Comparison of Solutions to SDEs

Question

Let $B(t)$ be a Brownian motion. If we assume that $b(t, x), b_i(t, x)$, $\sigma(t, x)$, and $\sigma_i(t, x)$ are Lipschitz functions for $i = 1, 2$ then it is known that if $X_i(t)$ is the unique solution to $$ dX_i(t) = b_i(t, X_i(t)) dt + \sigma(t, X_i(t)) dB(t) $$ with $X_1(0) = X_2(0) = x_0$ and $b_1(t, x) \leq b_2(t, x)$, then $X_1(t) \leq X_2(t)$ almost surely. The intuition with this case is that $X_2(t)$ always moves at least as much 'upwards' as $X_1(t)$ will.

I'm looking to consider the case where diffusion terms are different instead. The inequality we expect will not be quite as simple. My question is: If $Z(t)$ uniquely solves $$ dZ(t) = b(t,Z(t)) dt + \sigma(t, Z(t)) dB(t) $$ and for $i = 1,2$ the process $Z_i(t)$ unique solves $$ dZ_i(t) = b(t,Z_i(t)) dt + \sigma_i(t, Z_i(t)) dB(t) $$ with $Z(0) = Z_1(0) = Z_2(0) = z_0$ and $\sigma_1(t, x) \leq \sigma(t, x) \leq \sigma_2(t, x)$, then is it true that $$ |Z(t)| \leq \max (|Z_1(t)|, |Z_2(t)|)? $$

The intuition behind this statement is that, if the direction of movement caused by Brownian motion is upward (or downward), the process $Z(t)$ should be bounded by a process with stronger (or weaker) diffusion term, namely $Z_2(t)$ (or $Z_1(t)$).

Answer 1

There is a relatively simple counter example. Let $\sigma<\mu$ be constant diffusion coefficients and consider the two SDEs

\begin{align} dZ_t&=\sigma Z_t\,dB_t\,,&dY_t&=\mu Y_t\,dB_t\,, \end{align} with initial conditions $Z_0=Y_0=1\,.$ Their explicit solutions are the two GBMs \begin{align} Z_t&=\exp(\sigma B_t-\sigma^2t/2)\,,&Y_t&=\exp(\mu B_t-\mu^2t/2)\,. \end{align} For these, the conjecture $$\tag{1} |Z_t|\le |Y_t|\quad{ a.s. } $$ is equivalent to $$\tag{2} \frac{Z_t}{Y_t}\le 1\quad{ a.s. } $$ But this cannot be true. Proof: $$ \frac{Z_t}{Y_t}=\exp\Big(\sigma B_t-\mu B_t-\sigma^2t/2+\mu^2t/2\Big)\,. $$ Therefore, $$\tag{3} \mathbb E\Big[\frac{Z_t}{Y_t}\Big]=\exp(-\sigma\mu t+\mu^2t)\,. $$ It is easy to see that since $\sigma<\mu$ the right hand side is strictly greater than one. Therefore (2) cannot hold.

Let $$ dX_t=\nu X_t\,dB_t\,,\quad \nu<\sigma\,,X_0=1\,. $$ In this notation your conjecture reads $$ Z_t\le \max(X_t,Y_t)\quad\text{ a.s. }\,. $$ To refute this I will show now that with positive probability $$\tag{4} X_t<Y_t<Z_t $$ holds. Proof. \begin{align} \mathbb P\Big\{\frac{Y_t}{X_t}\le 1\Big\}&=\mathbb P\Big\{\exp\big(\mu B_t-\nu B_t-\mu^2t/2+\nu^2t/2\big)\le 1\Big\}\\ &=\Phi\Bigg(\frac{\mu^2 t-\nu^2t}{2\sqrt{\mu^2t+\nu^2t-2\mu\nu t}}\Bigg)\\ &=\Phi\Big(\frac{\mu+\nu}{2}\sqrt{t}\Big)\,,\tag{5} \end{align} and similarly, \begin{align} \mathbb P\Big\{\frac{Z_t}{Y_t}\le 1\Big\} &=\Phi\Bigg(\frac{\sigma^2 t-\mu^2t}{2\sqrt{\sigma^2t+\mu^2t-2\sigma\mu t}}\Bigg)\\ &=\Phi\Big(-\frac{\sigma+\mu}{2}\sqrt{t}\Big)\,.\tag{6} \end{align} The minus sign here comes from $\sigma<\mu\,.$

For example, when $t=1,\nu=0.01,\sigma=0.1,\mu=0.11$ the sum of these probabilities (5) and (6) is strictly less than one.

Then however $$ \mathbb P\Big(\big\{X_t<Y_t\big\}\cap\big\{Y_t<Z_t\big\}\Big)>0\,. $$