Suppose $X_n$ is a uniformly integrable submartingale and $T$ a stopping time. I wanna show that $X_T$ is integrable. To this end, I write $X_T=\lim_{n\to \infty}\sum_{k=1}^n X_k \mathbb{1}_{\{T=k\}}$. Now I would like to apply the dominated convergence theorem, but I can't find a clearly integrable function that would upper bound each of the partial sums. Perhaps $\sup_n X_n^+$: but is the uniform integrability of $(X_n)$ enough to guarantee that $\sup_n X_n^+$ is integrable?
Integrable submartingale
martingalesprobabilityprobability theorystopping-timesuniform-integrability
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$$\sup_{n} E[|X_n| 1_{|X_n|>M} ] \leq \sup_{n} E[|X_n|] \leq \sup_{n} E[|X_n|^p],$$ using Jensen.
You didn't apply Jensen's inequality correctly; it should read
$$\sup_{n} E[|X_n| 1_{|X_n|>M} ] \leq \sup_{n} E[|X_n|] \leq \sup_{n} \left( E[|X_n|^p] \right)^{\color{red}{\frac{1}{p}}}.$$
[...] and the claim follows by letting $M \rightarrow \infty$.
No, it's not that simple. Letting $M \to \infty$ you get
$$\lim_{M \to \infty} \sup_n \mathbb{E}(|X_n| 1_{|X_n|>M}) \leq \sup_{n \in \mathbb{N}} \|X_n\|_p,$$
but that's not good enough; you have to show that the limit equals $0$. Hint for this problem: Use Markov's inequality, i.e.
$$\mathbb{E}(|X_n| 1_{\{|X_n|>M}) \leq \frac{1}{M^{p-1}} \mathbb{E}(|X_n|^p 1_{|X_n|>M}) \leq \frac{1}{M^{p-1}} \mathbb{E}(|X_n|^p).$$
Define $$M_0:=\max_{n \in N} |X_n|.$$ Then we have $$E[|X_n| 1_{|X_n|>M_0}]= E[|X_n|\cdot 0 ] = 0,$$
No this doesn't work, because $M_0$ depends on $\omega$. Unfortunately, this means that your approach fails. Hint for this one: Using e.g. the dominated convergence theorem check first that the set $\{f\}$ is uniformly integrable. Extend the approach to finitely many integrable random variables.
When $E[\sup_n |X_n|] < \infty$, then the sequence is uniformly integrable.
Hint: By assumption, $Y := \sup_n |X_n|$ is integrable and $|X_n| \leq Y$ for all $n \in \mathbb{N}$. Consequently,
$$\mathbb{E}(|X_n| 1_{|X_n|>M}) \leq \mathbb{E}(|Y| 1_{|Y|>M}) \qquad \text{for all $M>0$ and $n \in \mathbb{N}$.}$$
Now use the fact that $\{Y\}$ is uniformly integrable (see question nr. 2).
The martingale $M_t$ is not uniformly integrable.
If it were, then by standard martingale facts, it would converge a.s. and in $L^1$ to some $M_\infty$. (An $L^1$-bounded martingale always converges a.s., and if it's uniformly integrable, then the convergence is also $L^1$ by Vitali.) In particular we would have $E[M_\infty] = \lim E[M_t] = 1$.
However, this martingale converges a.s. to zero, and that's a contradiction.
One way to see $M_t \to 0$ is to note, after simplifying, that $$M_t = \exp\left(\theta \sigma W_t - \frac{\theta^2 \sigma^2 t}{2}\right).$$ By the strong law of large numbers, $W_t / t \to 0$ a.s., and hence $$\frac{1}{t} \left(\theta \sigma W_t - \frac{\theta^2 \sigma^2 t}{2}\right) \to -\frac{\theta^2 \sigma^2}{2} < 0$$ which implies $$\theta \sigma W_t - \frac{\theta^2 \sigma^2 t}{2} \to -\infty.$$
Best Answer
Note that by uniform integrability, there exists a random variable $X_\infty$ such that $X_n\to X_\infty$ almost surely and in $L^1$. Moreover, for each non-negative integers $\ell$ and $k$, the submartingale property gives $$ \mathbb E\left[\lvert X_T\rvert\mathbf{1}_{\{T=k\}}\right]=\mathbb E\left[\lvert X_k\rvert\mathbf{1}_{\{T=k\}}\right]\leqslant \mathbb E\left[\lvert X_{k+\ell}\rvert\mathbf{1}_{\{T=k\}}\right] $$ and the convergence of $X_{k+\ell}$ to $X_\infty$ in $L^1$ shows that $$ \mathbb E\left[\lvert X_T\rvert\mathbf{1}_{\{T=k\}}\right]\leqslant \mathbb E\left[\lvert X_\infty\rvert\mathbf{1}_{\{T=k\}}\right].$$ Summing all the inequalities over $k$ and using integrability of $X_\infty$ finishes the proof.