Context:
We consider a geometric brownian motion: $$dS_t = rS_t dt + \sigma S_t dW_t$$ with $S_0 = x$.
And we are interested in the function $$ u(x) = E(g(S_T)) = \int_{\mathbb{R}^+} g(s) p(s,x) ds $$
Given that the stochastic process is log-normal we have: $$p(s,x) = {1\over{s\sqrt{2\pi \sigma^{2}T}}}\exp\left(-{\left(\log(s/x)-\left( r - {1\over{2}}\sigma^{2}\right)T \right)^{2}\over{2\sigma^{2}T}} \right)$$ $$S_t = x e^{(r-\frac{\sigma^2}{2})t + \sigma W_t}$$ Problem:
We are interested in computing the first and second derivative of $u$ of the form: $$ u'(x) = E(g(S_T) \delta(T,W_T)) $$ $$ u''(x) = E(g(S_T) \gamma(T,W_T)) $$
where we need to find $\delta$ and $\gamma$.
Using the likelihood ratio method, we have:
$$ u'(x) = \int g(s) \partial_x p(s,x) ds = \int g(s) \partial_x \log(p(s,x)) p(s,x) ds = E \left( g(S_T) \partial_x \log(p(S_T, x))\right) $$ $$ \begin{align*} \partial_x \log p(S_T,x) &= \partial_x \log \left( {1\over{S_T\sqrt{2\pi \sigma^{2}T}}}\exp\left(-{\left(\log(S_T/x)-\left( r - {1\over{2}}\sigma^{2}\right)T \right)^{2}\over{2\sigma^{2}T}} \right) \right)\\ &= - \frac{1}{2\sigma^2T}\partial_x \left({\left(\log(S_T/x)-\left( r - {1\over{2}}\sigma^{2}\right)T \right)^{2}} \right)\\ &= - \frac{\log(S_T/x)-\left( r - {1\over{2}}\sigma^{2}\right)}{\sigma^2T} \partial_x \log(S_T/x)\\ &= \frac{\log(S_T/x)-\left( r - {1\over{2}}\sigma^{2}\right)}{x \sigma^2T}\\ &= \frac{W_T}{x \sigma T}\end{align*}$$
This result is right and can be verified, for example using Malliavin calculus.
Now I have an issue when computing the second derivative. I tried to be smart and avoid bruteforcing it:
$$ u''(x) = \int g(s) \partial_{xx} p(s,x) ds = \int g(s) \frac{\partial_{xx}p(s,x)}{p(s,x)} p(s,x) ds = E\left(g(S_T)\frac{\partial_{xx}p(S_T,x)}{p(S_T,x)}\right)$$
$$ \begin{align*} \partial_{xx}\log(p(S_T,x)) &=\partial_x (\partial_x \log(p(S_T,x))) = \partial_x \frac{\partial_x p(S_T,x)}{p(S_T,x)} \\ &= \frac{\partial_{xx}p(S_T,x)}{p(S_T,x)} - \left(\frac{\partial_x p(S_T,x)}{p(S_T,x)}\right)^2 \\ \end{align*} $$
So, $$ \begin{align*} \frac{\partial_{xx}p(S_T,x)}{p(S_T,x)} &= \partial_{xx}\log(p(S_T,x)) + \left(\partial_x \log(p(S_T,x))\right)^2\\ &= \partial_x \left(\frac{W_T}{x \sigma T}\right) + \left(\frac{W_T}{x \sigma T}\right)^2\\ &= -\frac{W_T}{x^2 \sigma T} + \frac{W_T^2}{x^2 \sigma^2 T^2}\\ \end{align*} $$
However, this is not right. The right answer is:
$$\gamma=-\frac{W_T}{x^2\sigma T} -\frac{1}{x^2\sigma^2T}+ \frac{W_T^2}{x^2\sigma^2 T^2}$$
Again, this could be proven using Malliavin calculus.
Can someone tell me the issue with my calculations?
We in fact have that $W_T=\psi(S_T,x)=\frac{\ln(S_T/x)-(r-\sigma^2/2)T}{\sigma}$. So note $$\frac{d}{dx}\ln p(s,x)=\frac{\psi(s,x)}{x\sigma T},\qquad\frac{d\psi}{dx}(s,x)=-\frac{1}{x\sigma}$$ and we conclude: $$\begin{aligned}\frac{1}{p(s,x)}\frac{d^2p}{dx^2}(s,x)&=\frac{d}{dx}\bigg(\frac{\psi(s,x)}{x\sigma T}\bigg)+\bigg(\frac{\psi(s,x)}{x\sigma T}\bigg)^2\\ &=-\frac{\psi(s,x)}{x^2\sigma T}-\frac{1}{x^2\sigma^2 T}+\bigg(\frac{\psi(s,x)}{x\sigma T}\bigg)^2 \end{aligned}$$