Let $W$ be Brownian motion. Let $b_t$ and $\sigma_t$ be adapted to $\mathcal{F}_t^W$. Consider the SDE $$dx_t=b_tdt+\sigma_tdW_t.$$ Assume that $b$, $\sigma$ are such that $x$ stays non-negative. Fix $a>0$. Suppose we are interested in $\mathbb{P}[\sup_{t\in [0,T]}x_t\geq a]$.

If it is possible to find adapted processes $B_t$ and $\Sigma_t$ such that $b_t\leq B_t$ and $\sigma_t^2\leq \Sigma_t^2$, then is it true that $$\mathbb{P}[\sup_{t\in [0,T]}x_t\geq a]\,\,\leq\,\, \mathbb{P}[\sup_{t\in [0,T]}y_t\geq a]$$ where $y_t$ is governed by $$dy_t=B_tdt+\Sigma_tdW_t, \qquad y_0=x_0\,\,\,?$$


No, in general we cannot expect monotonicity.

Counterexample: It is not difficult to see that

$$x_t := \frac{1}{2} +t \qquad \quad y_t := \frac{1}{2} + t + W_{t \wedge \tau}$$


$$\tau := \inf\{t \geq 0; y_t = 0\}$$

are solutions to the SDEs

$$dx_t = \, dt \qquad x_0 = \frac{1}{2}$$ and $$dy_t = \, dt + 1_{(0,\infty)}(y_{t \wedge \tau}) \, dW_t, \qquad y_0 = x_0 = \frac{1}{2},$$

respectively. If we choose $a=1$, then

$$\mathbb{P}\left[ \sup_{t \leq \frac{1}{2}} x_t \geq 1 \right] = 1$$


$$\mathbb{P}\left[ \sup_{t \leq \frac{1}{2}} y_t \geq 1 \right] < 1.$$

  • $\begingroup$ The diffusion coefficient for $y$ should be $1_{(0,\infty)}(y_{t\wedge \tau})$. Otherwise for $t>\tau$, $y_t>0$ and satisfies $dy=dt$. $\endgroup$ – galan Oct 5 '14 at 8:12
  • $\begingroup$ @galan Right, thanks for your comment. $\endgroup$ – saz Oct 5 '14 at 8:56

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