Solving the SDE [itex]dX(t) = udt + \sigma X(t)dB(t)[/itex]

[operationsres]

**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.

 

[mighty2000]

Thank you in advance for your help.

**Reply from Scientist**

Hi there, great job so far on setting up the problem and using Ito's formula. To find d\langle X,B\rangle_t, we can use the definition of quadratic variation:


d\langle X,B\rangle_t = X(t)d\langle B\rangle_t + B(t)d\langle X\rangle_t + d\langle X\rangle_td\langle B\rangle_t


Since d\langle B\rangle_t = dt, we can simplify this to:


d\langle X,B\rangle_t = X(t)dt + B(t)d\langle X\rangle_t + d\langle X\rangle_tdt


Now, we can use the definition of co-variation to find d\langle X\rangle_t:


d\langle X\rangle_t = \sigma^2X(t)^2dt


Plugging this back into our previous equation, we get:


d\langle X,B\rangle_t = X(t)dt + B(t)\sigma^2X(t)^2dt + \sigma^2X(t)^2dtdt


Simplifying further, we get:


d\langle X,B\rangle_t = X(t)dt + \sigma^2X(t)^2dt


I hope this helps you progress further with the problem. Good luck!