TITLE: Absence of x and t dependence in a Ito SDE
QUESTION [1 upvotes]: I have the following SDE in the conventional Ito form
$$
dx(t) = a[x(t),t]dt+b[x(t),t]dW(t)
$$
where $W(t)$ is the Wiener process. For all $t$ and $t_0$, the equation can be simply integrated, giving
$$
x(t)-x(t_{0})=\int_{0}^{t}dt^{\prime}a[x(t^{\prime}),t^{\prime}] + \int_{t_{0}}^{t}dW(t^{\prime})b[x(t^{\prime}),t^{\prime}]
$$
Suppose $b$ is no longer $x$ nor $t$ dependent, how does one evaluate the Wiener integral term? Naively, I'd write
$$
x(t)= x(t_{0}) +\int_{0}^{t}dt^{\prime}a[x(t^{\prime}),t^{\prime}]+\int_{t_{0}}^{t}dW(t^{\prime})b \\
= x(t_{0}) +\int_{0}^{t}dt^{\prime}a[x(t^{\prime}),t^{\prime}]+b\left(W(t)-W(t_{0})\right)
$$
but how should one solve the equation with such Wiener terms in the equation? Alternately, how can I interpret the term $W(t)-W(t_{0})$?

REPLY [0 votes]: The Ito differential $dW_t$ is merely informal short hand for the Ito integral, say, $\int_0^t dW_u=W_t-W_0=W_t$ since $W_0=0$. Thus $\int_0^t dW_u \sim \mathcal{N}(0, t)$. So the SDE
$$ dX_t = a(t,X_t)dt+b dW_t$$
means
$$X_t= X_0 +\int_0^t a(u, X_u)du +b W_t.$$
In general if $a$ is linear, i.e. $a(t,x)=b(t)x+c(t)$ for some functions $b,c$, then an explicit solution is in principle available. Briefly, see here.
If the drift does not depend on time and can be written as $-U'(x)$ for some nice function $U$ then we can find some information about the long-term behavior of the process. Indeed, we can show the steady state density of $X_\infty$ must be proportional to $e^{-2U(x)/b^2}$. Indeed, the infinitesimal generator of $X$ is
$$\mathscr{L}f = -U'(x)f'(x)+\frac12 b^2 f''(x)$$
and its adjoint is
$$\mathscr{L}^*f = \frac{d}{dx}[U'(x)f(x)]+\frac12 b^2 f''(x).$$
The steady state PDF $f$ must satisfy $\mathscr{L}^*f = 0$. We can factor out a derivative and use some regularity facts to find out that the probability "flux" must vanish:
$$U'(x)f(x)+\frac12 b^2 f'(x)=0$$
Rearrange this into
$$ \frac{1}{f(x)} f'(x) = -\frac{2}{b^2}U'(x),$$
i.e. $f(x) = e^{-2U(x)/b^2}$ solves the ODE. To get a PDF just normalize, provided $U$ is such that the total integral of $f$ is finite. If harmonic potential is used, $U(x)=\frac12 x^2$, then you will get a specific Ornstein Uhlenbeck process.
I'm not sure what else you can say in general for non-autonomous drifts $a(t,x)$. Hopefully someone more knowledgeable will stop by. Comment if you have clarifications, or corrections.
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