# Prove $X(\omega) = \sup\{y \in \mathbb{R}: F(y) < \omega\}$ is a random variable.

Let F be a distribution function. On $(\Omega, \mathfrak{F}, P)=((0,1), \mathfrak{B}(0,1),\lambda)$ where $\lambda$ denotes Lebesgue measure.

Define X: $\Omega \to \mathbb{R}$ by $X(\omega) = \sup(y \in \mathbb{R}: F(y) < \omega)$.

1 Show that $\forall x \in \mathbb{R}, (\omega: X(\omega) \leq x) = (\omega: \omega \leq F(x))$

2 Show that X is a RV in $(\Omega, \mathfrak{F}, P)$ and that $F_X = F$.

1 LHS = $(\omega: X(\omega) \leq x)$

= $(\omega: X(\omega) \in (-\infty, x])$

= $(\omega: \sup(y \in \mathbb{R}: F(y) < \omega) \in (\infty, x])$

RHS = $(\omega: \omega \leq F(x))$

= $(\omega: \omega \in (-\infty, F(x)])$

= $(\omega: \omega \in (-\infty, P(X^{-1}( \ (-\infty,x] \ ))))$

= $(\omega: \omega \in (-\infty, P(LHS)))$

• A distribution function is in principle a function on $\mathbb R$, or more general on $\mathbb R^n$. A measurable function $X:\Omega\rightarrow\mathbb R$ will induce a distribution function $F_X$. – drhab Aug 5 '14 at 12:21
• You should start here just with: 'Let $F$ be a distribution function.' Not a distribution function on $(\Omega,...)$. – drhab Aug 5 '14 at 12:33
• @drhab 1 Edited 3rd statement. sorry for confusion. 2 Do you mean that the existence of a distribution function implies the existence of a random variable inducing it? I think this is precisely what is being attempted to be shown. – BCLC Aug 5 '14 at 12:34
• @drhab I edited exactly as it was given to me. "Let F...function. On...measure." The second statement "On...measure." is a fragment. – BCLC Aug 5 '14 at 12:35
• To be shown is: if $F$ is some distribution function, then we can always construct some rv $X$ such that its distribution function $F_X$ will satisfy $F_X=F$. A start has been made with this construction and you are asked to prove that the constructed $X$ here satisfies this condition. – drhab Aug 5 '14 at 12:36

If $X\left(\omega\right)\leq x<z$ then if follows immediately from the definition $X\left(\omega\right):=\sup\left\{ y\mid F\left(y\right)<\omega\right\}$ that $F\left(z\right)\geq\omega$.

The fact that $F$ is continuous on the right then allows the conclusion that also $F\left(x\right)=\lim_{z\rightarrow x+}F\left(z\right)\geq\omega$.

In conversely $F\left(x\right)\geq\omega$ then $X\left(\omega\right)=\sup\left\{ y\mid F\left(y\right)<\omega\right\} \leq x$ because $F$ is non-decreasing.

Proved is now that $$X\left(\omega\right)\leq x\iff\omega\leq F\left(x\right)$$

This equation gives us the second part:

$X:\Omega\rightarrow\mathbb{R}$ is a measurable function and this with: $$F_{X}\left(x\right)=P\left\{ \omega\mid X\left(\omega\right)\leq x\right\} =\lambda\left\{ \omega\in\left(0,1\right)\mid\omega\leq F\left(x\right)\right\} =\lambda\left((0,F\left(x\right)]\right)=F\left(x\right)$$

• THANK YOU SO MUCH. Question on last line: Why is it $F(x)]$ and not $F(x))$ ? I was thinking that there was a possibility that $F(x)=1$ so we would have to exclude it. I know it does not matter for Lebesgue measure so the answer is the same regardless. I was just thinking that the LHS of the last inequality should be piecewise or something: ] if $F(x) < 1$ and ) if $F(x) \leq 1$ – BCLC Aug 6 '14 at 4:22
• Mwah.. I think that $\Omega=\left(0,1\right)$ is a somewhat stupid choice here. To avoid confusion on this it is better to choose for $\Omega=\left[0,1\right]$. Then everythings goes fine too and issues like you mention do not arise. – drhab Aug 6 '14 at 6:59
• Thanks. I'll ask my prof. Fingers crossed figuring that out isn't part of the homework. If it is though...piecewise then? – BCLC Aug 6 '14 at 9:29
• If you wish, but I wouldn't choose for that. Elegance is lost. You can also it do with a last line that simply leaves out $=\lambda\left((0,F\left(x\right)]\right)$. Then the disputable interval $(0,F\left(x\right)]$ is kept out of sight. – drhab Aug 6 '14 at 9:43