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Suppose that $F$ is a distribution function that is absolutely continuous with respect to Lebesgue measure on $\mathbb{R}$ with density $f$. Let $F^{-1}$ be the associated quantile function and assume that $F^{-1}(t)$ is a singleton for all $t \in (0,1)$ (e.g. as in a normal distribution).

I am interested in properties of the derivative of $F^{-1}$, call it $G$. From calculus we expect that $G(t) = 1/f(F^{-1}(t))$ as long as $f(F^{-1}(t)) \neq 0$.

However, $f$ is only unique a.e. with respect to Lebesgue measure. So supposedly I could choose $f$ at $F^{-1}(t)$ to make $G(t)$ any value including non-existent by choosing $f(F^{-1}(t)) = 0$. This seems absurd. Can you explain to my why it is not absurd and/or clear up my confusion regarding measure-theoretic probability that is leading me to this conclusion?

evencoil
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1 Answers1

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From calculus we expect that $G(t)=1/f(F^{-1}(t))$ as long as $f(F^{−1}(t))\ne 0$.

We don't "expect" this; we know (from the inverse function theorem) that the derivative of $F^{-1}$ is equal to $1/F'(F^{-1}(t))$ as long as $F'(F^{−1}(t))\ne 0$.

This is what the calculus theorem says, and there is no $f$ in the above statement. Sure, if you denote the derivative $F'$ by $f$, you can write the above formula as $1/f(F^{-1}(t))$. But if $f$ is some other function, you cannot do that.

When people say "redefining a function on a set of measure zero does not change anything", they mean for the purpose of integration. Differentiation is another thing.

user127096
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  • I'm not sure that I understand your exposition, so let me rephrase what I think your point is and maybe you can tell me if I got it: $F$ is unique, therefore $F'$ is unique (assuming it exists), therefore $G'(t)$ is unique assuming that $F'(F^{-1}(t)) \neq 0$. And so, even though $F$ may admit many densities $f$ that are different on sets of measure zero, all this means is that any given $f$ that it admits need not satisfy $f(F^{-1}(t)) = F'(F^{-1}(t)) = G'(t)$. But that does not make $G'(t)$ non-unique or not well-defined. – evencoil Feb 23 '14 at 03:11
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    @evencoil Yes, that's what I meant. Put another way: the inverse function theorem requires $f=F'$ which is a stronger requirement than $F=\int f$. – user127096 Feb 23 '14 at 03:14
  • That clarification is also helpful. Thanks! – evencoil Feb 23 '14 at 14:43