Solution of two (first) SDEs.

Question

I'm about to study SDE's for the first time and I'm kinda having troubles "guessing"/"finding" solutions. Also I don't really know how and when analogies to simple ODEs are allowed (e.g. to get a first glimps how the solution could look like, later testing with Ito).

I got two SDE's that bother me atm:

$\mathbb{d}X_t=|X_t|\mathbb{d}B_t+\frac{X_t}{2}\mathbb{d}t, X_0=0$ and

$\mathbb{d}X_t=\sqrt{1+X_t^2}\mathbb{d}B_t+\frac{X_t}{2}\mathbb{d}t, X_0=x$

I tried to get rid of the "dt"-parts with the common ODE trick (multiply with exp(..)), with no effect. Guess the $dB_t$ part should be linear in $X_t$ for that trick to work?

I would very much appreciate any tips and/or solutions, in particular some intuition how to e.g. make educated guesses, use analogies from ODEs,...

Answer 1

Let's start with the SDE

$$dX_t = |X_t| \, dB_t+ \frac{X_t}{2} \, dt \qquad X_0=0$$

This one is pretty easy: As the coefficients are globally Lipschitz continuous, there exists a unique solution. Obviously, $X_t := 0$ is a solution, and so we are done.

Now consider the more interesting SDE

$$dX_t = \sqrt{1+X_t^2} \, dB_t + \frac{X_t}{2} \, dt, \qquad X_0=x \tag{2}$$

Without loss of generality, we may assume $x=0$. First of all, we note that the SDE is of the form

$$dX_t = \sigma(X_t) \, dB_t + b(X_t) \, dt \tag{3}$$

i.e. the coefficients do not depend (explicitely) on the time $t$. For this type of SDEs there exist conditions for a transformation into a linear SDEs - and these can be solved explicitely. Since you are interested in the intuition, I will try to motivate the Ansatz and state the corresponding result below:

Clearly, it would be nice to simplify the stochastic-integral term. Itô's formula shows that for $$Z_t := f(t) \cdot \int_0^{X_t} \frac{1}{\sigma(y)} \, dy$$ we get $$Z_t-Z_0 = \int f(s) \, dB_s + \int_0^t \ldots \, ds,$$

so (at least) the stochastic integral is fine. It remains to make the $ds$-term as nice as possible. Obviously, this term depends on our choice of $f$.The function $f(t)=e^{ct}$ might be a good choice (as it "reproduces" itself if we differentiate with respect to $t$). Indeed, there is the following (general) result:

Suppose that the coefficients $b$, $\sigma$ satisfy $$0 = \frac{d}{dx} \left( \sigma(x) \cdot \left( - \frac{d}{dx} \frac{b(x)}{\sigma(x)} + \frac{1}{2} \sigma''(x) \right) \right).$$ Then the transformation $$Z_t = e^{c t} \int_0^{X_t} \frac{1}{\sigma(y)} \, dy\tag{4}$$ converts $(3)$ into a linear SDE (for $(Z_t)_t$).

(see René L. Schilling/Lothar Partzsch: Brownian motion-An Introduction to Stochastic Processes, Example 18.7)

It is not difficult to check that for the SDE $(2)$, the condition is satisfied. Moreover, some straight-forward calculations show that $c=0$ is the best choice and we obtain

$$Z_t = \int_0^t \, dB_s = B_t$$

By definition,

$$B_t = Z_t \stackrel{(4)}{=} \int_0^{X_t} \frac{1}{\sqrt{1+y^2}} \, dy.$$

As $y \mapsto \frac{1}{\sqrt{1+y^2}}$ is the derivative of $y \mapsto \text{arsinh} \, y$, we get

$$B_t = \text{arsinh}(X_t)$$

i.e.

$$X_t = \sinh(B_t)$$

Let me finally remark that there are different approaches but at least the ones I know do not work out for the given SDE.