Expected value of general diffusion

Question

lets take measurable $b,\sigma:\mathbb{R}^+\times \mathbb{R}\to\mathbb{R}$ and consider the SDE $$dX_t=b(t,X_t)dt+\sigma(t,X_t)dB_t$$ with $X_0=x$. How can i use Itô's Lemma to show $$E_x[X_t-x]=t\cdot b(t,x)+o(t)\\ E_x[(X_t-x)^2]=t\cdot\sigma^2(t,x)+o(t)$$ as $t\downarrow 0$?

Answer 1

Write $X_t-x=\int_0^t b(s,X_s) ds + \int_0^t \sigma(s,X_s) dB_s$, now the first fact can be obtained by using the fundamental theorem of calculus on the first term and then noting that the second term has exactly zero expectation even when $t$ is finite.

The second fact relies on some scaling arguments that are rooted in Ito's lemma, which essentially tell you that the three terms in the square scale as $t^2,t^{3/2}$ and $t^1$ as $t \to 0$, with the last (asymptotically largest) one being the square of the stochastic term. Then you actually calculate this expectation by using the Ito isometry, and then again use the fundamental theorem of calculus to get the small $t$ asymptotic.

Answer 2

Ciao, I will just do explicitly the computation suggested by Ian.

So we have to prove that $$ E[X_t-x] = tb(t,x) + o(t). $$

Let's assume that $b(t,x)$ differentiable w.r.t. $t$ then: $$ b(t,x) = b(0,x) + t \partial_tb(0,x) + o(t). $$ and so: $$ b(0,x)-b(t,x) = - t \partial_tb(0,x) + o(t). $$ We are now ready for the straightforward computation:

\begin{align} E[X_t-x] & = tE[b(0,x)] + o(t) \\ & = tE[b(t,x)] + tE[b(0,x)-b(t,x)] + o(t) \\ & = tE[b(t,x)] - t^2 \partial_tb(0,x) + to(t) + o(t) \\ & = tE[b(t,x)] + o(t) \end{align}

Here we are.

Ciao