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 $Z_1(t)$).