1}(c) \in \mathfrak{F}$

It is 1 because...c=$sup \Omega$ ?

And then I have to prove that c has to be finite. Am I then to show that $P(X=c) \neq 0 nor 1$?

Community
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BCLC
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3 Answers3

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Let $F_X(\alpha) = p\{ \omega | X(\omega) \le \alpha \}$. It is not hard to show that $F_X$ is non-decreasing, continuous from the right, and $\lim_{\alpha \to \infty} F_X(\alpha) = 1$. Furthermore, we have $F_X(\alpha) \in \{0,1\}$ for all $\alpha$. Consequently, there is some finite $\hat{\alpha}$ such that if $\alpha \ge \hat{\alpha}$, we have $F_X(\alpha) = 1$.

Let $c = \inf F_X^{-1} \{1\}$. Since $F_X$ is continuous from the right, we have $F_X(c) = 1$, and $F_X(c-{1 \over n}) = 0$ for all $n$. It follows that $p\{ \omega | X(\omega) = c \} = p \cap_{n=1}^\infty \{ \omega | c - {1 \over n} < X(\omega) \le c \} = \lim_n ( F_X(c)- F_X(c-{1 \over n})) = 1$.

Edit: changed f to F

copper.hat
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    @BCLC: I'm pretty sure that he meant to write ${0,1}$ since $P(F)\in{0,1}$ for all $F$ by assumption. In particular it holds for $F={X\leq \alpha}$. – Stefan Hansen Aug 05 '14 at 11:10
  • Thanks but is it okay to replace the first equality of the last line with $\lim_{x \to c^{-}}(F_{X}(c)- F_{X}(x))$? – BCLC Aug 05 '14 at 11:48
  • Also why must be c be a finite real number? – BCLC Aug 05 '14 at 11:52
  • @StefanHansen Did you mean $\Omega \cap (X \leq \alpha)$ ? – BCLC Aug 05 '14 at 12:44
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    What's the difference? – Stefan Hansen Aug 05 '14 at 13:10
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    @BCLC: In response to your comment above, you can replace the limit, however, since we are dealing with measures, I prefer to use explicit countable operations where possible. In this case, it amounts to the same thing. – copper.hat Aug 05 '14 at 14:30
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    @BCLC: Since $F_X(\alpha) \to 1$, and $F_X(\alpha)$ can only take the values ${0,1}$ (this was an assumption), you must have $F_X(\alpha) = 1$ for some finite $\alpha$ (otherwise the limit would be zero). – copper.hat Aug 05 '14 at 14:31
  • @stefanhansen I don't think $(-\infty, \alpha]$ is an event/meas set? – BCLC Aug 05 '14 at 14:32
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    @BCLC: For a real valued function $f$ to be a random variable, we must have that $f^{-1} (-\infty, x]$ is measurable for all $x$ (there are lots of equivalent definitions, this is typically the most convenient). – copper.hat Aug 05 '14 at 14:38
  • If a preimage of a set is measurable, the set is then measurable? – BCLC Aug 05 '14 at 15:12
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    Typically $f$ is a random variable if $f^{-1}(-\infty,x]$ is measurable for all $x$. Hence it holds on the Borel sets generated by intervals. The events for a real values random variable are the Borel sets. – copper.hat Aug 05 '14 at 15:18
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    But, to answer your question, no. Take the function $f(x) = 1$ and any non-measurable set $A \subset \mathbb{R}$. Then $f^{-1} A $ is either the empty set or the whole space, both of which are measurable. – copper.hat Aug 05 '14 at 15:21
  • Ah. Thanks. Btw for the earlier comment, why would the limit be zero? – BCLC Aug 05 '14 at 15:28
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    $F_X(\alpha)$ can only be zero or one. If it is zero for all $\alpha$, then the limit must also be zero. – copper.hat Aug 05 '14 at 15:29
  • What if $c=-\infty$? I can say that this makes $F = 1 on \mathbb{R}$ since if the lowest value (loosely speaking) that makes F become 1 is $-\infty$ which contradicts $lim_{x \to -\infty} F_{X}(x) = 0$. But that doesn't seem very precise, and I would much rather try to say this in the way of @drhab. I can't seem to justify that $\inf(x:F_{X}(x)=1) = \mathbb{R}$ if $c=-\infty$. After all $\inf A= -\infty$ <=/=> $A = \mathbb{R}$ unlike when $\inf A= \infty$ (which is necessary and sufficient that $A = \emptyset$). – BCLC Aug 07 '14 at 12:05
  • Never mind found it on Wiki. Hahaha. "In analysis the infimum or greatest lower bound of a subset S of real numbers is denoted by inf(S) and is defined to be the biggest real number that is smaller than or equal to every number in S. If no such number exists (because S is not bounded below), then we define inf(S) = −∞. If S is empty, we define inf(S) = ∞ (an extended real number line)." – BCLC Aug 07 '14 at 12:14
  • @StefanHansen Definitely no difference. Not sure what I was thinking then hahaha – BCLC Aug 05 '15 at 20:30
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CDF $F$ can only have values in $\{0,1\}$ here.

