# Proving $X_t = 1 + \int_0^t X_s \, dN_s$ is a supermartingale

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?

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.