Let
- $(\Omega,\mathcal{A},\operatorname{P})$ be a probability space
- $\mathbb{F}$ be a Filtration on $(\Omega,\mathcal{A})$
- $X=(X_n)_{n\in\mathbb{N}_0}$ be a nonnegative $\mathbb{F}$-supermartingale
- $\tau$ be a $\mathbb{F}$ stopping time and $\operatorname{P}$-almost surely $\tau<\infty$
Moreover, let $$X_\tau(\omega):=X_{\tau(\omega)}(\omega)\;\;\;\text{for all}\; \omega\in\Omega$$ and $$\tau\wedge n:=\min\left\{\tau,n\right\}\;\;\;\text{for all}\;n\in\mathbb{N}_0$$
I want to prove $$\operatorname{E}\left[X_\tau\right]\le \operatorname{E}\left[X_0\right]<\infty$$
My ideas:
- $\operatorname{P}\left[\tau<\infty\right]=1\Rightarrow\operatorname{P}$-almost surely there is a $T\in\mathbb{N}$ with $\tau\le T$
- So, we've got for $\operatorname{P}$-almost every $\omega\in\Omega$ $$\left|X_{\tau(\omega)\wedge n}-X_{\tau(\omega)}\right|=0\;\;\;\text{for all }n\ge T$$ and thereby $\operatorname{P}$-almost surely $$\lim_{n\to\infty}X_{\tau\wedge n}=X_\tau$$
- Additionaly, we've got $\operatorname{P}$-almost surely $\tau\wedge n\le T$ and thereby $$\operatorname{E}\left[X_{\tau\wedge n}\right]\le\operatorname{E}\left[X_0\right]$$ (which can be easily proven)
- Since $X_\tau\ge 0$ and $\operatorname{P}$-almost everywhere $$X_\tau\le\max_{0\le n\le T}X_n\in\mathcal{L}^1(\operatorname{P})$$
- Thus, by Lebesgue's dominated convergence theorem and the lemma of Fatou, we've got $$\operatorname{E}\left[X_\tau\right]=\operatorname{E}\left[\lim_{n\to\infty}X_{\tau\wedge n}\right]\le \liminf_{n\to\infty}\operatorname{E}\left[X_{\tau\wedge n}\right]\le\operatorname{E}\left[X_0\right]<\infty$$
Is this a correct proof? If not, can you explain to me what I did wrong and come up with a correct proof?