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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 $t\downarrow 0$?

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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.

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  • $\begingroup$ Thanks so far, but i do not really understand why it is possible to apply the fund. thm of calc and how since $b$ ist just measurable. $\endgroup$ – stievie May 21 at 18:41
  • $\begingroup$ If $b$ is just measurable then the result is not exactly true as stated since $b$ is not really defined pointwise. But the suitable generalization still follows from the Lebesgue FTC. $\endgroup$ – Ian May 21 at 18:50
  • $\begingroup$ Okay, let's assume that $b$ is continuous. Then FTC says $$E(X_t)=E\int_0^tb(s,X_s)ds=\int_0^t E(b(s,X_s))ds$$ so $\frac{d}{dt}E(X_t)=E(b(t,X_t))$. Then i stuck. Could you give me a hint? $\endgroup$ – stievie May 21 at 19:41
  • $\begingroup$ Now differentiate at $t=0$ and take the first order Taylor approximant. $\endgroup$ – Ian May 21 at 21:41
  • $\begingroup$ I get then via Taylor $$EX_t-x=E(b(0,x))\cdot (t-0)+o(t)= t\cdot b(0,x)+o(t)$$. This is not ... $b(t,x)$.... Did i do something completely wrong? $\endgroup$ – stievie May 22 at 7:36
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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

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