Suppose you have a continuous-time stochastic process $$\textbf{X}_t$$, that takes values in $$\mathbb{R}^d$$.

Now assume that we define a second stochastic process $$\textbf{Y}_t$$, also in $$\mathbb{R}^d$$ by an Ito diffusion:

$$d\textbf{Y}_t = \textbf{q}\left(\textbf{X}_t\right)\,dt + \pmb\sigma(\textbf{X}_t)\cdot d\textbf{W}_t.$$

assume that $$\pmb\sigma,\pmb q$$ are smooth bounded functions. Let's say that at time $$t=0$$, $$X_t=\mathbf{r}$$. Is it true that

$$\pmb\sigma\left(\mathbf{r}\right)\pmb\sigma\left(\mathbf{r}\right)^\dagger = \lim_{h\rightarrow0^+} \frac{\mathbb{E} \left[(\textbf{Y}_{h}-\textbf{Y}_0)(\textbf{Y}_{h}-\textbf{Y}_0)^{\dagger}\right]}{h}$$

This would be true if $$\textbf{X}_t$$ was an Ito diffusion as well, but would this expression hold regardless of what kind of stochastic process $$\textbf{X}_t$$ is? (probably not, but are there other instances in which it would hold?)

• Did you mean $\mu(X_t)$ instead of $q(X_t)$ in your definition of $dY_t$? Jan 31 at 14:57

This holds as long as $$X$$ is right-continuous at $$0$$ (assuming you meant $$h$$ instead of $$2h$$ in the denominator). I'll just show it for $$d=1$$ and $$Y_0 = 0$$, but the general case works exactly the same way (just with a bunch of sums involved). In this case, we have $$Y_t = \int_0^t \mu(X_s) ds + \int_0^t \sigma(X_s) dW_s,$$ so \begin{align*} \mathbb{E}[Y_t^2] &= \mathbb{E}\left[\left(\int_0^t \mu(X_s) ds\right)^2\right] + 2 \mathbb{E}\left[\left(\int_0^t \mu(X_s) ds\right)\left(\int_0^t \sigma(X_s) dW_s\right)\right] + \mathbb{E}\left[\left(\int_0^t \sigma(X_s) dW_s\right)^2\right] \\ &= \mathbb{E}\left[\left(\int_0^t \mu(X_s) ds\right)^2\right] + 2 \mathbb{E}\left[\left(\int_0^t \mu(X_s) ds\right)\left(\int_0^t \sigma(X_s) dW_s\right)\right] + \mathbb{E}\left[\int_0^t \sigma(X_s)^2 ds \right], \end{align*} by Ito's isometry (which applies because $$\sigma$$ is a bounded function).

Now, we can compute the limit term-by-term. We can exchange the limits and expected values thanks to the dominated convergence theorem and the boundedness assumptions on $$\mu$$ and $$\sigma$$. First, \begin{align*} \lim_{h \downarrow 0} \frac 1h \left(\int_0^h \mu(X_s) ds\right)^2 &= \lim_{h \downarrow 0} \left( \frac 1h \int_0^h \mu(X_s) ds\right) \cdot \lim_{h \downarrow 0} \left(\int_0^h \mu(X_s) ds\right) \\ &= \mu(X_0) \cdot 0 = 0, \end{align*} where $$\lim_{h \downarrow 0} \left( \frac 1h \int_0^h \mu(X_s) ds\right) = \mu(X_0)$$ follows from the fundamental theorem of calculus and the assumption that $$X$$ is right-continuous (and that $$\mu$$ is smooth).

Now, \begin{align*} \lim_{h \downarrow 0} \frac 1h \left(\int_0^h \mu(X_s) ds\right)\left(\int_0^h \sigma(X_s) dW_s\right) &= \lim_{h \downarrow 0} \left( \frac 1h \int_0^h \mu(X_s) ds\right) \cdot \lim_{h \downarrow 0} \left(\int_0^h \sigma(X_s) dW_s\right) \\ &= \mu(X_0) \cdot 0 = 0. \end{align*}

Finally, \begin{align*} \lim_{h \downarrow 0} \frac 1h \left(\int_0^h \sigma(X_s)^2 ds\right) &= \sigma(X_0) = \sigma(r). \end{align*}

Combining all of these, \begin{align*} \lim_{h \downarrow 0} \frac 1h \mathbb{E}[Y_h^2] &= \lim_{h \downarrow 0} \frac 1h \mathbb{E}\left[\int_0^h \sigma(X_s)^2 ds \right] \\ &= \mathbb{E}\left[\lim_{h \downarrow 0} \frac 1h \int_0^h \sigma(X_s)^2 ds \right] \\ &= \mathbb{E}[\sigma(X_0)] \\ &= \sigma(r). \end{align*}

To show that it doesn't hold if $$X$$ is not right-continuous at $$0$$, note that we could take $$X_0 = r$$ and $$X_t = r+1$$ for $$t > 0$$ (with $$\sigma(r) \ne \sigma(r+1)$$). Most o the proof would still go through, except that now \begin{align*} \lim_{h \downarrow 0} \frac 1h \left(\int_0^h \sigma(X_s)^2 ds\right) &= \sigma(r+1). \end{align*}

• Is it possible to relax the boundedness assumption on $\sigma$, $\mu$ and still recover the same result? For example, if $\sigma$ is square-integrable as a function of position, or some other reasonable condition on the way it blows up? Mar 19 at 17:13
• @Asasuser Possibly, but you would probably need to impose more conditions on $X$ as well. You would need to control $\int \sigma(X_s)^2 ds$ rather than just $\int \sigma(s)^2 ds$. You really need uniform integrability of the terms $\frac{1}{h} \int_0^h \sigma(X_s)^2 ds$. I don't know of any easy conditions to imply that. Mar 19 at 18:12