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 !

  • $\begingroup$ 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). $\endgroup$
    – James Yang
    Jul 14, 2019 at 13:49

1 Answer 1



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.

  • $\begingroup$ 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? $\endgroup$
    – Tithus248
    Jul 14, 2019 at 18:35
  • $\begingroup$ No, because Itô integral is a continuous martingale if the integrand is $L^2(\Omega \otimes [0,T])$. $\endgroup$
    – Surb
    Jul 14, 2019 at 18:36
  • $\begingroup$ Why is $F‘‘(X_t) = (\mu + \frac{1}{2} \sigma^2) e^{\sigma B_t + \mu t} $ ? $\endgroup$
    – Tithus248
    Jul 16, 2019 at 1:39
  • $\begingroup$ @Tithus248: Why should it be that ? I used itô formula on $f(B_t,t)$ for $f(x,t)=e^{\sigma x+\mu t}$. $\endgroup$
    – Surb
    Jul 16, 2019 at 7:18

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