if $M$ is a UI - martingale then $M_t \rightarrow M_{\infty}$ in $L^1$

I'm trying to prove the following:

Let $$M$$ be a uniformly integrable martingale. Then there exists a random variable $$M_{\infty}$$ such that $$M_t \rightarrow M_{\infty}$$ in $$L^1$$.

This is what I have so far:

A UI martingale $$M$$ is clearly a $$L^1$$-martingale. Take, for example $$\epsilon = 1$$. Then, by definition (of UI-martingale), it exists $$K_1$$ such that $$\sup_{t \geq 0} E|M_t|< 1+K_{1}.$$ Hence, by the martingale convergence theorem, there exists $$M_{\infty} \in L^1$$ such that $$M_t \rightarrow M_{\infty}$$ a.s. Now, to show $$E|M_t-M_{\infty}| \rightarrow 0$$ as $$t \rightarrow \infty$$, I guess I have to use the dominated convergence theorem but I can't find any bound. If it was $$L^2,$$ I could use Doob's $$L^p$$-inequality to find the bound, but we are in $$L^1,$$ so I don't know how to continue. How can I finish the proof? Is there another way to prove it?

• If $Y_n \to Y$ a.s and $(Y_n)$ is uniformly integrable then $E|Y_n-Y| \to 0$. Ref. Chung's book. Commented Jan 2, 2020 at 0:05
• Thanks, I know the result is true but is there a proper proof in that book?
– UBM
Commented Jan 2, 2020 at 0:26
• It is amazing that we can take $\epsilon=1$ to claim something that has no apparent connection to $\epsilon$. (Similarly: "Take $\epsilon=1$. Then, we know my glasses are in the bathroom.") Commented Jan 2, 2020 at 1:47
• To show @KaviRamaMurthy proposition, we can start by noticing that $X_n = |Y_n - Y|$ is again u.i. then we have $E[X_n] = E[X_n1_{X_n < R}] + E[X_n1_{X_n \geq R}]$. For all $\epsilon > 0$ and for $R$ big enough, the 2nd term will be lower than $\epsilon/2$ by u.i. And again with that $R$, we have that the first term is lower than $\epsilon/2$ using the probablity convergence for $n$ big enough. Commented Jan 2, 2020 at 2:21
• @Sesame: Thank you. It may be a silly question but, why does $Y_n \rightarrow Y$ (in probability) implies $E[X_n 1_{X_n < R}]< \frac{\epsilon}{2}$?
– UBM
Commented Jan 2, 2020 at 19:04

Truncate, using UI, to be able to use DCT. In more detail, given $$\epsilon>0$$ use the fact that $$(M_n-M_\infty)$$ is UI (why?) to choose $$K$$ so large that $$E[|M_n-M_\infty|; |M_n-M_\infty|>K]<\epsilon$$. By DCT and pointwise convergence, $$\lim_nE[|M_n-M_\infty|; |M_n-M_\infty|\le K]=0$$. Therefore $$\limsup_nE|M_n-M_\infty|\le\epsilon$$.

• So given $\epsilon,$ from the fact that $|M_t-M_{\infty}|$ is UI, we get two things: 1) $E[|M_t-M_\infty|; |M_t-M_\infty|>K_{\epsilon}] < \frac{\epsilon}{2};$ 2) provides the bound $K_{\epsilon}$ that allows me to use DCT in order to get $E[|M_t-M_\infty|; |M_t-M_\infty| \leq K_{\epsilon}] < \frac{\epsilon}{2}.$ Hence, $E[|M_t-M_\infty|]=E[|M_t-M_\infty|; |M_t-M_\infty|>K_{\epsilon}]+E[|M_t-M_\infty|; |M_t-M_\infty| \leq K_{\epsilon}]< \epsilon.$ Is my understanding correct?
– UBM
Commented Jan 2, 2020 at 18:53
• Almost. The quantity $E[|M_t-M_\infty|; |M_t-M_\infty|\le K_\epsilon]$ is small for large $t$, because of DCT. Commented Jan 2, 2020 at 20:16
• Ok, thank you very much. One last thing, I've just found in Rogers and Williams a proof of this result. They use the fact that $M_t \rightarrow M_{\infty}$ a.s. implies convergence in probability and they use later what they called the bounded-convergence theorem. Their proof seems unnecessary long comparing with yours. I wonder why. It may be because we haven't proved that $|M_t - M_{\infty}|$ is UI? Any idea of how to prove $|M_t - M_{\infty}|$ is UI?
– UBM
Commented Jan 2, 2020 at 20:34
• First use Fatou to show that $M_\infty$ is integrable. Then just use triangle inequality: because $|M_t-M_\infty|\le|M_t|+|M_\infty|$ you have $$\{|M_t-M_\infty|>K\}\subset\{|M_t|>K/2\}\cup\{|M_\infty|>K/2\},$$ so $$E[|M_t-M_\infty|; |M_t-M_\infty|>K]\le E[|M_t|; |M_t-M_\infty|>K]+E[|M_\infty|; |M_t-M_\infty|>K].$$ Each term on the right can be made small (uniformly in $t$) by choosing $K$ large: the first because $(M_t)$ is UI, the second because $M_\infty$ is integrable. Commented Jan 3, 2020 at 18:00
• Shouldn't it be $E[|M_t - M_{\infty}|;|M_t - M_\infty|>K] \leq E[|M_t - M_{\infty}|;|M_t|>\frac{K}{2}] + E[|M_t - M_{\infty}|;|M_\infty|> \frac{K}{2}]$ instead?
– UBM
Commented Jan 3, 2020 at 19:34

