# Approximation of stochastic differential equations

Consider the two following real Stochastic Differential Equations (SDE) starting from the same initial condition: $$dx_t = f(x_t)dt + \sigma dB_t$$ $$dy_t = f(y_t)g_{\epsilon}(y_t)dt + \sigma dB_t$$

where $f$ and $g_{\epsilon}$ are such that there exists strong solutions to both SDEs (typically local Lipschitz assumptions on the coefficients).

We assume that $|1-g_{\epsilon}(y)|\leq \epsilon$ for all $y\in \mathbb{R}$.

I want to prove the following convergence: for all finite time $T>0$,

$$\lim_{\epsilon \to 0} \mathbb{E}\left[\sup_{0\leq t \leq T} |x_t-y_t|\right]=0$$

I would like to know how to prove it without a global Lipschitz assumption (think for instance that there may be some quadratic terms in $f$).

Can anyone explain me how to do it rigorously or point me to some article/book where it is already done ?

Thanks !

• @ Unknown 161:Hi isn't there anything that can be said starting from the sde followed by $x_t-y_t$ (from Itô's formula)? Best regards. Oct 23, 2012 at 20:54

As formulated, there are some difficulties for the local Lipschitz case. For quadratic $f(x)=x^2$, consider the corresponding ODE for $x$ (i.e. the SDE with $\sigma = 0$). Then there exists a locally unique solution, due to the local Lipschitz property. This solution has the exact expression $x(t) = x_0/(1- t x_0)$, as may be verified. We see that we have blow-up in finite time for $x_0 > 0$. In general, with only a local Lipschitz condition you need linear growth as extra condition to ensure existence for all $t$.
Let us work under the assumption $f$ globally Lipschitz and bounded, with Lipschitz constant and supremum norm equal to $K_1$ and $K_2$, respectively. Then $|x_t- y_t| = \left|\int_0^t \{ f(x_s) - f(y_s)g_{\varepsilon}(y_s) \} \ d s \right| \leq \int_0^t \left\{|f(x_s) - f(y_s)| + |f(y_s)| |1 - g_{\varepsilon}(y_s)| \right\} \ d s \leq \int_0^t \left\{ K_1 |x_s - y_s| + \varepsilon K_2 \right\} \ d s$.
By the Gronwall inequality, $|x_t - y_t| \leq \varepsilon t K_2 \exp(K_1 t)$.