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! 



total variation distance between two solutions of SDE
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 $\mu,\nu \ll \lambda$.


