Example 1.1.12. (Exponential Martingale) Suppose that $N$ is a semi-martingale on $\mathbb{R}$ with $N_0 = 0$. Consider the equation $$ X_t = 1 + \int_0^t X_s \, dN_s. $$ The solution is $$ X_t = \exp \left( N_t - \frac{1}{2} [N, N]_t \right). $$ If $N$ is a local martingale, then $X$ is called an exponential >martingale.

We leave the proof of the following properties of exponential martingales as an exercise:

1) $X$ is a nonnegative supermartingale; hence $\mathbb{E}[X_t] \leq 1$ for all $t \geq 0$.

2) $X$ is a martingale if and only if $\mathbb{E}[X_t] = 1$ for all $t \geq 0$.

3) If $\mathbb{E}[\exp(\alpha [N, N]_t)]$ is finite for some $\alpha > \frac{1}{2}$, then $\mathbb{E}[X_t] = 1$.

Regarding point 1) It is straightforward to see that if N is a martingal then

$$ E(X_t) = E(1 + \int_0^t X_s \, dN_s)=1+0=1. $$

However, when dealing with a semimartingale, we want to prove according to point one that X_t is a supermartingale.

If s<t:

$$E(X_t|\mathscr{F}_s)=E(1 + \int_0^t X_i \, dN_i|\mathscr{F}_s)=E(1 + \int_0^t X_i \, dA_i+\int_0^t X_i \, dM_i|\mathscr{F}_s)=?$$

I am using the Doob decomposition in the second equality. I know that since we want a super martingale as a result $A_t$ should be a decreasing process. The question now is how to proceed. It is not intuitive to me how one gets a super martingale once that would depend on $A_t$ being increasing or decreasing.


Can someone help me solve this problem and provide some insight into this kind of proof with semimartingales?

Thanks in advance!


1 Answer 1


First, note that these are all about the case where $X$ is an exponential martingale, so $N$ is a local martingale. This also implies $X$ is a local martingale.

For 1), let $(\tau_n) \rightarrow \infty$ be a localizing sequence for $X$. Since $X$ is non-negative, we can use the conditional Fatou lemma to conclude \begin{align*} \mathbb{E}[X_t|\mathcal F_s] &= \mathbb{E}[ \lim_{n \rightarrow \infty} X^{\tau_n}_t | \mathcal F_s] \\ &\le \lim_{n \rightarrow \infty} \mathbb{E}[ X^{\tau_n}_t | \mathcal F_s] \tag{conditional Fatou} \\ &= \lim_{n \rightarrow \infty} X_s^{\tau_n} \tag{$X^{\tau_n}$ is a martingale} \\ &= X_s, \end{align*} so $X$ is a supermartingale. Note that this actually shows that any non-negative local martingale is a supermartingale.

For 2), if $X$ is a martingale implies $\mathbb{E}[X_t] = X_0 = 1$ instantly, so we focus on the other direction. Fix $T > 0$ and suppose $\mathbb{E}[X_T] = 1$. Let $\tau \le T$ be a stopping time. By Doob's stopping theorem, we have $X_{\tau} \ge \mathbb{E}[X_T|\mathcal F_{\tau}]$, so \begin{align*} \mathbb{E}[X_\tau] &\ge \mathbb{E}[\mathbb{E}[X_T|\mathcal F_{\tau}]] \tag{Doob} \\ &= \mathbb{E}[X_T] \tag{Tower property} \\ &= 1. \end{align*}

On the other hand, since $X$ is a supermartingale, we know $\mathbb{E}[X_\tau] \le X_0 = 1$, so we conclude $\mathbb{E}[X_\tau] = 1$. Since $\tau$ was an arbitrary stopping time (on $[0,T]$), this proves $X$ is a martingale on $[0,T]$. Since $T>0$ was arbitrary, we have that $X$ is a martingale.

The third question is more involved and should be asked as a separate question.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .