How to show the convergence of a random variable does not imply convergence of expectation

69 Views Asked by At

Suppose that the expectation operator is defined by

$$E(X) = \lim_{D\to \infty} \frac{1}{2D} \int_{-D}^{D} X(\omega)d\omega$$

Where

$$X_n\left(\omega\right) = \left\{ \begin{array}{lr} 1 & : |\omega| \le n\\ 0 & : \text{otherwise} \end{array} \right.$$

for $n = 1, 2, ...$

I need to show that $E(X)$ satisfies all of the axioms of expectation except the following:

$\mathbf{Axiom}\;\mathbf{5:}$

If a sequence of random variables $\{X_n(\omega)\}$ increases monotonically to a limit $X(\omega)$ then

$$E(X) = \lim E(X_n)$$

I've been able to show that this formula does satisfy the other axioms of expectation but I'm having trouble showing it doesn't satisfy the fifth axiom stated above.

What I want to show is that even though $\{X_n(\omega)\}$ converges to $X(\omega)$

$$\lim_{n\to \infty}E(X_n) = \lim_{n\to \infty}[\lim_{D\to \infty}\frac{1}{2D}\int_{-D}^{D} X_n(\omega)d\omega]\neq E(X)$$

I'm not really sure how to proceed here. My thinking is that maybe I could show that the limits for $D$ and $n$ are not interchangeable or lead to different results when interchanged and computed. Or perhaps show a lack of continuity as $E(X) = \lim E(X_n)$ is essentially a continuity demand. I was also thinking perhaps Fatou's Lemma might be useful here.

Does anyone have a tip or hint on how I can proceed? Thank you kindly.