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Hello

I am trying to follow the derivation of the Generalized Cornish Fisher Expansion from this paper.

I dont see how the author goes from equation (21) to (22).

For one to be able to do this consider an arbitrarily differentiable function $G(v)$.

Make the change of variable $v = \Phi(u)$ which implies $u=\Phi^{-1}(v)$. Note that $\frac{du}{dv}=\frac{1}{\phi(u)}$. Let $D_x=\frac{d}{dx}$ the differential operator.

Now equation (21) would imply (22) if

$D_v^r[G(v)]= \left(D_u\frac{\text{du}}{\text{dv}}\right){}^r=D_u^r\left(\frac{\text{du}}{\text{dv}}\right)^r$

Is this true? and if so how?

It is clearly true for $r=1$ by the chain rule. But for $r=2,3,...$ I don't see how.

Going from (22) to (23) is also a mystery to me.

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    $\begingroup$ I took a quick look, and it appears to me that (22) is obtained by a direct substitution of the change of variable formulas into (21) without any use of any rules or formulas from calculus. Please note the order in which everything is written and the parentheses in the equation. Going from (22) to (23) is a little more elaborate but nothing fancy is going on there, either. I think you need to go through this slowly and together in person with someone else. $\endgroup$
    – Deane Yang
    Commented May 15, 2011 at 2:45
  • $\begingroup$ @Deane Yang, Thanks ... the second equality is certainly a bit more tricky but with your advice i see it now $\endgroup$
    – AUK1939
    Commented May 18, 2011 at 8:14

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