Let $X_0^i$ be iid random variables with law $\mu_0$ for $i=1,\dots, N$. Let $\{X_t^i\}_{1\le i\le N}$ be satisfied as the following OU process: for $i=1,\dots, N$: $$ dX_t^i=-X_t^idt+\sqrt{2}dB_t^i. $$

Why can we say $$ \exp\left(\sum_{i=1}^N\left(\int_0^t X_s^i dB_s^i-\frac{1}{2}\int_0^t (X_s^i)^2ds\right)\right) $$ is a martingale?

It seems that it follows from the result that

[textbook by Le Gall] for $Z_t=\int_0^t f(s)dB_s$ where $f(s)\in L^2$, then $$ \exp(Z_t-\frac{1}{2}[Z]_t)=\exp(\int_0^t f(s)dB_s-\frac{1}{2}\int_0^t f^2(s)ds) $$ is a martingale.

But I have no idea why the first display is true because that is to take $$ Z_t=\sum_{i=1}^N\int_0^t X_t^i dB_t^i. $$

By https://en.wikipedia.org/wiki/Ornstein%E2%80%93Uhlenbeck_process or Ito formula, we can get $$ X_t^i=X_0^ie^{-t}+\sqrt{2}\int_0^t e^{-(t-s)}dB_s^i. $$

  • $\begingroup$ Use the multi dimensional Ito formula. In general those exponentials are only local martingales. $\endgroup$
    – Kurt G.
    Mar 4 at 6:12
  • $\begingroup$ @KurtG. Yes. But I read the lecture note, and it says that they are martingale... I am not sure the reason. $\endgroup$
    – Hermi
    Mar 4 at 7:07
  • $\begingroup$ I do not want to give a lecture here from scratch on stochastic calculus. I said in general. For a stochastic integral w.r.t. a BM to be a martingale the integrand has to satisfy certain conditions. Standard stuff you can find in zillions of books. My favourite: Karatzas & Shreve. $\endgroup$
    – Kurt G.
    Mar 4 at 7:33
  • $\begingroup$ @KurtG. So how about the specific in my question? Is that a martingale? $\endgroup$
    – Hermi
    Mar 4 at 7:44
  • $\begingroup$ @KurtG. Here is a result from WIKI: $$ X_t^i=X_0^ie^{-t}+\sqrt{2}\int_0^t e^{-(t-s)}dB_s^i. $$ $\endgroup$
    – Hermi
    Mar 4 at 8:50

1 Answer 1


As mentioned in the comments you need to verify whether the Novikov condition is met. This automatically guarantees that that exponential is indeed a martingale.

In this case, or in general for a OU process, we can reason like this: by Jensen inequality we have

$$ \mathbb{E} \bigg[ \exp{ \bigg( \frac{1}{2} \int_{0}^{T} X_{t}^{2} dt \bigg)} \bigg] \le \int_{0}^{T} \mathbb{E} \bigg[ \exp{ \bigg( \frac{X_{t}^{2}}{2} \bigg)} \bigg] dt = \int_{0}^{T} \mathbb{E} \bigg[ \exp{ \bigg( \frac{1}{2} (\mu_t + \sigma_{t}^{2} Z)^2 \bigg)} \bigg] dt $$

where $ Z \sim \mathcal{N}(0,1), \mu_t = X_0 e ^{-t}, \sigma_{t}^{2} = 1-e^{-2t}$

Since that last integral is convergent (compute it) the quantity is finite. Hence the NC is fulfilled.

----- Continuation

We can notice that the process we want to prove to be a martingale can be rewritten as

$$ Z_t = \exp\bigg( \int \sum_i X_i dB_s - \frac{1}{2} \int\sum X_{i}^{2} ds\bigg) $$

hence we need to prove the NC for the process $Y := \sum_i X_i \sim \mathcal{N}(A,B) $ with

$$ A := e^{-t} \sum_i X_{0}^{i}, \quad B:= n(1-e^{-2t}) $$

Building on the procedure I suggested above, we have a similar argument with a new guassian variable and we can test that.

  • $\begingroup$ But what am I asked is the sum of $X_t^I$. So dose your inequality still work? $\endgroup$
    – Hermi
    Mar 5 at 5:30
  • $\begingroup$ @Hermi I expanded my answer, I hope it is now clearer. $\endgroup$
    – finch
    Mar 5 at 12:03

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