Show that $X$ is finite and $E(|X|)<\infty$

Question

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 $X=\lim _{n \rightarrow \infty} X_{n}$ (and it is finite) and $E(|X|)<\infty$

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I have shown that the martingale is uniformly integrable and thus $X=\lim _{n \rightarrow \infty} X_{n}$ with the convergence happening in $L^1$ (even though I also have shown it is bounded in $L^2$). Now, I do not know how to argue for that $X$ is finite a.s and that $E|X|<\infty$.

For the first one: I am thinking since the convergence happens in $L^1$, then $X$ must be finite?

For the second one: I have a result saying that if $L$ is a vector space and expectation is a linear map on it then $|X|\in L \Leftrightarrow X \in L$ but I am not sure if that helps me in any way.

I think a finite random variable implies a finite expectation, but the reverse implication is not necessarily true but if I could show that the absolut value of it is finite then part 2 is solved.

Answer 1

I will answer your question and along the way I will try to clear some of the confusions you have regarding basic properties of random variables and the $L^{p}$ norms.

Firstly , do you know the Martingale Convergence theorem?

It says that if $(X_{n})$ is a submartingale with $\sup_{n\in\Bbb{N}}E(X_{n}^{+})<\infty$ , then $X_{n}\xrightarrow{a.s.} X$ for some $X$ and $X\in L^{1}$ .

So in particular uniform integrability of $(X_{n})$ directly gives that $\sup_{n\in\Bbb{N}}E(X_{n}^{+})<\infty$ . But since you also have uniform integrability , then use the uniform integrability theorem to conclude $L^{1}$ convergence too from almost sure convergence.

Uniform Integrability Theorem

A sequence of rvs $X_{n}\xrightarrow{L^{1}} X$ iff and only if $X_{n}$ converges in probability to $X$ and $(X_{n})$ is uniformly integrable.

And yes of course, a random variable has to be finite $a.s.$ for $E(|X|)<\infty$ .

Here finiteness is not the same as boundedness. A bounded random variable has finite expectation. A finite $a.s.$ random variable means that $X$ cannot take the value infinity with positive probability . For example consider your favourite random variables like Normal variates or exponential variates , Poisson variates or even Cauchy variates . Bounded random variables are those which are bounded almost surely. That is there exists some $M>0$ and some set $A$ with $P(A)=0$ such that $\sup_{\omega\in A^{c}}|X(\omega)|<M$ . Examples of these are Bernoulli random variables as they take value only $0$ and $1$. There are many many others and are easy to think of.

But finiteness almost surely does NOT guarantee finite expectation.

Cauchy variates are a good example of this or consider any real valued function from $[0,1]$ whose integral is not finite like $\frac{1}{x}$ .

On the other hand if expectation is finite, then the Random variable has to be finite almost surely.

That is $P(\{\omega:X(\omega)=\infty\})=0$ Because otherwise if the above probability is positive say $=p>0$ , you consider sets of the form $A_{n}=\{\omega:|X(\omega)|>n\}$ , then all $A_{n}$'s will have positive measure and in particular $E(|X|)\geq np$ for all $n\in\Bbb{N}$. Hence $E(|X|)=\infty$ which would contradict finiteness of expectation.

Now what you have shown is convergence in $L^{1}$ . This in particular implies that $E(|X|)=||X||_{L^{1}}<\infty$ .

By the way, the result that you are using to conclude $L^{2}$ boundedness is given explicitly here

What this does is show that $(X_{n})$ is uniformly bounded in $L^{2}$ . Which means that $\sup_{n\geq 1}E(X_{n}^{2})<\infty$. If $\sup_{n\geq 1}E(|X_{n}|^{1+p})<\infty$ then $(X_{n})$ is uniformly integrable. So in particular $L^{2}$ boundedness(uniformly) implies the sequence is uniformly integrable.

Answer 2

If you has convergence in $L^2$ so you has convergence in Probability. So you can use the following result: Suppose that $\mathbb{E}[|X_{n}|]<\infty$ for all $n$. If $X_{n}\to X$ in probability , then the following are equivalent:

1.$\{X_{n}:n\geq 0\}$ is uniformly integrable.

2.$X_{n} \to X$ in $L^{1}$.

3.$\mathbb{E}[|X_{n}|] \to \mathbb{E}[|X|] < \infty$