Bounded $L^2$ expectation implies convergence in $L^2.$

Question

In probability at least four different types of convergence are considered:

$a)$ Almost sure convergence

$b)$ Convergence in probability

$c)$ Convergence in $L^p$

$d)$ Convergence in distribution

It is known (see e.g. the book of Karr: "Probability", Springer) that $$a)\,\Rightarrow \, b)\,\Rightarrow \, d)$$ and that $$c)\,\Rightarrow \, b).$$ All the other implications are false and it is possible to find counterexamples to them here: Convergence types in probability theory : Counterexamples

Moreover, convergence in $L^p$ implies convergence in $L^q$ if $p>q$ and the space is finite.

Now, let ${x_n}$ be a family of bounded random variables defined over a given probability space that converge a.s. to a bounded random variable $x.$

The following statements are true or false?

1) If $\mathbb E(x_n^2)\leq \mathbb E(x^2)$ then $x_n$ converges to $x$ in $L^2.$

2) If $\mathbb E(x_n^2)=1$ $\forall n$ then $lim_{n\to \infty}\mathbb E(x_n)=\mathbb E(x).$

3) If $\mathbb E(x_n^2)=1$ and $\mathbb E(x_n^3)=c\in \mathbb R\,$ $\,\forall n$ then $x_n$ converges to $x$ in $L^2.$

Answer 1

I proved that 1) and 2) are true in the following way.

Since $x_n^2$ is a sequence of non-negative random variables, we can apply the Fatou Lemma getting:

$$\int_{\Omega}x^2=\int_{\Omega}lim\,inf_{n\to \infty}x_n^2\leq lim\,inf_{n\to \infty}\int_{\Omega}x_n^2 \leq lim\,inf_{n\to \infty} \int_{\Omega}x^2=\int_{\Omega}x^2 $$ where the first equality is due to the fact that $x_n^2$ converges a.s. to $x^2.$

Hence we have that $lim_{n\to \infty}\mathbb E(x_n^2)= \mathbb E(x^2).$ This implies that $x_n$ converges to $x$ in $L^2$ by the following well known proposition (see e.g. Jang: "Large Sample Techniques for Statistics", Springer, pag. 33)

"Suppose that $\mathbb E(|x_n|^p)<\infty$ $\forall n$ and that $x_n$ converges to $x$ in probability. Then the following are equivalent:

(i) $x_n,$ $n = 1, 2, . . .,$ is uniformly integrable in $L^p$;

(ii) $x_n$ converges to $x$ in $L^p$ with $\mathbb E(|x|^p)<\infty$;

(iii) $lim_{n\to\infty}\mathbb E(|x_n|^p)=\mathbb E(|x|^p) < \infty$."

This proves 1).

As a Corollary to the just stated proposition (see Jang: "Large Sample Techniques for Statistics", Springer, pag. 35), the author shows that if $x_n$ converges to $x$ in probability and if $\mathbb E(|x_n|^q)$ is bounded for some $q$ and for all $n$ then $x_n$ converges to $x$ in $L^p$ for all $0<p<q.$

We can apply this Corollary to my second question with $q=2,p=1$ getting that $x_n$ converges to $x$ in $L^1.$

As a consequence, $$0=lim_{n\to\infty}\mathbb E(-|x-x_n|)\leq lim_{n\to\infty}\mathbb E(x-x_n)\leq lim_{n\to\infty}\mathbb E(|x-x_n|)=0$$ This imply that $lim_{n\to\infty}\mathbb E(x_n)=\mathbb E(x)$ so 2) is also true.

For 3) I found the following counterexample. Consider the space $\Omega=[0,1]$ with the usual Lebesgue misure. Let $$x_n(t)=\begin{cases}0& \mbox{ if }0\leq t\leq 1-\frac{1}{n}\\ \sqrt{n}& \mbox{ if } 1-\frac{1}{n}<t\leq 1-\frac{0.5}{n}\\ -\sqrt{n}& \mbox{ if }1-\frac{0.5}{n}< t\leq 1\\ \end{cases}.$$ Now, $x_n$ converges a.s. to the random variable $x$ identically zero over $[0,1].$ Moreover $\mathbb E(x_n^3)=0,$ $\mathbb E(x_n^2)=1,$ $\mathbb E(x)=0.$ Hence the hypotesis are satisfied but $x_n$ does not converges to $x$ in $L^2.$ Note that in 3) it is not possible to use the same technique used in 2) with $q=3$ and $p=2$ because here we know $\mathbb E(x_n^3)=c\in\mathbb R$ and not $\mathbb E(|x_n|^3)=c\in\mathbb R.$ Actually in the previous counterexample we have $\mathbb E(|x_n|^3)=\sqrt{n}\to \infty.$

Answer 2

For 1) you may consider on $((0,1],B((0,1]),λ_{|(0,1]})$ the sequence$(X_n)$ defined by $\forall n, \ X_n = \imath n 1_{(0,\frac{1}{n}]}(\omega)$. It converges almost surely to $0$, but not in $L^2$ and $\mathbb{E}[X_n^2] = -n^2 \frac{1}{n} = -n \lt 0$.