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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 $F$?

JimmyP
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1 Answers1

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

drhab
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    Could you please explain 2 points: 1. Why "CDF F is continuous from the right" leads to $F^{-1}(x) \in $ {$ {y| x \leq F(y)}$} 2. Why this fact leads to the statement "This tells us that conversely $F^{-1}(x) \leq y \Rightarrow x \leq F(y)$". Thank you very much for your help! I'm struggling to understand this – InTheSearchForKnowledge Mar 10 '21 at 21:37
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    @InTheSearchForKnowledge For a fixed $x\in\left(0,1\right)$ let $A_{x}:=\left{ y\in\mathbb{R}\mid x\leq F\left(y\right)\right} $. Then $A_{x}\neq\varnothing$ and $A_{x}\neq\mathbb{R}$ and $y\in A_{x}\implies\left[y,\infty\right)\subseteq A_{x}$ for every $y\in\mathbb{R}$. That leaves two options for $A_{x}$. It must be of shape $\left[r,\infty\right)$ or of shape $\left(r,\infty\right)$ for some $r\in\mathbb{R}$. Here $r=\inf A_{x}$. – drhab Mar 11 '21 at 08:42
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    ....Now note that we have $F\left(y\right)\geq x$ whenever $y>r$. That implies that $\lim_{y\downarrow r}F\left(y\right)\geq x$. Here the monotonicity of $F$ ensures us that this limit exists. The fact that $F$ is continuous from the right tells us that $\lim_{y\downarrow r}F\left(y\right)=F\left(r\right)$ so actually we found that $F\left(r\right)\geq x$ or equivalently that $r\in A_{x}$. So it cannot be that $A_{x}=\left(r,\infty\right)$ and we conclude that $A_{x}=\left[r,\infty\right)$. Then $r\leq y\implies x\leq F\left(y\right)$. Finally be aware that $r=F^{-1}(x)$. – drhab Mar 11 '21 at 08:42