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Suppose I have an odd, increasing function $h$ with $h(0)=0$ and an unknown increasing function $f(D)$, $f(0)=0$.


$$\phi(h(f(D)) h(f(D)) h'(f(D)) f'(D)=D$$

where $\phi$ is the standard normal pdf

From the above equation, we can see that $D$ spans the real line and the when the RHS is $> 0$ so is the LHS and vice versa.

But, I can rewrite the above equation as:


$$\implies 0.5 D^2=-\phi(h(f(D)) $$

which does not make sense since the LHS is $> 0$ and the RHS is $< 0$

share|cite|improve this question
a) You forgot the integration constant, which could change the sign. b) Why would it indicate a mistake if you derived a contradiction? You've assumed an arbitrary equation; if you then show that it leads to a contradiction, that just disproves the assumption. – joriki Apr 9 '12 at 23:54
Hmm, so the integration constant would be $\phi(0)$ and since this is the maximum value of the standard normal pdf, the RHS is now positive...? – Greg Apr 10 '12 at 0:23
^ Yes. (Well, nonnegative technically.) – anon Apr 10 '12 at 0:26

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