I am asked to solve the following SDE:

$$dX_t = (a-bX_t)dt + cdB_t,\ \text{ where }X(0) = x.$$

($(B_t)_{t\ge0}$ is a brownian motion.)

For constants $a$, $b$ and $c$ and $X$ is a random variable independent of brownian motion. Also, find $m$ and $\sigma$ and a condition on $b$ so that if $x$ is normal with mean $m$ and variance $\sigma$ then $X(t)$ is distributed as $x$.

Here is what I have done for solving the equation.

Let $\alpha_t$ be a (deterministic) process solution of

$$d\alpha_t = (a-b\alpha_t)dt,\ \text{ where }\alpha_0 = 1.$$

This is an ODE and solving it yields the answer $\alpha_t = \frac{a-e^{-bt}}{b}$.

Let us write $X_t = \alpha_t Y_t$ and search for an equation for $Y_t$. By the integration by parts formula (in differential form) we have :

$$dX_t = d\alpha_t Y_t+\alpha_t dY_t.$$
(The other term is zero since $\alpha$ has bounded variation.)

Substituting the expression for $\alpha_t$ we get :

$$dX_t = e^{-bt} Y_t + \alpha_t dY_t = (a-b\alpha_t)Y_tdt + \alpha_tdY_t.\quad (I)$$

On the other hand:

$$dX_t = (a-bX_t)dt + cdB = (a-b\alpha_t Y_t)dt + cdB\quad(II).$$

Equating $(I)$ and $(II)$ we conclude that

$$\alpha_tdY_t = cdB,$$ in other words $$dY_t = \frac{c}{\alpha_t}dB_t.$$

With Y_0 = X_0/alpha_0 = x_0. This implies that:

$$Y_t = x_0 + c \int_0^t \frac{1}{\alpha_s}dB_s = x_0 + c\int_0^t\frac{b}{a-e^{-bt}}dB_s. $$

Therefore the solution $X_t = \alpha_t Y_t$ can be obtained by multiplying $\alpha_t$ by $Y_t$. Is this correct? the final answer obtained looks a bit messy.

Also how should approach the second part of the question? I appreciate any help.


1 Answer 1


Your solution to the ODE is incorrect. Indeed, $$ \mathrm d\alpha_t = (a-b\alpha_t)\mathrm dt,\quad\alpha_0=1, $$ has solution $$ \alpha(t)={\mathrm e}^{-bt}+\frac ab\left(1-{\mathrm e}^{-bt}\right),\quad t\ge0. $$ Additionally you made a mistake when equating (I) and (II), namely you should get $$ aY_t\mathrm dt+\alpha_t\mathrm dY_t=a\mathrm dt+c\mathrm dB_t, $$ so this approach does not work. (Since this S.D.E. is no easier to solve than the initial one.)

You should rather find the solution to $$ \mathrm d\alpha_t = -b\alpha_t\mathrm dt,\quad\alpha_0=1, $$ and apply the variation of the constant method that you propose.

Once you have done this, in order to solve question $2$ start by finding the distribution of $X_t$ when $X_0\sim\mathcal N(m,\sigma^2)$ is independent of $X_t$. At this point, it should be clear what the required conditions are.

  • $\begingroup$ I see, thank you very much. $\endgroup$
    – Alex
    Feb 16, 2015 at 16:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .