# Showing that two random variables defined in different probability spaces have the same distribution

Let the distribution functions of $$X_n$$ be $$F_n$$ and assume that $$F_n(x)$$ is a continuous and strictly increasing function in $$x$$. Note that this implies that the inverse function of $$F_n$$ exists and is strictly increasing and continuous on $$(0,1)$$. Let $$\Omega' = (0,1)$$ be the unit interval and $$F'$$ be the Borel $$\sigma$$-algebra on $$\Omega'$$ and $$P'$$ be the uniform measure on $$\Omega'$$. Let $$Y_0, Y_1, ...$$ be random variables defined on $$\{\Omega', F', P'\}$$ such that $$Y_n(\omega') = F_n^{-1}(\omega')$$ for $$n \in \{0,1,2,...\}$$ and $$\omega' \in \Omega$$.

Show that for $$n \in \{0,1,2,...\}$$ the random variable $$Y_n$$ has the same distribution as $$X_n$$.

I don't really know how to interpret that $$Y_n(\omega') = F_n^{-1}(\omega')$$, where does this random variables map to? By definition I know that $$F_{X_n} = P(\{\omega \in \Omega; X_n(\omega) \leq x\})$$ How can I define the distribution function of $$Y_n$$?

I would appreciate any advice.

## 1 Answer

So $$Y_n$$ would be defined on $$(0,1)$$ and take values in $$(0,1)$$ since $$F_n$$ maps into $$(0,1)$$. As for its distribution, $$P'(Y_n\leq t)$$, we compute: $$P'(Y_n\leq t) =P'(\{\omega : \omega \leq F_n(t)\})=F_n(t),$$ by definition of $$P'$$.