# Comparison of Solutions to SDEs

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 $$Z_1(t)$$).

• I'm guessing you probably want something like $0 < c \le \sigma_1(t,x)$ for some constant $c$ too, because the solutions to SDEs can get weird if the diffusion term approaches $0$. Feb 2 at 4:11
• @user6247850 This is true. If a result like this can be shown with even some restrictive assumptions that would be great, I'm just not sure what is the best way to approach this. Feb 2 at 5:09

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 $$\tag{4} X_t holds. Proof. \begin{align} \mathbb P\Big\{\frac{Y_t}{X_t}\le 1\Big\}&=\mathbb P\Big\{\exp\big(\mu B_t-\nu B_t-\mu^2t/2+\nu^2t/2\big)\le 1\Big\}\\ &=\Phi\Bigg(\frac{\mu^2 t-\nu^2t}{2\sqrt{\mu^2t+\nu^2t-2\mu\nu t}}\Bigg)\\ &=\Phi\Big(\frac{\mu+\nu}{2}\sqrt{t}\Big)\,,\tag{5} \end{align} and similarly, \begin{align} \mathbb P\Big\{\frac{Z_t}{Y_t}\le 1\Big\} &=\Phi\Bigg(\frac{\sigma^2 t-\mu^2t}{2\sqrt{\sigma^2t+\mu^2t-2\sigma\mu t}}\Bigg)\\ &=\Phi\Big(-\frac{\sigma+\mu}{2}\sqrt{t}\Big)\,.\tag{6} \end{align} The minus sign here comes from $$\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 $$\mathbb P\Big(\big\{X_t0\,.$$

• Thanks for shedding some light onto the problem. If we had $dX_t = \nu X_t dB_t$ also with $\nu < \sigma$, then wouldn't $\mathbb{E}(Z_t/X_t) < 1$? So while we have $|Z_t| \leq |Y_t|$ almost surely, this is not clear for $|Z_t| \leq |X_t|$. Please let me know if I am confusing myself here. Feb 4 at 19:23
• Yes. With the same method you can show that $\mathbb E[Z_t/X_t]<1\,.$ You can however not conclude from this that $Z_t<X_t$ a.s. I am a bit confused about your sentence "So while we have $|Z_t|≤|Y_t|$ almost surely ..." This is exactly what I showed to be false. Feb 4 at 20:20
• In fact, the truth of $\mathbb E[Z_t/X_t]<1$ shows in a totally symmetric fashion that the conjecture $X_t\le Z_t$ a.s. is false as well. Feb 4 at 20:28