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Suppose we have two stochastic differential equations with the same initial conditions: $$d X_t^1= b_1(t,X_t^1)dt + dW_t$$ $$d X_t^2= b_2(t,X_t^2)dt + dW_t,$$ $X_0^1=X_0^2=x_0$; $W_\cdot$ is a standard one-dimensional Brownian motion, the functions $b_1$ and $b_2$ are uniformly bounded and Lipschitz.

Using Gronwall's inequality one can easily derive bounds on $|X_t^1-X_t^2|$ and hence it is possible to estimate the Wasserstein distance between $X_t^1$ and $X_t^2$.

My question is, how one can estimate the total variation distance between $X_t^1$ and $X_t^2$? Intuitively, because the processess $X_t^1$ and $X_t^2$ start from the same point, the total variation distance between them should be small (at least for small $t$). So, is it possible to obtain the bounds of the form $$d_{TV} (X_t^1,X_t^2)\le \phi(t) $$ for some function $\phi$? (here $d_{TV}$ means the total variation distance between two measures)

Thanks!

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Isn't a bound on $|X_t^1 - X_t^2|$ considerably stronger than a bound in total variation distance? One is strong and the other is weak? –  weakstar Dec 1 '12 at 17:55
    
I don't think so: if I have two delta measures $\mu:=\delta_0$ and $\nu:=\delta_{0.01}$ then the total variation distance between them is 2, whereas the Wasserstein distance is just 0.01. Thus, a bound in the total variation distance is stronger than a bound on $|X^1_t−X^2_t|$. –  Oleg Dec 1 '12 at 18:11
    
Yeah but your construction of $\mu$ and $\nu$ is a weak one, in that it says nothing about dependence on particular values of $\omega \in \Omega$. Using Grownwall's inequality you can get a convergence rate for (strong) $L^1$ convergence, $E \left|X_t^1 - X_t^2 \right|$ tends to zero in some sense. This implies total variation convergence. –  weakstar Dec 1 '12 at 19:38
    
Weakstar, thanks for your comment, but I'm not sure I can agree with it. Indeed, consider the following sequence of random variables: $X_n:=1/n$ and $Y:=0$. Then $E|X_n-Y|=1/n$ and tends to zero as $n\to\infty$, whereas total variation distance between $X_n$ and $Y$ equals 2 for any $n$. –  Oleg Dec 1 '12 at 21:17
    
Those random variables have distribution functions which are discontinuous. –  weakstar Dec 2 '12 at 3:57
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1 Answer

up vote 2 down vote accepted

Such a bound can be derived with Girsanov's theorem and Pinsker's inequality. Let $X_t = x_0 + W_t$. Supposing $b$ has linear growth in $x$, we may define measures $P_i$, $i=1,2$ by

$\frac{dP_i}{dP} = \exp\left(\int_0^Tb_i(t,X_t)dW_t - \frac{1}{2}\int_0^T|b_i(t,X_t)|^2dt\right)$.

Then, under $P_i$, $W^i_t := W_t - \int_0^tb_i(s,X_s)ds$ is a Brownian motion, and so $X$ is a weak solution of SDE (i). Let $P^i \circ X^{-1}$ denote the $P^i$-law of the entire process, a measure on the space of continuous functions. Then

$\frac{dP_2}{dP_1} = \exp\left(\int_0^T(b_2(t,X_t) - b_1(t,X_t))dW^1_t - \frac{1}{2}\int_0^T|b_2(t,X_t) - b_1(t,X_t)|^2dt\right)$.

Let $\mathcal{H}(\cdot | \cdot)$ denote the relative entropy. By Pinsker's inequality,

$ \begin{align} d_{TV}^2(P_1 \circ X^{-1}, P_2 \circ X^{-1}) &= d_{TV}^2(P_1, P_2) \le 2\mathcal{H}(P_1 | P_2) \newline &= -2\mathbb{E}^{P_1}\left[\log \frac{dP^2}{dP^1}\right] \newline &= \mathbb{E}^{P_1}\left[\int_0^T|b_1(s,X_s) - b_2(s,X_s)|^2ds\right] \newline &= \mathbb{E}^{P}\left[\frac{dP_1}{dP}\int_0^T|b_1(s,X_s) - b_2(s,X_s)|^2ds\right] \end{align} $

since the stochastic integral term is a true martingale. This is actually a much stronger control than you requested, since it is easy to see that $d_{TV}(P_1 \circ X_t^{-1}, P_2 \circ X_t^{-1}) \le d_{TV}(P_1 \circ X^{-1}, P_2 \circ X^{-1})$ for any $t$. If you have a uniform bound on $|b_1 - b_2|$, you have a good bound on the $TV$ distance between your processes. You can probably get away with an $L^2$ bound, but you'll then need to fuss with the $dP_1/dP$ term a bit. Hope this helps!

EDIT 1: It's interesting to note that this approach breaks down if you have two different volatility coefficients in your SDE, because the laws of the processes are then singular. But you could still bound the TV distance between the time-$t$ laws, as requested, probably via Malliavin calculus.

EDIT 2: I should elaborate on the first of the string of equalities above. Since $X$ and $W$ generate the same $\sigma$-fields, $dP^i/dP$ is $X$-measurable, and so

$\frac{dP^i \circ X^{-1}}{dP \circ X^{-1}} (X) = \mathbb{E}\left[\frac{dP^i}{dP}|X\right] = \frac{dP^i}{dP}$.

From this it is clear that the TV-distances above are the same, from the formula $d_{TV}(\mu,\nu) = \int d\lambda|d\mu/d\lambda - d\nu/d\lambda|$ for $\mu,\nu \ll \lambda$.

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