By Ito's formula we have that for any suitable function $v(t,x)$,
$$ v(T, X_T) = v(t,X_t) + \int_t^T\left( v_s(s, X_s)+ b(s, X_s)v_x(s,X_s)+\frac{1}{2}\sigma^2(s, X_s)v_{xx}(s, X_s) \right)ds+\int_t^T\sigma(s, X_s)v_x(s, X_s)dW_s $$
where $X_t$ satisfies the SDE,
$$ dX_t=b(t, X_t)dt + \sigma(t, X_t)dW_t $$
Taking expectations conditional on $X_t =x$ we have,
$$ \mathbb E_t(v(T, X_T))-v(t, x)=\int_t^T\mathbb E_t\left( v_s(s, X_s)+ b(s, X_s)v_x(s,X_s)+\frac{1}{2}\sigma^2(s, X_s)v_{xx}(s, X_s) \right)ds $$
Let $v(t, x)=0$ and $v(s, x) \to 0 $ as $ s \to T $, uniformly in x. Then, $$ 0=\int_t^T\mathbb E_t\left( v_s(s, X_s)+ b(s, X_s)v_x(s,X_s)+\frac{1}{2}\sigma^2(s, X_s)v_{xx}(s, X_s) \right)ds $$ Now write in terms of the transition density and integrate by parts twice to get that the density $p(t, x;T, y )$ satisfies,
$$ \frac{\partial p}{\partial T} + \frac{\partial (bp)}{\partial y} - \frac{1}{2}\frac{\partial^2(\sigma^2 p)}{\partial y} = 0 $$
I'm not sure how proceed with the integration by parts step, which would start with something like this?
$$ \int_t^T\mathbb \int_{-\infty}^{\infty}\left( v_s(s, X_s)+ b(s, X_s)v_x(s,X_s)+\frac{1}{2}\sigma^2(s, X_s)v_{xx}(s, X_s) \right)p(s, x)\ dx\ ds $$
How does partial differentials with respect to T and y come out from integrating this, can someone kindly enlighten ?
Thanks!
I think one of the ways to proceed is as follows...
We consider each of the three terms inside the integrals separately (suppressing arguments) and we have,
$$ \begin{align} I_1 & = \int_t^T\mathbb \int_{-\infty}^{\infty} v_sp\ dx\ ds\\ & = \int_{-\infty}^{\infty} \left[ \left. vp\right|_{t}^{T}-\int_t^T vp_s\ ds\right] dx\\ I_2 & = \int_t^T\mathbb \int_{-\infty}^{\infty} bv_xp\ dx\ ds\\ & = \int_{t}^{T} \left[ \left. bvp\right|_{-\infty}^{\infty}-\int_{-\infty}^\infty v\frac{\partial{(bp)}}{\partial x}\ dx\right] ds\\ I_3 & = \int_t^T\mathbb \int_{-\infty}^{\infty} \frac{1}{2}\sigma^2v_{xx}p\ dx\ ds\\ & = \int_{t}^{T} \left[ \left. \frac{1}{2}\sigma^2v_xp\right|_{-\infty}^{\infty}-\int_{-\infty}^\infty \frac{1}{2}v_x\frac{\partial{(\sigma^2p)}}{\partial x}\ dx\right] ds\\ & = \int_{t}^{T} \left[ \left. \frac{1}{2}\sigma^2v_xp\right|_{-\infty}^{\infty}-\left[ \left. \frac{1}{2}v\frac{\partial{(\sigma^2p)}}{\partial x}\right|_{-\infty}^{\infty}\ -\int_{-\infty}^\infty \frac{1}{2}v\frac{\partial^2{(\sigma^2p)}}{\partial x^2} dx\right]\right] ds \end{align} $$
If by suitable choice of $v(t, x)$ we choose the class of functions with compact support ( and with earlier conditions ), all the terms with $\left. \right|_{-\infty}^{\infty}$ in $I_1, I_2, I_3$ vanishes and we get,
$$ \begin{align} 0 &=\int_t^T \int_{-\infty}^{\infty}vp_s\ dx\ ds + \int_t^T \int_{-\infty}^\infty v\frac{\partial{(bp)}}{\partial x}\ dx\ ds - \int_t^T \int_{-\infty}^\infty \frac{1}{2}v\frac{\partial^2{(\sigma^2p)}}{\partial x^2} dx\ ds\\ & = \int_t^T \int_{-\infty}^{\infty}v \left[ \frac{\partial p}{\partial s }+ \frac{\partial{(bp)}}{\partial x}-\frac{1}{2}\frac{\partial^2{(\sigma^2p)}}{\partial x^2}\right] dx\ ds \end{align} $$
Since $v(t, x)$ is arbitrary it means that we must have ( after renaming variables ),
$$ \frac{\partial{p}}{\partial T} + \frac{\partial{(bp)}}{\partial y}-\frac{1}{2}\frac{\partial^2{(\sigma^2p)}}{\partial y^2} = 0 $$
which gives the Kolmogorov Forward Equation