# Geometric Brownian motion, Itô formula

Let $$B_t$$ be a Brownian motion, then

$$X_t:=e^{\sigma B_t + \mu t}$$ is a martingale iff $$\mu = - \frac{\sigma^2}{2}$$.

I already know how to compute this claim, but I am trying to solve it via Itô formula. Here I give the Itô formula:

$$F : \mathbb{R} \to \mathbb{R}$$ twice continuously differentiable and $$X$$ a continuous semimartingale. Then

$$F(X_t) - F(X_0) = \int_0^t F'(X_s) dX_s + \frac{1}{2} \int_0^t F ''(X_s) d[X]_s$$ a.s. for all $$t \geq 1$$.

How is it possible ? Thanks in advance !

• Look at $X_t$ as $f(t, B_t)$ and apply Ito on $f$. Then try to find a condition where the finite variation part becomes $0$ (the $dt$ part). Jul 14, 2019 at 13:49

Hint

Using Itô formula, if $$f(x,t)=e^{\sigma x+\mu t}$$ $$X_T=1+\int_0^T \left(\mu+\frac{1}{2}\sigma ^2\right)e^{\sigma B_t+\mu t}\,\mathrm d t+\int_0^T\sigma e^{\sigma B_t+\mu t}\,\mathrm d W_t.$$

So, $$X_t$$ is a martingale if $$\int_0^T \left(\mu+\frac{1}{2}\sigma ^2\right)e^{\sigma B_t+\mu t}\,\mathrm d t=0\quad \text{and}\quad \sigma e^{\sigma B_{\cdot }+\mu .}\in L^2(\Omega \otimes [0,T]).$$

I let you conclude.

• I understand that under this conditions we can conclude now that $\mu$ has to be equal to $- \frac{1}{2} \sigma^2$, but why it follows that $X_t$ is a martingale? Is it because $\sigma e^{\sigma B_t + \mu}$ is square integrable? Jul 14, 2019 at 18:35
• No, because Itô integral is a continuous martingale if the integrand is $L^2(\Omega \otimes [0,T])$.
– Surb
Jul 14, 2019 at 18:36
• Why is $F‘‘(X_t) = (\mu + \frac{1}{2} \sigma^2) e^{\sigma B_t + \mu t}$ ? Jul 16, 2019 at 1:39
• @Tithus248: Why should it be that ? I used itô formula on $f(B_t,t)$ for $f(x,t)=e^{\sigma x+\mu t}$.
– Surb
Jul 16, 2019 at 7:18