# Can we prove directly that $M_t$ is a martingale

Suppose we define the stochastic process $$M_t:=e^{\int_0^t\phi_s dW_s -\frac{1}{2}\int_0^t\phi_s^2ds}$$ where $\phi\in L^2[0,T]$, $t\in [0,T]$. Note that $M_t$ is just the stochastic exponential of $\int_0^\cdot \phi_s dW_s$, where $W$ is a Brownian Motion. I am able to prove that $M_t$ is a (UI)-martingale using two different approaches:

1. Novikovs Condition
2. Using elementary measure theoretic induction

However I was wondering if we can prove this directly, i.e.

$$E[M_t|\mathcal{F}_s]=M_s$$

with using that $W$ is a Brownian Motion w.r.t to $(\mathcal{F}_t)$?

-

This may not be exactly what you're looking for, but here's a proof which at least uses the specific form of the process (being the exponential local martingale of an integral of a square-integrable deterministic process with respect to a Brownian motion):

First off, in order to make things fit better with the standard framework, I'll assume that $\phi$ is a mapping from $[0,\infty)$ to $\mathbb{R}$ with the property that $\int_0^t \phi(s)^2ds$ is finite for all $t\ge0$. Define $N_t = \int_0^t \phi(s)dW_s$, we then have $M_t = \mathcal{E}(N)_t$, the exponential local martingale, and the objective is to prove that $\mathcal{E}(N)$ is a martingale (instead of just a local martingale). By some classical results, $\mathcal{E}(N)$ is a nonnegative supermartingale, and it is a martingale if and only if $E\mathcal{E}(N)_t = 1$ for all $t\ge0$. I don't know where you actually can find a proof of these claims, but they follow from applications of the optional sampling theorem and Fatou's lemma. The conclusion is that we need to show $E\mathcal{E}(N)_t=1$ for all $t\ge0$.

To do so, first note that $[N]_t = \int_0^t \phi(s)^2 ds$. We apply Itô's formula:

$\mathcal{E}(N)^2_t = 1 + 2\int_0^t \mathcal{E}(N)_sd\mathcal{E}(N)_s+[\mathcal{E}(N)]_t\\ =1 + 2\int_0^t\mathcal{E}(N)_sd\mathcal{E}(N)_s + \int_0^t \mathcal{E}(N)^2_sd[N]_s\\ =1 + 2\int_0^t\mathcal{E}(N)_sd\mathcal{E}(N)_s + \int_0^t \mathcal{E}(N)^2_s\phi(s)^2ds.$

As $\mathcal{E}(N)$ and $\mathcal{E}(N)\cdot \mathcal{E}(N)$ are continuous local martingales, they are locally bounded. Let $(T_n)$ be a localising sequence of stopping times such that both $\mathcal{E}(N)^{T_n}$ and $(\mathcal{E}(N)\cdot \mathcal{E}(N))^{T_n}$ are bounded martingales. By the martingale property of $(\mathcal{E}(N)\cdot\mathcal{E}(N))^{T_n}$, we then obtain

$E\mathcal{E}(N)_{t\land T_n}^2 = 1 + E\int_0^{t\land T_n}\mathcal{E}(N)_sd\mathcal{E}(N)_s + E\int_0^{t\land T_n}\mathcal{E}(N)^2_s\phi(s)^2 ds\\ =1 + E \int_0^t \mathcal{E}(N)^2_s\phi(s)^21_{[0,T_n]}(s)ds \le 1 + \int_0^t E\mathcal{E}(N)^2_{s\land T_n}\phi(s)^2ds$

where we also applied Tonelli's theorem and nonnegativity of $\mathcal{E}(N)$. Now consider a fixed $n\ge1$ and define $g_n(t) = E\mathcal{E}(N)^2_{t\land T_n}$. The above then states that

$g_n(t) \le 1+ \int_0^t g_n(s)\phi(s)^2ds$.

By a classical analysis lemma, Gronwall's lemma (See the Wikipedia article on Gronwall's inequality), we then obtain $g_n(t) \le \exp(\int_0^t\phi(s)^2ds)$. In other words, we have now shown

$E\mathcal{E}(N)^2_{t\land T_n}\le \exp(\int_0^t \phi(s)^2ds)$

for all $n\ge1$ and $t\ge0$. Now fix $t\ge0$. By the above, the family $(\mathcal{E}(N)_{t\land T_n})_{n\ge1}$ is then bounded in $\mathcal{L}^2$, therefore uniformly integrable. Furthermore, $\mathcal{E}(N)_{t\land T_n}$ converges almost surely to $\mathcal{E}(N)_t$. Combining this with uniform integrability, $\mathcal{E}(N)_{t\land T_n}$ converges in $\mathcal{L}^1$ to $\mathcal{E}(N)_t$, and so the means also converge. And as $\mathcal{E}(N)^{T_n}$ is a martingale, $\mathcal{E}(N)_{t\land T_n}=1$. We conclude that $E\mathcal{E}(N)_t = 1$, and so $\mathcal{E}(N)$ is a martingale.

-
Very nice! +1, thanks for this different proof. – user20869 Feb 8 '13 at 11:12

Another possibility in this context if $\phi$ is only $L^2$ and not a stochastic process is to use the fact that $\int_0^t\phi_sdW_s$ is a Wiener integral and so it is a gaussian process with known second moment equal to $\int_0^T \phi^2_sds$ from this the martingality follows.

Best regards

Edit with respect to the comment : Wiener integrals are indeed Gaussian processes (with indepdenent increments), the gaussian process here is $X_t =\int_0^t\phi_sdW_s$, not $M_t$, and $M_t=\mathcal{E}(X_t)$ (the Doléans-Dade exponential). Conditionning up to $s$ gives :

$E[M_t|\mathcal{F}_s]=E[M_s.(M_t/M_s)|\mathcal{F}_s]=M_s.E[(M_t/M_s)|\mathcal{F}_s]=M_s.E[\mathcal{E}(X_t-X_s)|\mathcal{F}_s]$ (1)

The law of $X_t-X_s$ given $X_s$ is a Gaussian random variable $N(s,t)$ with nul first moment and second moment equal to $\int_s^t \phi^2_udu$ so that you get $E[\mathcal{E}(X_t-X_s)|\mathcal{F}_s]=E[e^{N(s,t)}.exp(-\int_s^t \phi^2_udu)]=exp(-\int_s^t \phi^2_udu).E[e^{N(s,t)}]=1$
Now reintegrating this into (1) gives the result. It is not very elegant but I think it does the trick.

-
Thanks for your answer. I think you do not know that it is a Gaussian process. You know that $M_t$ is Gaussian. About the joint distribution you don't know anything. And how does the martingale property follows from there? – user20869 Feb 7 '13 at 8:59
hulik: M_t is not gaussian. – Did Feb 7 '13 at 9:49
@Did I mean lognormal! Sorry for the mistake. – user20869 Feb 7 '13 at 9:51
@TheBridge The only thing I do not know is, that $X_t-X_s$ is independent from $\mathcal{F}_s$. Can you give me a reference for that? – user20869 Feb 7 '13 at 15:36
@ hulik : As the process $X$ is Markovian conditionning with respect to $\mathcal{F}_s$ is equivalent to conditionning with respect to $X_s$, and from the fact that it is gaussian (wit hgausiian increment) process you only have to check that $E[X_s.(X_t-X_s)]=E[X_s].E[(X_t-X_s)](=0)$ I beleive. – TheBridge Feb 7 '13 at 16:23