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Solve the SDE: $dX(t) = udt + \sigma X(t)dB(t)$

Provided Question

The SDE is $dX(t) = udt + \sigma X(t)dB(t)$. Find $X(t)$, where $X(t)$ is some stochastic process and $B(t)$ is a Wiener process. Both $u$ and $\sigma$ are constants.

Hint: Multiply both sides by the "integrating factor" $e^{-\sigma B(t) + \frac12 \sigma^2 t}$.

Current Progress

Multiplying both sides by the appropriate integrating factor:

$$ \exp{\big( -\sigma B(t) + \frac12 \sigma^2t\big)}dX(t) = \exp{\big(-\sigma B(t)+\frac12\sigma^2t\big)}(udt + \sigma X(t)dB(t)) $$

Then set $f(t,x,b):=\exp{\big( -\sigma B(t) + \frac12 \sigma^2t\big)}X(t)$ and apply Ito's formula. Some of the required results before actually applying Ito's formula:

$ \frac{df}{dt} = \frac12\sigma^2 X(t)e^{-\sigma B(t) + \frac12 \sigma^2t} \\ \frac{df}{dx} = e^{-\sigma B(t) + \frac12 \sigma^2t} \\ \frac{df}{db} = -\sigma X(t)e^{-\sigma B(t) + \frac12 \sigma^2t} \\ \frac{d^2f}{dx^2} = 0 \\ \frac{d^2f}{db^2} = \sigma^2 X(t) e^{-\sigma B(t) + \frac12 \sigma^2t} \\ \frac{d^2f}{dxdb} = -\sigma e^{-\sigma B(t) + \frac12 \sigma^2t} $

Given this, we need to know the following derivatives of the quadratic variations and co-variations:

$ d\langle B\rangle_t = dt \\ d\langle X,B\rangle_t = ??? $

My Request

Please instruct me on what $d\langle X,B\rangle_t$ is equal to so that I may progress further with this problem.

share|cite|improve this question
$d\langle X,B\rangle_t = \sigma X dt$, in general for 2 $\int h dB, \int g dB$ the QcoV is $gh dt$ – mike Nov 9 '12 at 19:00
@mike That makes complete sense. But, are you 100% sure this is accurate in this instance? – Jase Nov 10 '12 at 3:31
this is prop ix.2.3 in revuz & yor. $e = \frac 1 f$ is a martingale that satisfies $de = edB$. I don't think you ever need $d \langle X,B \rangle$ bt I'm pretty sure about it. – mike Nov 10 '12 at 14:46
up vote 5 down vote accepted

As mike already wrote: $<X,B>_t = \sigma \cdot X_t \, dt$ - this fact follows straight from the definition of $<\cdot,\cdot>$.

Another approach to solve the SDE:

  1. Solve the homogeneous SDE $dX_t = \sigma \cdot X_t \, dB_t$. Idea: Set $Z_t := \log X_t$ and apply Itô's formula to $f(x) := \log x$, then $$\begin{align*} dZ_t &= \frac{1}{X_t} \, dX_t - \frac{1}{2} \frac{1}{(X_t)^2} (dX_t)^2 \\ &= \left(- \frac{1}{2} \sigma^2 \right) \, dt + \sigma dB_t \\ \Rightarrow X_t &= X_0 \cdot \exp \left(- \int_0^t \frac{1}{2} \sigma^2 \, ds \right) \cdot \exp \left(\int_0^t \sigma \, dB_s \right) \\ &= X_0 \cdot \exp \left(- \frac{1}{2} \sigma^2 \cdot t + \sigma \cdot B_t \right) \end{align*}$$
  2. Solve the inhomogeneous SDE: Let $$X_t^0 := \exp \left( \frac{1}{2} \sigma^2 \cdot t - \sigma \cdot B_t \right)$$ then we know that $\frac{1}{X_t^0}$ solves the homogeneous equation (and $X_0^0=1$). Now set $Z_t := X_t \cdot X_t^0$ and apply Itô's formula to $f(x,y) = x \cdot y$, then $$dZ_t = \ldots = u \cdot X_t^0 \, dt$$ thus $$X_t = \frac{Z_t}{X_t^0} = \frac{u \cdot \int_0^t \exp \left( \frac{1}{2} \sigma^2 \cdot s - \sigma \cdot B_s\right) \, ds}{\exp \left( \frac{1}{2} \sigma^2 \cdot t - \sigma \cdot B_t\right)}$$

This approach also works in a similar fashion for the (more general) SDE

$$dX_t = (a(t)+b(t) \cdot X_t) \, dt + (c(t)+d(t) \cdot X_t) \, dB_t$$

share|cite|improve this answer
When you apply Ito's formula to $xy$, where did the cross derivative term go? Seems like $dZ_t = uX_t^0dt - \sigma^2 X_t X_t^0 dt$ – bcf Apr 3 '15 at 5:38
@bcf I think it would be better to ask a new question. Please include your own calculations (i.e. what's $dX_t^0$, how you applied Itô's formula, ...). – saz Apr 8 '15 at 15:33
@saz could you elaborate the general case a bit more? Which function should one choose for the homogenous case? – Liealgebrabach Apr 15 at 14:32
@Liealgebrabach Actually the same. See this question: – saz Apr 15 at 17:26

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