Recall that if $$X_n\to X$$ in probability, then there exists a subsequence $$\{n_k\}$$ such that $$X_{n_k}\to X$$ a.s. For each positive integer $$k$$, we have that $$\lim_{n\to\infty} \mathbb P(|X_n-X|>2^{-k})=0$$. So for each $$k$$, we may find $$n_k$$ such that $$\mathbb P(|X_{n_k}| > 2^{-k})\leqslant 2^{-k}$$, and consequently $$\sum_{k=1}^\infty \mathbb P(|X_{n_k}-X|>2^{-k})\leqslant \sum_{k=1}^\infty 2^{-k}<\infty.$$ Then by the Borel-Cantelli lemma, $$\mathbb P\left(\limsup_{n\to\infty}\left\{|X_{n_k}-X|>2^{-k}\right\} \right) = 0,$$ from which it follows that $$X_{n_k}\to X$$ a.s.

Since $$X_{n_k}\to X$$ a.s. we have by Fatou's lemma $$\mathbb E[|X|] = \mathbb E\left[\liminf_{k\to\infty}|X_{n_k}|\right]\leqslant \liminf_{k\to\infty} \mathbb E[|X_{n_k}|].$$

A sequence of random variables $$\{X_n\}$$ is said to be uniformly integrable if $$\sup_n\lim_{K\to\infty} \mathbb E[|X_n|\mathsf 1_{\{|X_n|>K\}}] = 0.$$ This implies that $$\sup_n\mathbb E[|X_n|]<\infty$$. Now, we show that for every $$\varepsilon>0$$, there exists $$\delta>0$$ such that for any event $$E$$, $$\mathbb P(E)<\delta\implies \sup_n\mathbb E[|X_n|\mathsf 1_E]<\varepsilon.\tag1$$ Write $$E_n = \{|X_n|>K\}$$. Then $$\mathbb E[|X_n|\mathsf 1_E = \mathbb E[|X_n|(\mathsf 1_{E\cap E_n}+\mathsf 1_{E\setminus E_n})] \leqslant \mathbb E[|X_n|\mathsf 1_{E_n}] + K\mathbb P(E).$$ Given $$\varepsilon>0$$, there exists $$K>0$$ such that $$\sup_n\mathbb E[|X_n|\mathsf 1_{E_n}]<\frac\varepsilon2$$. Setting $$\delta=\frac\varepsilon{2K}$$, we see that $$(1)$$ holds.

Now from $$\mathbb E[|X|]\leqslant \liminf_{k\to\infty}\mathbb E[|X_{n_k}|]$$ and $$(1)$$, we have that $$\mathbb E[|X|]<\infty$$, i.e. $$X\in L^1$$. The inequality $$|X_n-X|^r \leqslant 2^r (|X_n|^r +|X|^r),\quad r>0$$ shows that the sequence $$\{|X_n-X|\}$$ is uniformly integrable (check this!) and so for for each $$\varepsilon>0$$, \begin{align} \mathbb E[|X_n-X|] & = \mathbb E[|X_n-X|\mathsf 1_{\{|X_n-X|>\varepsilon\}}] + \mathbb E[|X_n-X|\mathsf 1_{\{|X_n-X|\leqslant\varepsilon\}}]\\ &\leqslant \mathbb E[|X_n-X|\mathsf 1_{\{|X_n-X|>\varepsilon\}}] + \varepsilon. \end{align} Since $$\{|X_n-X|\}$$ is uniformly integrable, $$\lim_{n\to\infty} \mathbb E[|X_n-X|\mathsf 1_{\{|X_n-X|>\varepsilon\}}] = 0,$$ from which the result holds.

