
> 
> Let $u: \mathbb{R}\_+ \times \mathbb{R}^d$ be a bounded $C^2$ function whose first and second partial derivatives are uniformly bounded (or, more generally, have at most polynomial growth as $|x| \to \infty)$ on $[0, T] \times \mathbb{R}^d$, for any $0 \le T < \infty$. For any $t \ge 0$ and any $x \in \mathbb{R}^d$ it is not hard to see that,$$E^x u(t, W\_t) = u(0, x) + E^x \int\_0^y \left( {\partial\over{\partial s}} + {1\over2}\Delta\_s\right) u(s, W\_s)\,ds.$$
> 
> 
> 


I don't see why this equality is true? Could anyone explain?



Expectation equation, harmonic functions, do not understand why equation is true
This a simple consequence of Itô formula in stochastic calculus. I'll explain the one dimensional formula, the d-dimensional is similar.


First of all to answer the questions of Stefan and Wilie, $W\_t$ is a Brownian motion process which takes the value $x$ at $t=0$ so the dependence of $x$ is implicit here, most of the references will consider $W\_t$ a standard Brownian motion which means it takes the value $0$ at $t=0$ and in the above formula you'll have $x+W\_t$ instead. $\Delta\_s$ here means the sum of all the mixed derivatives.


Itô formula states that 
$$du(s,W\_s)=\frac{\partial u}{\partial s}(s,W\_s)ds+\frac{\partial u}{\partial x}(s,W\_s)dW\_s+\frac 12\frac{\partial^2 u}{\partial s^2}(s,W\_s)ds $$


Integrating this equation between $0$ and $t$ gives 


$$u(t,W\_t)=u(0,W\_0)+\int\_0^t\frac{\partial u}{\partial s}(s,W\_s)ds+\int\_0^t\frac{\partial u}{\partial x}(s,W\_s)dW\_s+\int\_0^t\frac12\frac{\partial^2 u}{\partial s^2}(s,W\_s)ds $$


now taking the expactation $E^x$ which means the conditional expactation knowing that $W\_0=x$ we get the results as we have 
$$ E^x[\int\_0^t\frac{\partial u}{\partial x}(s,W\_s)dW\_s]=0$$


by the way, you have a typo in your formula where you wrote $y$ instead of $t$.


All the conditions about the polynomial growth are made just to make the expectations converge.
Hope this is clear.


