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Some preliminary definitions:

Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space.

Let $\mathcal B(\mathbb{R})$ be the Borel $\sigma$-algebra on $\mathbb{R}$.

I define $\mathcal B (\mathbb{R})$ to be the smallest $\sigma$-algebra generated by the collection of sets $\{ (-\infty, a] : a \in \mathbb{R} \}$. See optional info

$X : \Omega \mapsto \mathbb{R}$ is a random variable if and only if for every $B \in \mathcal B(\mathbb{R})$, $X^{-1}(B) = \{ \omega \in \Omega : X(\omega) \in B\} \in \mathcal{F}$ or in other words, $X^{-1}(B)$ is measurable w.r.t probability measure $\mathbb P$.

Now, assume $X$ is a random variable as per the previous definition. Hence, every $B \in \mathcal B(\mathbb{R})$ is a measurable set and define the measure $\mathbb{P}_X(B) = \mathbb{P}(X^{-1}(B))$.

Define $Y: \mathbb{R} \mapsto [0,1]$ by $Y = F_X(X)$, where $F_X$ is CDF of $X$.

Show that $Y$ is a random variable.

My thoughts about solution:

  1. First form a $\sigma$-algebra on $[0,1]$, call it $\sigma_Y$
  2. Then show that for every $C \in \sigma_Y$, $Y^{-1}(C) \in \mathcal{B}(\mathbb{R})$ or equivalently, $Y^{-1}(C)$ is measurable w.r.t probability measure $\mathbb{P}_X$

EDIT:

consider $\sigma_Y$ to be the $\sigma$-algebra generated by the collection $\{ (0, a] : a \in (0,1) \}$

Define $F_X^{-1}(a) = \sup\{x \in \mathbb R : F_X(x) \leq a \}$ for $0 < a < 1$.

For $(0, a]$ where $0 < a < 1$, we have $Y^{-1} \left( (0, a] \right) = \left(-\infty, F_X^{-1}(a) \right ] $ which is clearly $\in \mathcal{B}(\mathbb{R})$.

Is this complete proof?

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1 Answer 1

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$F_X$ is Borel measurable because it is increasing. [$F_X^{-1}(I)$ is an interval whenever $I$ is an interval]. Hence $(F_X(X))^{-1}(B)=X^{-1}(F_X^{-1}(B))$ is measurable for any Borel set $B$.

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  • $\begingroup$ The only definition I know of measurable function is in terms of their preimages being measurable sets. If you still have time, can you read my EDIT above and help me complete the proof? Thanks. $\endgroup$
    – rims
    Feb 11, 2020 at 1:25
  • $\begingroup$ Your proof looks correct but I want to prove it with stuff I already know. $\endgroup$
    – rims
    Feb 11, 2020 at 1:33

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