Let $W = \{W_t : t\ge0\}$ be a standard Brownian motion on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, and let $f$ be a deterministic function such that $$ \int_0^tf^2(s)\,ds<\infty $$ for all $t\ge 0$. Show that the stochastic exponential $$ M_t = \exp\left(\int_0^tf(s)\,dW_s - \frac{1}{2}\int_0^tf^2(s)\,ds\right) $$ is a martingale.

I'm aware that this can be verified immediately using Novikov's criterion, for example, but I'm looking for a more direct proof than this. It's easy to demonstrate using Ito's lemma that $$ M_t = 1 + \int_0^tf(s)M_s\,dW_s, $$ and so the result can be boiled down to showing that $$ \mathbb{E}\left[\int_0^tf^2(s)M_s^2\,ds\right]<\infty. $$ This seems promising, but I'm unsure how to proceed. I should mention that the subsequent part of the exercise I'm trying to solve asks to use the martingale property of $M_t$ to show that $$ \int_0^tf(s)dW_s\sim N\left(0,\int_0^tf^2(s)\,ds\right). $$ This is straightforward to do, but it suggests that the martingale property can be demonstrated without appealing to these facts. Any suggestions or references would be greatly appreciated.

  • $\begingroup$ I guess this should not make much sense since you need $\mathbb EM_s^2ds<\infty$ to prove your last line, which is just equivalently to $M$ being a martingale. $\endgroup$ Aug 25, 2018 at 1:04
  • $\begingroup$ Related: math.stackexchange.com/questions/1529726/…. $\endgroup$ Aug 25, 2018 at 1:09
  • 1
    $\begingroup$ If @saz isn't sure then we probably don't have much hope. $\endgroup$
    – Alex W
    Aug 25, 2018 at 15:30
  • $\begingroup$ For a deterministic $f$, $\int_0^t f(s)\, dW_s$ is Gaussian with mean zero and variance $\int_0^t f(s) ^2 ds$. Nothing else is needed. $\endgroup$
    – zhoraster
    Aug 25, 2018 at 15:56
  • 2
    $\begingroup$ @AlexW you are doing me too much honour. $\endgroup$
    – saz
    Aug 25, 2018 at 17:43

1 Answer 1


Itô's formula shows that

$$M_t = 1+ \int_0^t f(s) M_s \, dW_s \tag{1}$$

and this implies, in particular, that $(M_t)_{t \geq 0}$ is a local martingale. (Note that $(M_t)_{t \geq 0}$ has continuous sample paths, and therefore the stochastic integral on the right-hand side is well-defined.) On the other hand, it follows from the very definition that $M_t \geq 0$ for each $t \geq 0$. Since any non-negative local martingale is a supermartingale (see e.g. this question for details), we conclude that $(M_t)_{t \geq 0}$ is a supermartingale. Thus,

$$\mathbb{E}(M_t) \leq \mathbb{E}(M_0) \leq 1$$

which implies

$$\mathbb{E} \exp \left( \int_0^t f(s) \, dW_s \right) \leq \mathbb{E} \exp \left( \frac{1}{2} \int_0^t f(s)^2 \, ds \right)$$

for each $t \geq 0$. Replacing $f$ by $2f$ we find in particular that

$$\mathbb{E} \left| \exp \left( \int_0^t f(s) \, dW_s \right) \right|^2 = \mathbb{E}\exp \left( \int_0^t 2f(s) \, dW_s \right) \leq \exp \left( 2 \int_0^t f(s)^2 \, ds \right) < \infty,$$

and so

$$\mathbb{E}(M_t^2) \leq \exp \left( \int_0^t f(s)^2 \, ds \right).$$

Using this estimate and the fact that $f$ is deterministic, we can easily check that the stochastic integral $\int_0^t f(s) M(s) \, dW_s$ is a true martingale, and now $(1)$ shows that $(M_t)_{t \geq 0}$ is a martingale.


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