How can we show that the Dolean Dade exponential is a supermartingale without using the Ito's lemma

Question

How can we show that the Doleans Dade exponential is a supermartingale without using Ito's lemma. I am trying to solve problem 2.2.8 in Chapter 3 of Karatzas and Shreve. Where $\mathcal{P}$ is the collection of equivalence classes of all measurable adapted processes $X=\{X_t, \mathcal{F}_t, 0 \leq t < \infty\}$ satisfying $$ P\bigg[\int_0^T X_t^2 d \langle M \rangle_t < \infty\bigg] =1 \text{ for every } T \in [0, \infty) $$

I tried to expand the exponential but i couldn't go very far. Any hints would be appreciated!

Answer 1

Hints:

1. Show the assertion for simple integrands of the form $$X(t,\omega) = \sum_{j=1}^n \varphi_j(\omega) 1_{[s_j,s_{j+1})}(t)$$ where $n \in \mathbb{N}$, $0 \leq s_1 \leq \ldots \leq s_{n+1}$ and $\varphi_j \in L^{\infty}(\mathcal{F}_{s_j})$.
2. Extend the assertion to progressively measurable processes $X$ satisfying the integrability condition $$\forall t \geq 0: \quad \mathbb{E} \left( \int_0^t X_s^2 \,ds \right) < \infty. \tag{1}$$ To this end, choose a sequence of simple processes $(X^n)_{n \in \mathbb{N}}$ such that $X^n \to X$ in $L^2(\lambda_T \otimes \mathbb{P})$. Then a suitable subsequence of $$M_t^n := \exp \left( \int_0^t X_s^n \, dW_s - \frac{1}{2} \int_0^t (X_s^n)^2 \, ds \right)$$ converges almost surely to $$M_t := \exp \left( \int_0^t X_s \, dW_s - \frac{1}{2} \int_0^t X_s^2 \, ds \right).$$ Conclude from Fatou's lemma and Step 1 that $$\mathbb{E}(M_t \mid \mathcal{F}_s) \leq M_s$$ for any $s \leq t$.
3. Extend the assertion to processes $X \in \mathcal{P}$: Pick a localizing sequence $(\tau_n)_{n \in \mathbb{N}}$ of stopping times such that $(X_{t \wedge \tau_n})_{t \geq 0}$ satisfies $(1)$ for each $n \in \mathbb{N}$. Use Step 2 and Fatou's lemma to show that $$N_t := \exp \left( \int_0^t X_s \, dW_s - \frac{1}{2} \int_0^t X_s^2 \, ds \right)$$ is integrable for each $t \geq 0$. Apply once more Fatou's lemma to prove the supermartingale property.