# Uniform integrability of a martingale

Let $$\left(X_{n}\right)_{n \geq 0}$$ is a martingale with $$X_{0}=0$$. Assume

$$\sum_{n=1}^{\infty} E\left(\left(X_{n}-X_{n-1}\right)^{2}\right)<\infty .$$

Show that the martingale is uniformly integrable.

I have a result that says that a martingale $$X=(X)_{n\geq 0}$$ is bounded ($$\sup_nE(X_n^2)<\infty$$) if and only if

$$\sum_{n=1}^{\infty} E\left(\left(X_{n}-X_{n-1}\right)^{2}\right)<\infty$$

However, this says nothing about the fact that if the martingale is uniformly integrable.

I have a theorem saying that if $$\sup_{X\in \mathcal{H}}E(|X|^p)<\infty$$ for some $$p>1$$, then the class of random variables $$\mathcal{H}$$ is uniformly integrable. Is this applicable here? Thanks.

Yes this is directly applicable!

Let us restate this useful criterion:

Basically boundedness in $$L^p$$ for $$p>1$$ (note that $$p=1$$ does not work) implies that a family of random variables $$(X_i)_{i \in \mathbb N}$$ is uniformly integratable, more precisely:

Lemma: Let $$(X_i)_{i \in \mathbb N}$$ be a family of random variables such that $$\sup_{i \in \mathbb N} \mathbb E[|X_i|^p] \leq C < \infty,$$ then $$(X_i)_{i \in \mathbb N}$$ is u.i.

Proof: Let $$p>1$$ and $$\sup_{i \in \mathbb N} \mathbb E[|X_i|^p] \leq C$$. Then observe that $$\mathbb E[|X_i| \mathbb 1_{|X_i| > K}] \leq \mathbb E\bigg[\frac{|X_i|^{p-1}}{K^{p-1}} |X_i| \mathbb 1_{|X_i|>K}\bigg] \leq \frac{1}{K^{p-1}} \mathbb E[|X_i|^p] \leq \frac{C}{K^{p-1}}.$$ Now taking the supremum over $$i \in \mathbb N$$ and $$K \to \infty$$ shows that $$\{X_i\}_{i \in \mathbb N}$$ is u.i.

So you are indeed already done:

Since you already know that your condition implies that $$\sup_{i \in \mathbb N} \mathbb E[|X_i|^2] \leq C < \infty$$ we know that $$\{X_i\}$$ is bounded in $$L^2$$ and by above this is sufficies.

Appendix on Boundedness: Let us first speak some basic observations:

1. If $$|X_n| \leq M$$ then $$\mathbb E[|X_n|] \leq M$$.
2. If $$M_1 \leq X_n \leq M_2$$ then $$\mathbb E[|X_n|] \leq |M_1| + |M_2|$$ (rewrite the integral in terms o positive and negative part of $$X_n$$
3. If $$M_1 \leq X_n \leq M_2$$ then $$\mathbb E[|X_n|] \leq (|M_1| + |M_2|)^p$$

For more I advise to check out the following question [https://math.stackexchange.com/questions/3433896/if-a-random-variable-is-bounded-does-it-mean-its-expectation-is-bounded].

Generally speaking one also knows that if $$X$$ is bounded in $$L^q$$ for $$1 \leq p \leq q$$ then it is also bounded in $$L^p$$ by applying Hölder's inequality. Usually the idea about this can maybe explained in terms of bounded second moments, i.e. that $$E[|X|^2] < \infty$$, since then not only the the absolute value $$|X|$$ is finite almost surely but also it's second moment $$E[|X|^2]$$, which in case that $$X \sim N(0,\sigma)$$ gives you automatically that also the variance of single random variable or the variance of a whole family of r.v. has finite variance. Does this help or do you have a more specific question about this?

• This makes sense however I am a bit lost on the boundedness. It is indeed bounded by my first argument but does the second argument imply that it is bounded in $L^2$? Commented Apr 4, 2023 at 22:04
• boundedness in general is really not intuitive for me at all. Commented Apr 4, 2023 at 22:16
• I am a bit confused, I believe in your text above you state already that you know that the given condition gives you $L^2$ boundedness. This is all you need for the exercise by above. I guess you are confused regarding why there are two notions of „boundedness“, i.e L^p boundedness and a random variable is bounded, shall I give some details on this? Commented Apr 5, 2023 at 8:48
• yes exactly that is what I am confused on. Commented Apr 5, 2023 at 10:47
• It is edited, I hope it is now a bit more clear, however the question is a bit vague, but I tried to shed some light on multiple aspects. Do you have a remaining question you can open a new question, and pin it here :) Commented Apr 5, 2023 at 15:44

Yes, it's applicable. You have $$\sup_n E(|X_n|^2)<\infty,$$ and $$2>1.$$

Perhaps the confusion is the "class" thing? Here the class in question is $$\{X_n: n\in \mathbb N\}.$$

Let A denote your condition $$\sum_{n=1}^{\infty} E\left[ (X_{n} - X_{n-1})^{2} \right] < \infty,$$ B(p) denote the condition $$\sup_{n} E\left( |X_{n}|^{p} \right) < \infty,$$ and C denote uniform integrability.

The results you have so far are: [A $$\iff$$ B(2)], and [B(p) $$\implies$$ C] for all $$p > 1$$. From the first result, we can get that [A $$\implies$$ B(2)] by ignoring the other direction. From the second result, we can get [B(2) $$\implies$$ C] by considering $$p = 2$$. Composing the implication, we get [A $$\implies$$ C].