1

Suppose $X\sim\mathcal{F}_X$ and denote $F_X(x)$ the cdf and $F_X^{-1}(x)$ the quantile function of $\mathcal{F}_X$ evaluated at $x$.

Now define:

$Y=\exp(X)$

and denote $\mathcal{F}_Y$ the distribution of $Y$. Since $\exp$ is a monotone increasing transformation, I (think I) know that:

$$(0)\quad F_Y(y)=F_X(\log(y))$$

my question is: can we also obtain an expression for $F^{-1}_Y$ in terms of $F^{-1}_X$? I looked online and could not find an equivalent to $(0)$ for inverse CDF's.

Sorry if the question is naive (or notation a bit off, I would certainly appreciate any comment on these): I'm not a professional mathematician and I'm trying to understand a derivation.

user42397
  • 399

2 Answers2

3

We have the rule: $$(f\circ g)^{-1}=g^{-1}\circ f^{-1}$$ wich in this case leads to: $$F_Y^{-1}(z)=\exp(F_X^{-1}(z))$$

drhab
  • 151,093
  • Thanks Sire! One question: what is $z$ in the last equation? – user42397 Jun 12 '15 at 10:54
  • 1
    You should look at is as a variable taking values in $(0,1)$. If CDF $F_Y$ is properly increasing and continuous then $x=F_Y^{-1}(z)\iff F_Y(x)=z$ for each $z\in(0,1)$. I have chosen $z$, but not for a particular reason. Often $u$ is practicized here, but that is not a must. – drhab Jun 12 '15 at 12:49
  • Ok Thanks. This is also what I thought but wanted to be sure;) – user42397 Jun 12 '15 at 13:22
  • How do $f$ and $g$ relate to $\exp$, $F_y$ and $F_x$? It looks like in the first equation you only have two functions ($f$ and $g$), while in the second there are three ($F_y, F_x, \exp$). Is $g = \exp$? If yes, then why is $g^{-1}$ not equal to $\ln$? – Confounded Aug 06 '21 at 12:13
  • @Confounded The rule is applied on $f:=F_X$ and $g:=\ln$. Note that then $F_Y=f\circ g$ so that this application gives $F_Y^{-1}=g^{-1}\circ f^{-1}. – drhab Aug 06 '21 at 19:27
2

Let $U$ be uniformly distributed on $[0,1]$. Tnen $F_Y^{-1}(U)$ is a random variable whose distribution is the same as that of $Y$. Also $F^{-1}_X(U)$ is distributed like X. So,

$$P(F^{-1}_X(U)<\ln(x))=F_X(\ln(x))=P(X<\ln(x))=P(Y<x)=P(F_Y^{-1}(U)<x).$$

zoli
  • 20,452