Is $\sup_n\mathbf{E}X_n = \mathbf{E}\sup_n X_n$?

Question

I was proving implications of Dominated Convergence Theorem, $\sup_n \mathbf{E}|X_n| < \infty$ if $X_n$ is uniformly integrable, and had this confusion:

If there's a finite collection of uniformly integrable rvs $\{X_n\}$, is $\sup_n\mathbf{E}X_n = \mathbf{E}\sup_nX_n$? I think it is.

I define $X^{\ast}=\sup_n X_n$; the collection of rvs can be ordered, i.e. $\{X_{a_1} \leq X_{a_2} \leq \ldots X^{\ast}\}$ for some sequence if indices $a_j$ and all $\omega \in \Omega$. Since $\mathbf{E}$ is order-preserving, this ordering applies to expectations $\{\mathbf{E}X_{a_1} \leq \mathbf{E}X_{a_2} \leq \ldots \mathbf{E}X^{\ast}\}$. By definition, $\mathbf{E}X^{\ast} = \mathbf{E}\sup_n X_n$. From the second ordering, $\mathbf{E}X^{\ast} = \sup_{a_j} \mathbf{E}X_{a_j} = \sup_n \mathbf{E}X_n$, proving the statement.

If this is correct, I can prove, using uniform integrability, for any number $\alpha \to \infty$ \begin{align} \mathbf{E}X^{\ast} &\leq \mathbf{E}|\sup_n X_n|\leq \color{red}{\mathbf{E}\sup_n|X_n| = \sup_n \mathbf{E} |X_n|} \\ &= \sup_n \mathbf{E}\big(|X_n|I_{|X_n|> \alpha} + |X_{n}|I_{|X_{n}| \leq \alpha} \big) \to_{\alpha} \sup_n \big( 0 + \mathbf{E}|X_{n}| \big)\\ &< \sup_n c = c <\infty \end{align} as required. The last inequality is by definition of integrability, $\int_{\mathbb{R}}|X|< c< \infty$. The step I proved is in red. Note I don't use convergence anywhere except the upper bound on the Lebesgue integral.

Answer 1

The best you can say is: $E\sup |X_n| \ge \sup E(|X_n|)$, even if you have a finite number of $X$'s. So unfortunately this inequality goes the wrong way for you.

Example: On the unit interval equipped with Lebesgue measure, take $n=2$, with $X_1:=I_{[0,\frac12]}$ and $X_2:=I_{[\frac12,1]}$. Compute $E(|X_n|)=\frac12$ for $n=1,2$, while $\sup |X_n|$ is identically 1.

Answer 2

In general $\sup_n E[X_n]\neq E[\sup_n X_n]$ for a bounded (in $L_1$) sequence of random variables:

Let $X_n=n\mathbb{1}_{\big[0,\tfrac{1}{n^2}\big]}$ on $([0,1],\mathscr{L},\lambda_1)$. $E[X_n]=\frac{1}{n}$ and so $\sup_nE[X_n]=1$, but $\sup_nX_n$ is not integrable.

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Even in the finite case: $Y_1=\mathbb{1}_{[0,1/2]}$, $Y_2=\mathbb{1}_{(1/2,1]}$. $\max\{E[Y_1],E[Y_2]\}=\frac12<E[\max\{Y_1,Y_2\}]$.

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Are there cases where the equality hold?

yes: for $\{X_n:n\in\mathbb{N}\}$ is monotone increasing, the equality follows from monotone convergence.