To answer @UBM's question, we have $$\sup_n \mathbb E[|X_n-X|] \leqslant 2( \sup_n\mathbb E[X_n] + \mathbb E[|X|)<\infty$$ using $$r=1$$. Pick $$M>0$$ such that $$2( \sup_n\mathbb E[X_n] + \mathbb E[|X|). Now for each $$\varepsilon>0$$ we may pick $$\delta<\frac\varepsilon M$$ such that for any event $$E$$, $$\mathbb P(E)<\delta\implies \sup_n \mathbb E[|X_n-X|\mathsf 1_E]\leqslant M\mathbb P(E) < M\frac\varepsilon M = \varepsilon.$$ This implies that $$\{|X_n-X|\}$$ is uniformly integrable.

• Note that we did not need the assumption that the process was a martingale. Commented Jan 2, 2020 at 2:51
• The comment by sesame is almost an entire proof. Commented Jan 2, 2020 at 5:20
• @Math1000: Thank you very much for all the detail but if you didn't assume that $M$ is UI martingale, what claim are you proving exactly?
– UBM
Commented Jan 2, 2020 at 19:32
• Also, what fact did you use to justify that $\mathbb E[|X_n-X|\mathsf 1_{\{|X_n-X|\leqslant\varepsilon\}}]< \varepsilon$?
– UBM
Commented Jan 2, 2020 at 19:36
• I'm a little bit lost because we already knew that $X_n \rightarrow X$ a.s. and that $X \in L^1.$ It seems that most of your answer is about proving those two facts. Maybe I'm missing something?
– UBM
Commented Jan 2, 2020 at 19:43

From Rogers and Williams (1st Volume).

We will need the following two results:

Proposition 1. Suppose that $$X \in L^1$$. Let $$\epsilon > 0.$$ Then there exists $$K$$ such that $$E[|X|;|X|>K] < \epsilon.$$

Theorem 2. (Bounded-Convergence Theorem) Let $$(X_n)$$ be a sequence of of random variables, and let $$X$$ be a random variable. Suppose that $$X_n \rightarrow X$$ in probability and that, for some $$K \in [0, \infty),$$ we have for every $$n$$ and $$\omega,$$ that $$|X_n(\omega)| \leq K.$$ Then $$E[ |X_n -X |] \rightarrow 0.$$

A UI martingale $$M$$ is clearly a $$L^1$$-martingale. Take, for example $$\epsilon = 1$$. Then, by definition (of UI-martingale), for all $$t \geq 0,$$ there exists $$K_1$$ such that $$E|M_t| = E[|M_t|;|M_t|>K_1] + E[|M_t|;|M_t| \leq K_1]= 1 + K_1.$$ Hence, $$\sup_{t \geq 0}E|M_t| \leq 1+ K_1$$ and $$M$$ is a $$L^1$$-martingale. By the martingale convergence theorem, there exists $$M_{\infty} \in L^1$$ such that $$M_t \rightarrow M_{\infty}$$ a.s., which implies that $$M_t \rightarrow M_{\infty}$$ in probability.
Next, for $$K \in [0,\infty),$$ define the functions $$g_K: \mathbb R \rightarrow [-K,K]$$ as follows: $$g_K(x):= \begin{cases} K \quad \text{ if } x>K; \\ x \quad \text{ if } |x| \leq K; \\ -K \quad \text{ if } x
Now, using the family of functions $$g_K,$$ we will prove that $$M_t \rightarrow M_\infty$$ in $$L^1$$.
Let $$\epsilon > 0$$ and choose $$K$$ large enough so \begin{align*} E|g_K(M_t)-M_t| &< \frac{\epsilon}{3} \tag*{(since M is a UI-martingale)} \\ E|g_K(M_\infty)-M_\infty| &< \frac{\epsilon}{3} \tag*{(by Proposition 1)} \end{align*}
Moreover, note that the functions $$g_K$$ satisfy that for all $$x,y \in \mathbb R,$$ $$|g_K(y)-g_K(x)| \leq |y-x|.$$ Hence, given $$K$$ from the step before, we have that for all $$t \geq 0$$ $$|g_K(M_\infty)-g_K(M_t)| \leq |M_\infty-M_t|,$$ which implies that $$g_K(M_t) \rightarrow g_K(M_\infty) \text{ a.s. }$$ and also, $$g_K(M_t) \rightarrow g_K(M_\infty)$$ in probability. Hence, by Theorem 2, for large enough $$t$$ we have $$E|g_K(M_\infty)-g_K(M_t)|< \frac{\epsilon}{3}.$$ Therefore, by the triangular inequality \begin{align*} E|M_\infty - M_t| &= |M_t - g_K(M_t) + g_K(M_t) - g_K(M_\infty) + g_K(M_\infty) - M_\infty| \\ &\leq |M_t - g_K(M_t)| + |g_K(M_t) - g_K(M_\infty)| + |g_K(M_\infty) - M_\infty| \\ &< \epsilon. \end{align*}