Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Let:

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

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:

$$\frac{-\partial\phi(h(f(D))}{\partial{D}}=D$$

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

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

share|improve this question
5  
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
1  
^ Yes. (Well, nonnegative technically.) –  anon Apr 10 '12 at 0:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.