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Let $W^1$ be a Brownian motion and $\sigma(\cdot)$ be a positive, bounded, continuous function. Define \begin{align*} V_t=\int_0^t\sigma(Y_s)dW_s, \end{align*} where $(Y_t)_{t\geq 0}$ is a deterministic function, so \begin{align*} t^{-1/2}V_t\thicksim N\left(0,\frac{1}{t}\int_0^t\sigma^2(Y_s)ds\right)\to N\left(0,\sigma^2(Y_0)\right),\text{ as }t\to 0. \end{align*} Does the Gaussian limit as $t\to 0$ still hold if $Y_t$ is no longer deterministic? Assume \begin{align*} dY_t & =\alpha(Y_t)dt+\gamma(Y_t)\left(\rho dW_t+\sqrt{1-\rho^2}dZ_t\right),\\ Y_0 & =y_0, \end{align*} where the initial value $y_0$ is deterministic, $\rho\in[-1,1]$, and $Z$ is a Brownian motion, independent of $W$. Then, if I'm correct, \begin{align*} \mathbb{E}\left(t^{-1/2}V_t\right)&=0,\\ \text{Var}\left(t^{-1/2}V_t\right)&=\frac{1}{t}\int_0^t\sigma^2(Y_s)ds\to\sigma^2(Y_0) \end{align*} but is the limiting distribution as $t\to 0$ Gaussian (under any circumstances)?

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Yes, and it doesn't depend on $\alpha$, $\gamma$ or $\rho$. Fix $\theta\in\mathbb{R}$. Then, define $M$ for $t\ge 0$ by $$ M_t = \exp\left[i\theta\int_0^t\sigma(Y_s)dW_s+\frac{\theta^2}2\int_0^t\sigma(Y_s)^2ds\right].$$

As $\sigma$ is bounded, $M$ is a martingale (with $M_0=1$). Fix $t>0$ and $u\in\mathbb{R}$. Set $\theta = \frac{u}{\sqrt{t}}$ to get $$\mathbb{E}\left(\exp\left[\frac{iu}{\sqrt{t}}\int_0^t\sigma(Y_s)dW_s + \frac{u^2}{2t}\int_0^t\sigma(Y_s)^2ds\right]\right) = 1.$$

By Cauchy-Schwarz, or other estimates, $$\mathbb{E}\left(\exp\left[\frac{iu}{\sqrt{t}}\int_0^t\sigma(Y_s)dW_s\right]\left(\exp\left[\frac{u^2}{2t}\int_0^t\sigma(Y_s)^2ds\right] - \exp\left[\frac{u^2}{2}\sigma(y_0)^2\right]\right)\right)$$

converges to $0$ as $t\to 0$.

So $$\lim_{t\to 0} \mathbb{E}\left(\exp\left[\frac{iu}{\sqrt{t}}\int_0^t\sigma(Y_s)dW_s + \frac{u^2}2\sigma(y_0)^2\right]\right) = 1.$$

Hence we are done by Lévy's Continuity Theorem.

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    $\begingroup$ @ Ben Derret: Very nice I thought it wasn't true. I really learned something today Sometimes thank you. Best regards $\endgroup$ – TheBridge Dec 27 '13 at 14:05
  • $\begingroup$ @TheBridge Thanks! $\endgroup$ – Ben Derrett Dec 27 '13 at 14:30
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Hi your first assertions are true because $\sigma$ is bounded so that $V_t$ is a martingale, this gives the null expectation result and then by Itô's isometry you get the convergence result for the "averaged" variance.

Regarding the distribution here I am quite convinced that it is false, to prove it under milder hypothesis than yours, I would suggest that you try a smooth $\sigma$ and use a euler scheme to approximate $Y$ in small time and look at a Taylor expansion to try to see how the integral behaves in that case, I am quite sure (but I can be wrong) that you might be able to find nice (or bad) $\sigma, \alpha, \gamma$ that makes $V_t$ non gaussian even in small times.

Best regards

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