Proof of backward heat equation and brownian martingales

Question

I am currently studying the relation between brownian motion and PDE's in the textbook of Karatzas and Shreve: Brownian motion and stochastic calculus. When reading the proof that if $v$ satisfies the backward heat equation $\partial_{t}v + \frac{\partial_{x}^{2}v}{2}=0$, that then $M_t = v(t,W_{t})$ is a martingale (where $W$ is the standard brownian motion) I got confused. First of all: to prove that $M_{t}$ is a martingale Itô's lemma is invoked on the stopped $M_{t + (s \wedge T_{a} \wedge T_{b})}$ where $T_{a}, T_{b}$ are the hitting times of $W$ for $a, b$ with $a<b$ and $0 \leq s \leq T-t$ where $T$ is the time on which we define the terminal conditions as usual for the backward heat equation. As this is done Karatzas and Shreve obtain.

$M_{t + (s \wedge T_{a} \wedge T_{b})} = v(t,W_{0}) + \int_{0}^{s \wedge T_{a} \wedge T_{b}}\partial_{x}v(t+q, W_{q})dW_{q} + \int_{0}^{s \wedge T_{a} \wedge T_{b}} (\partial_{t} + \frac{\partial_{x}^{2}}{2})v(t+q, W_{q})dq$

Now I have two questions:

1. Where does the $\partial_{t}$ term come from in the second integral above? Indeed, it is very desirable as it causes the whole integral to vanish due to $v$ satisfying the backward heat equation. However: it is not present in Itô's lemma, one would only have the $ \frac{\partial_{x}^{2}}{2}v(t+q, W_{q})$ part, what's going on there?
2. Is it truly necessary to work with the stopped version $M_{t + (s \wedge T_{a} \wedge T_{b})}$? I understand that it allows one to immediately say that the first integral $\int_{0}^{s \wedge T_{a} \wedge T_{b}}\partial_{x}v(t+q, W_{q})dW_{q}$ its expectation is then zero due to boundedness of $\partial_x v$ on $[a,b]$, however the whole argument one then constructs using fatou's lemma and the strong markov property to first get that $v(t,x) \geq \mathbb{E}^{x}[v(t+s,W_{s})] $ and later $v(t+s_{1}, W_{s_{1}}) \geq \mathbb{E}^{x}[v(t+s_{2},W_{s_{2}})| \mathcal{F}_{s_{2}}] $ with later one again proving everything is an equality is frankly slightly annoying

Answer 1

I will try to answer the two questions

1. I think you have to consider the Ito's formula for the process $s\mapsto (s+t,W_{s+t})$ and so it will give ( we denote $W_{s+t}=W'_{s}$ for convenience) $$M_{t+(s\wedge T_a \wedge T_b)}=v(t+(s\wedge T_a \wedge T_b),W_{t+(s+\wedge T_a \wedge T_b)})=v(t,W_t)+\underbrace{\int_{0}^{s+\wedge T_a \wedge T_b}\partial_{t}(v)(t+s,W'_s)ds}_{s\mapsto s+t \text{ part}}+\underbrace{\int_{0}^{s+\wedge T_a \wedge T_b}\partial_{x}(v)(t+s,W_s)dW'_{s}+\frac{1}{2}\partial^{2}_{x}(v)(s+t,W'_s)}_{s\mapsto W_{s+t} \text{ part}}ds$$

and, so it gives $$M_{t+(s\wedge T_a \wedge T_b)}=v(t,W_t)+\int_{0}^{s\wedge T_a \wedge T_b} \partial_{x}(v)(t+s,W'_s)dW'_{s}+\int_{0}^{s+\wedge T_a \wedge T_b} (\frac{\partial_{x}^{2}}{2}+\partial_{t})(v)(t+s,W'_s)ds$$

2. I think stopping times are necessary that ensure that the stochastic integration is well defined. I do not see another way to prove the following except this fatou lemma's tricks.

I hope it is clear enough