It is said that if $S_T$ is log-normal, then its pdf is given by $$ g(x) = \frac{1}{x\sigma\sqrt{T}} \phi(\zeta(x)) $$ where $$ \zeta(x) = \frac{\log(x/S_0) - (r - \sigma^2/2)T}{\sigma \sqrt{T}} $$ I'm trying to get the score function of this by taking the derivative w.r.t $S_0$. So, when I take the derivative, I get $$ \frac{\partial g(x)}{\partial S_0} = \frac{1}{x\sigma\sqrt{T}} \frac{\partial \phi(\zeta(x))}{\partial \zeta(x)}\frac{\partial \zeta(x)}{\partial S_0} $$

But, what is weird is that the score function is given as $$ \frac{\dot{g}(x)}{g(x)} = - \zeta(x) \frac{ \partial \zeta(x)}{\partial S_0} $$ This doesn't make sense to me because it omits $\phi'(x)$ and $\phi(x)$ in the equation. Can anyone explain this to me?


What about this is odd? It is also true of the normal distribution (because $\phi '(x)=-x \phi(x)$).

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  • $\begingroup$ Is that true for any parameters of $\zeta(x)$ like $S_0$ or $\sigma$? $\endgroup$ – Astaboom May 11 '16 at 8:15

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