Since it is not constant it will take both values.

$F$ is non-decreasing so $c=\inf\{x\mid F(x)=1\}\in\mathbb R$.

Also $F$ is right-continuous so $F(c)=\lim_{x\rightarrow c+}F(x)=1$.

For $x<c$ we have $F(x)=0$.

Proved is now that $F$ is the characteristic function of set $[c,\infty)$ wich is the CDF of a rv that is constant at $c$.

drhab
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  • What happens if c is +/- $\infty$? I know $c \in \mathbb{R}$, but that is the 2nd question. – BCLC Aug 05 '14 at 12:14
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    That cannot happen. There at least a point $x\in\mathbb R$ with $F(x)=1$ so that $c\leq x$ and there is at least a point $y\in\mathbb R$ with $F(y)=0$ so that $y\leq c$. – drhab Aug 05 '14 at 12:16
  • And if one were to claim that $c = x = \infty$? – BCLC Aug 05 '14 at 12:23
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    $c=+\infty$ can only be achieved here if $F$ is constant and only takes value $0$. A CDF is never constant. – drhab Aug 05 '14 at 12:27
  • " c=+∞ can only be achieved here if F is constant and only takes value 0." because P(F)=0 or 1 for F={X≤α}? – BCLC Aug 05 '14 at 12:57
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    Statement $c=+\infty$ is exactly the same statement as: $\left{ x\mid F\left(x\right)=1\right} =\emptyset$ or $\forall x; F\left(x\right)<1$. Here $F\left(x\right)<1$ implies directly $F\left(x\right)=0$ since $F\left(x\right)$ can only take the values $0$ and $1$. So statement $c=+\infty$ is here exactly the same statement as: $\forall x; F\left(x\right)=0$. However a function defined like this is not a CDF. – drhab Aug 05 '14 at 13:15
  • You mean c=+∞ is exactly the same statement as: {x∣F(x)=1 or ∀x 0<F(x)<1}=∅ ? – BCLC Aug 05 '14 at 13:17
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    No, it is the same statement as $\left{ x\mid F\left(x\right)=1\right} =\emptyset$ of $\left{ x\mid F\left(x\right)<1\right} =\mathbb{R}$. In general if $A\subset\mathbb{R}$ then $\inf A=+\infty\iff A=\emptyset$. This is my last comment on this. – drhab Aug 05 '14 at 13:23
  • If by "this" you mean the $+\infty$ part, then how about $-\infty$ please? – BCLC Aug 07 '14 at 12:10
  • Never mind found it on Wiki. Hahaha. "In analysis the infimum or greatest lower bound of a subset S of real numbers is denoted by inf(S) and is defined to be the biggest real number that is smaller than or equal to every number in S. If no such number exists (because S is not bounded below), then we define inf(S) = −∞. If S is empty, we define inf(S) = ∞ (an extended real number line)." – BCLC Aug 07 '14 at 12:13
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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 $\forall \ d < c$,

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

BCLC
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