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

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 $$X_t^i=X_0^ie^{-t}+\sqrt{2}\int_0^t e^{-(t-s)}dB_s^i.$$

• Use the multi dimensional Ito formula. In general those exponentials are only local martingales. Mar 4 at 6:12
• @KurtG. Yes. But I read the lecture note, and it says that they are martingale... I am not sure the reason. Mar 4 at 7:07
• 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. Mar 4 at 7:33
• @KurtG. So how about the specific in my question? Is that a martingale? Mar 4 at 7:44
• @KurtG. Here is a result from WIKI: $$X_t^i=X_0^ie^{-t}+\sqrt{2}\int_0^t e^{-(t-s)}dB_s^i.$$ Mar 4 at 8:50

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.

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