Can anyone help me with the following SDE?

Solve the following stochastic differential equation: $$dY_t=aY_tdt+(b(t)+cY_t)dB_t$$ with $Y_0=0$.

Hint: Try a solution of the form $Z_tH_t$ where $Z_t = exp(cB_t+(a-\frac{1}{2}c^2t))$ and $dH_t=F(t)dt+G(t)dB_t$ for some adapted process F and G which need to be determined.

Thanks gt6989b for your input!

Please correct me if I'm wrong anywhere.

Since, $$Z_t = exp(cB_t+(a-\frac{1}{2}c^2t))$$ Hence we get: $dZ_t = Z_t(cdB_t-\frac{1}{2}c^2)dt$

and we also have $dH_t=F(t)dt+G(t)dB_t$

Then we apply Ito's Lemma on $Z_tH_t$, which yields \begin{eqnarray*} d(Z_tH_t) &=& Z_tdH_t+H_tdZ_t+dH_tdZ_t \\ &=& Z_t(F(t)dt+G(t)dB_t)+Z_tH_t(cdB_t-\frac{1}{2}c^2dt)+Z_tcG(t)dt \\ &=& Z_t(F(t)-\frac{1}{2}c^2H_t+cG(t))dt+(Z_tG(t)+cZ_tH_t)dB_t \end{eqnarray*}

By letting $Y_t = Z_tH_t$, we compare between the expressions $dY_t$ and $d(Z_tH_t)$ in the $dt$ and $dB_t$ terms respectively. And we get the following:

\begin{eqnarray} Y_t &=& Z_tH_t \\ G(t) &=& \frac{Z_t}{b(t)} \\ F(t) &=& (a+\frac{1}{2}c^2)H_t - c\frac{Z_t}{b(t)} \end{eqnarray}

Note: $F(t) = P$ and $G(t) = Q$ for your P and Q respectively.

But now, I don't really understand how would this result help me in solving the original $dY_t$ SDE.

  • 1
    $\begingroup$ Well, apply Ito's Lemma to $A_t = Z_t H_t$ that will tell you what $F,G$ need to be to match those of $Y_t$. $\endgroup$
    – gt6989b
    Mar 20, 2013 at 19:50
  • $\begingroup$ what he said ^ follow the hint.... $\endgroup$
    – Lost1
    Mar 20, 2013 at 19:59

2 Answers 2


Here is an idea to get you started.

Note that $$ dZ_t = cZ_t dB_t + \frac{c^2}{2}Z_t dt = Z_t \left(cdB_t + \frac{c^2}{2}dt\right) $$

and let $dH_t = P dt + Q dB_t$.

Consider $A_t = Z_t H_t$. Then, by Ito's Lemma, and expansion for $dZ_t$,

$$\begin{split} dA_t &= Z_t dH_t + H_t dZ_t + dZ_t dH_t \\ &= Z_t (P dt + Q dB_t) + H_t Z_t \left(cdB_t + \frac{c^2}{2}dt\right) + cQdt \\ &= dt\left[ ??? \right] + dB_t \left[ ??? \right] \end{split} $$

Now match this to $dY_t$, what can you say about $P,Q$?

  • $\begingroup$ Thx gt6989b, I've updated my question. I hope you can help me in establishing the link between P,Q and the original SDE. Thanks in advance. $\endgroup$
    – Justin K
    Mar 21, 2013 at 0:22

A linear SDE $$\text{d}Y_{t} = (\alpha(t)+\beta(t)Y_{t})\,\text{d}t+(\gamma(t)+\delta(t)Y_{t})\,\text{d}W_{t}$$ has an explicit (strong) solution which can be found on Wikipedia . Here $\alpha$, $\beta$, $\gamma$ and $\delta$ denote deterministic functions and $W$ a one-dimensional Wiener process. The solution, $Y\ ,$ is expressed in terms of the stochastic exponential, $X\ ,$ which is the strong solution to the linear SDE $$\text{d}X_{t} = \beta(t)Y_{t}\,\text{d}t+\delta(t)Y_{t}\,\text{d}W_{t}\ .$$


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