I have the following stochastic equation:

$u(x,t)=e^{\rho t}\mathbb{E}[f(x+W(t))]$

For some function $f(x)$ and $W(t)$ is a Brownian motion. In my notes it says that this equation satisfies the heat equation:

$\frac{\partial u}{\partial t}=\rho u+\frac{1}{2}\frac{\partial^2 u}{\partial x^2}$

I believe this identity comes from the backward Kolmogorov equation, but can't seem to find exactly this result. In particular, my signs seem to be wrong when just applying the backward Kolmogorov equation. Any ideas?

## Best Answer

If $f$ is sufficiently smooth, we may exchange integration and differentiation to obtain that $\frac{\partial^2 u}{\partial x^2} = e^{\rho t} E(\partial_x^2 f(x+W_t)) = e^{\rho t} E[f^{\prime \prime} (x+ W_t)]$.

In particular, assuming that $f \in C^2$ with $f^\prime \in L^2$, we get by ItÃ´'s lemma that $$f(x + W_t) = f(x) + \int_0^t f^\prime (x + W_s) dW_s + \frac{1}{2} \int_0^t f^{\prime \prime}(x + W_s) ds$$ Since $f^\prime \in L^2$, we observe that $\int_0^t f^\prime (x + W_s) dW_s$ is a martingale that starts at zero, hence: $$E[f(x + W_t)] = f(x) + \frac{1}{2}\int_0^t E[f^{\prime \prime}(x+W_s)]ds$$ This implies that $$\partial_tE[f(x + W_t)] = \frac{1}{2} E[f^{\prime \prime}(x+W_t)]$$ The product rule applied to $u$ gives us: $$\frac{\partial u}{\partial t} = \rho e^{\rho t} E[f(x+W_t)] + e^{\rho t} \partial_t E[f(x+W_t)]= \rho u + \frac{1}{2}e^{\rho t}E[f^{\prime \prime}(x+W_t)] = \rho u + \frac{\partial^2 u}{\partial x^2}$$ Which shows that $u$ satisfies the heat equation.