Measure Theory: Convergence of Random Variables

Question

I just came out of my midterm exam of my measure theory course, and I could not answer the following question.

"Let $(\Omega,F,P)$ be a probability space and consider the following random variables $Z,X_1,X_2,\dots$, all in $L^2(\Omega,F,P)$. If

$\lim_\limits{{n\to \infty}}E[(X_n-Z)^2]=0$

Show that:

1. $E[X_n] \rightarrow E[Z]$
2. $E[|X_n|] \rightarrow E[|Z|]$
3. $E[(X_n)^2] \rightarrow E[(Z)^2]$"

My attempt

From the question we can see that it follows:

$\lim_\limits{{n\to\infty}}E[(X_n)^2]$ = 2 $\lim_\limits{{n\to\infty}}E[X_nZ]$-$E[(Z)^2]$

So, if we show that $\mbox{Lim}_{n\rightarrow\infty}E[X_nZ]= E[(Z)^2]$, then part (3) could be solved. However, I could not successfully prove this.

Parts (1) and (2) were less clearer to me. I tried using the fact that if the random variables belong to $L^2$, then they will also be in $L^1$.

Thanks for the responses, I just really wanted to know the answer!

Answer 1

To show that

$$\mathbb{E}(X_n Z) \to \mathbb{E}(Z^2)$$

you can use the Cauchy-Schwarz inequality:

$$|\mathbb{E}(X_n Z)-\mathbb{E}(Z^2)| = |\mathbb{E}((X_n-Z)Z)| \leq \sqrt{\mathbb{E}(|X_n-Z|^2)} \sqrt{\mathbb{E}(Z^2)}.$$

By assumption, the right-hand side converges to $0$ as $n \to \infty$.

$\mathbb{E}(X_n) \to \mathbb{E}(Z)$ and $\mathbb{E}(|X_n|) \to \mathbb{E}(|Z|)$.

By Jensen's inequality, we have

$$\mathbb{E}(|X_n-Z|) \leq \sqrt{\mathbb{E}(|X_n-Z|^2)} \xrightarrow[]{n \to \infty} 0.$$

This implies

$$|\mathbb{E}(X_n)-\mathbb{E}(Z)| = | \mathbb{E}(X_n-Z)| \leq \mathbb{E}(|X_n-Z|) \xrightarrow[]{n \to \infty} 0$$

as well as

$$|\mathbb{E}(|X_n|) -\mathbb{E}(|Z|) \leq \mathbb{E}(||X_n|-|Z||) \leq \mathbb{E}(|X_n-Z|) \xrightarrow[]{n \to \infty} 0$$

where we have used the reverse triangle inequality, i.e.

$$||x|-|y|| \leq |x-y|.$$

Answer 2

Have you tried using Cauchy-Schwarz? For example, for part (1), we may observe: $$ \left\vert E[X_n] - E[Z] \right\vert \leq \int_\Omega \left(1.|X_n - Z |\right)dP \leq \left( \int_\Omega 1 \ dP \right)^{\frac 1 2}\left( \int_\Omega |X_n - Z|^2\right)^{\frac 1 2}=\left(E[(X_n - Z)^2] \right)^{\frac 1 2}$$ and the result follows by sending $n \to \infty$ on both sides.

[Notice that the inequality I wrote down is merely the statement that $Var[X_n - Z] \geq 0$.]

Part (2) can be proved in the same way, by virtue of the fact that $\left\vert |X_n| - |Z|\right\vert \leq |X_n - Z |$.

For part (3), we can finish off your argument by observing that $$ \left\vert E[X_n Z] - E[Z^2] \right\vert = \int_\Omega |(X_n - Z). Z |dP\leq E[X_n - Z]^{\frac 1 2} . E[Z]^{\frac 1 2},$$ then applying the result of part (1).