If $X$ be a random variable with cdf $F$, and $F$ is continuous and strictly increasing on $\mathbb{R}$ how do you show $F$ has a well-defined inverse function $G : (0,1) \to \mathbb{R}$, which is strictly increasing.


closed as off-topic by jameselmore, Davide Giraudo, Harish Chandra Rajpoot, Em., Yagna Patel Jan 26 '16 at 20:10

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To put you on the right track: for $u\in(0,1)$ "define" $G(u):=\inf\{x\in\mathbb R\mid F(x)\geq u\}$.

Now it remains to check:

  • Is this indeed a well-defined function $\mathbb (0,1)\to\mathbb R$?
  • Is this function strictly increasing?
  • Does this function serve as an inverse of $F$?

Can you do that?

Alternatively you can show that for every $u\in(0,1)$ there is a unique $x_u\in\mathbb R$ such that $F(x_u)=u$. In order to prove that $F(x)=u$ for some $x\in\mathbb R$ make use of the continuity of $F$ and apply the intermediate value theorem. Its uniqueness follows from the fact that $F$ is strictly increasing. Now define $G$ by prescription $u\mapsto x_u$.


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