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Let $F$ be a distribution function and $X$ a random variable which is uniformly distributed on the interval $(0,1)$. Let $F^{-1}$ be the inverse defined by $F^{-1}(y)=\inf\{x:F(x)\geq y\}$. How do I show that the random variable $Y=F^{-1}(X)$ has distribution function $F$?

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Let $x\in(0,1)$.

$F^{-1}\left(x\right):=\inf\left\{ y\mid x\leq F\left(y\right)\right\} $ tells us directly that $x\leq F\left(y\right)\Rightarrow F^{-1}\left(x\right)\leq y$.

Since CDF $F$ is continuous from the right we also have $F^{-1}\left(x\right)\in\left\{ y\mid x\leq F\left(y\right)\right\} $.

This tells us that conversely $F^{-1}\left(x\right)\leq y\Rightarrow x\leq F\left(y\right)$.

So we have: $$F^{-1}\left(x\right)\leq y\iff x\leq F\left(y\right)$$ leading to:

$$F^{-1}\left(X\right)\leq y\iff X\leq F\left(y\right)$$

and: $$P\left(F^{-1}\left(X\right)\leq y\right)=P\left(X\leq F\left(y\right)\right)=F\left(y\right)$$

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