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.
Question:
Can someone help me solve this problem and provide some insight into this kind of proof with semimartingales?
Thanks in advance!
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.