SDEs: Bounding the variance of a solution I've been thinking about something that would seem intuitive, but I haven't really been able to dig a direct answer to. This is a rough draft of it.
Let
$$X_t = \mu_{X,t} \mathrm{d}t + \sigma_{X,t} \mathrm{d}B \quad \text{and} \quad Y_t = \mu_{Y,t} \mathrm{d}t + \sigma_{Y,t} \mathrm{d}B, $$
where the $\mu$'s and $\sigma$'s are "nice".
If we had $0<\mu_{X,t} < \mu_{Y,t}$ and $0<\sigma_{X,t} < \sigma_{Y,t}$ for all $t$. Would we have $\text{Var}[X_t] < \text{Var}[Y_t]$ for all $t$?
I suspect this is not the case, expect for a very limited choice of $\mu$'s and $\sigma$'s, since it would probably otherwise be a widely cited fact.
 A: I don't think this can be true in general.  Let $\sigma_{X,t} = \sigma_{Y,t} = 0$ (or if you insist that they be nonzero, take $\sigma_{X,t} = \epsilon$ and $\sigma_{Y,t} = 2\epsilon$ for $\epsilon$ very small).  Let $Z$ be a random variable with $P(Z = 2) = P(Z=4) = 1/2$.  Note $\operatorname{Var}(Z) = 1$.  Set $\mu_{X,t} = Z$ and set $\mu_{Y,t} = 6$, so that $0 < \mu_{X,t} < \mu_{Y,t}$ surely.  Then $X_t = Zt$ and $Y_t = 6t$, so $\operatorname{Var}(X_t) = t^2$ but $\operatorname{Var}(Y_t) = 0$.
Given that variance is a measure of "spread", it's not reasonable to expect it to respect pointwise ordering, since a random variable can be very large but be concentrated near a constant.
A: Suppose the $\mu$ and $\sigma$ are constants, i.e., there are constants $\mu_X$, $\mu_Y$, $\sigma_X$, $\sigma_Y$ such that for all $t$,
$$
\mu_{X_t} = \mu_X,\qquad \sigma_{X_t} = \sigma_X,
$$
$$
\mu_{Y_t} = \mu_Y,\qquad \sigma_{Y_t} = \sigma_Y.
$$
Then both $\{X_t\}$ and $\{Y_t\}$ are examples of Brownian motion with drift hence for all $t$
$$\operatorname{Var}[X_t]= \sigma_X^2t < \sigma^2_Yt=\operatorname{Var}[Y_t]$$
as desired.
(In this case don't need the assumption that $\mu_X<\mu_Y$.)
