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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?)

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

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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*}

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  • $\begingroup$ 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? $\endgroup$
    – Asasuser
    Mar 19 at 17:13
  • $\begingroup$ @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. $\endgroup$ Mar 19 at 18:12

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