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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.

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  • $\begingroup$ If I had to guess, I would expect that you would need additional monotonicity conditions on $\sigma$ (e.g. $\sigma_{X,t}(x)$ and $\sigma_{Y,t}(x)$ are monotonic in $t$ and $x$) to make something like this work. Then I would look for a coupling proof. $\endgroup$
    – Anthony Quas
    Commented Apr 12, 2015 at 21:38
  • $\begingroup$ Are $\sigma_{X,t}$ and $\mu_{X,t}$ assumed to be non-random? $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Apr 13, 2015 at 0:57
  • $\begingroup$ @bjørn : This post is an "is there anything like this out there"-type of post and all contributions are very welcome. $\endgroup$
    – Matias Heikkilä
    Commented Apr 13, 2015 at 5:44

2 Answers 2

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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.

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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 $\mu_X<\mu_Y$.)

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  • $\begingroup$ I think this will also work if you let the $\mu$, $\sigma$ be time dependent, as long as they are nonrandom, since we should get $\operatorname{Var}(X_t) = \int_0^t \sigma_{X,t}^2\,dt$ and the same for $Y$, which preserves the pointwise ordering. $\endgroup$
    – Nate Eldredge
    Commented Apr 13, 2015 at 1:28
  • $\begingroup$ Actually based on the constant-case alone I was inclined to guess that $\sigma_{X,t}<\sigma_{Y,t}$ would instantly result in greater variance in $Y_t$. However, if μ's are allowed to be for example $X_t$-dependent one may consider $X_t= c X_t \mathrm{d}t+\sigma\mathrm{d}B$. Having only thought about the constant case up to that point I was very surprised to see an exponential growth of variance when simulating this $X_t$. I presume that this kind of process amplifies the variance accumulated at previous times by $c$, hence the exponential growth of variance. $\endgroup$
    – Matias Heikkilä
    Commented Apr 13, 2015 at 6:07

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