# Geometric Brownian Motion (Physics)

I am visiting a lecture for stochastic methods for physicists and for the current exercise sheet i need to solve the following SDE $$dX=\mu X dt +\sigma XdW(t)$$

with initial condition $$X(0)=X_0$$ for Ito and Stratonovich. I know that there are already a lot of topics about the same question in this forum but I can't apply the solutions that are used there.

To solve the above SDE I need an Ansatz $$f(X,t)=\log X$$ and apply Ito's formula. In The lecture we defined this as $$df(X,t)= \left( \frac{\partial f}{\partial t}+ a\frac{\partial f}{\partial x} + \frac{b^2}{2}\frac{\partial^2 f}{\partial x^2}\right) dt + b\frac{\partial f}{\partial x}dW$$

Applying this to $$f=\log X$$ i get $$d(\log X) = \frac{a}{X}dt - \frac{b^2}{2X^2}dt + \frac{b}{X}dW$$

which looks completely different from other results that were this discussed in this forum (for example https://math.stackexchange.com/q/873728). There they have a $$dX$$-term after applying the lemma for which they then substitute the initial SDE.

Could someone please explain to me what I am doing wrong.

• The formula from your lecture for $df(X,t)$ is for the case $dX = \mu dt +\sigma dW(t)$. There must be a more general formula in your notes. – Raskolnikov Jul 12 at 11:07
• We derived this formula for $dX=a(x,t)dt+b(x,t)dW$. If i compare that with the SDE $a=\mu X, b=\sigma X$ i get $d(log X) = \mu dt - \frac{\sigma^2}{2}dt + \sigma dW = (\mu-\frac{\sigma^2}{2})dt+\sigma dW$. I cant solve this because i dont know the Integral of $\int_0^t dW$ – Miradius Jul 12 at 11:17
• If im using Ito calculus then $\int_{t_0}^t dW= \lim_{n\to\infty}\sum_{i=n}^n (W(t_i)-W(t_{i-1}))$. Im note sure how to compute this sum.. – Miradius Jul 12 at 11:52
• Surely $\int_0^t dW = W(t) - W(0)$ ? – Raskolnikov Jul 12 at 12:38

You are not doing anything wrong and your derivation is correct so far. I think you're facing difficulties to understand the steps which were used from application of Ito's lemma to the derivation of the intermediate result which you have shared.

Below is an intermediate result after some simplifications, but not the actual form of the Ito's lemma.

$$df(X,t) = \left(\frac{\partial f}{\partial t} + a \frac{\partial f}{\partial x} + \frac{b^2}{2} \frac{\partial^2 f}{\partial x^2}\right)dt + b \frac{\partial f}{\partial x}dW$$

Here is the raw form of Ito's formula. Also review the concept of quadratic variation of a process from your notes. $$[X,X]_t$$ denotes the quadratic variation process of the stochastic process $$X_t$$.

$$df(X,t) = \frac{\partial f}{\partial t}dt + \frac{\partial f}{\partial x} dX_t +\frac{\partial^2 f}{\partial x^2} d[X,X]_t$$

Above form is what you referred to in your question which has $$dX$$-term. For the case where $$X_t$$ follows the below dynamics,

$$dX_t = a(x,t)dt+b(x,t)dW$$

you can plug this to the original form of Ito's formula along with the fact that $$d[X,X]_t = (b(X_t,t))^2d[W,W]_t = (b(X_t,t))^2dt$$ , to get the intermediate result.

Also, as you derived in the comments, for the particular problem, you have,

$$d(\ln X)=(\mu−\frac{\sigma^2}{2})dt+\sigma dW$$

You can now integrate above, between two limits of integration, $$t=0$$ and $$t=T$$ to get your desired result.