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Preamble: This question is similar to the one in total variation distance between two solutions of SDE . The difference is that in my case the drift is the same but there are different diffusion coefficients.

Question: Consider the two SDE: \begin{align} dX_t &= f(X_t) dt + \frac{\sigma(X_t)}{\sqrt{n}} dW_t\\ dY_t &= f(Y_t) dt + \frac{\bar\sigma_n(Y_t)}{\sqrt{n}} dW_t, \end{align} where $t \in [0,T]$, $X_0 = Y_0 = 0$, $W_t$ is the standard 1-dimensional Brownian motion. The hypotheses are that the drift $f$ and the diffusion coefficient $\sigma > 0$ are bounded and Lipschitz, and the $\bar\sigma_n$ are defined as the piecewise constant approximations of $\sigma$, constant on the intervals of the form $\big[\frac{j}{n}T,\frac{j}{n}T\big)$.

Can we conclude that the total variation distance between the two solution processes tends to zero as $n$ tends to infinity?

Considerations There should be no problem proving that there is $L^2$ convergence, thanks to Theorem 10, chapter V of Protter's "Stochastic Integration and Differential Equations", but I think that this is not enough to obtain the convergence in total variation. In the answer to the question linked above, there is a mention to a possible solution of a similar problem via Malliavin calculus, but since I am not at ease with this theory I would gladly accept any suggestion or reference in that direction.

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up vote 2 down vote accepted

No. the total variation distance is always one, since the quadratic variation of the processes is different, and so the measures are mutually absolutely singular.

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