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.

  • 1
    $\begingroup$ The error in your argument comes from the fact that the $a_j$s are themselves random variables (since the ordering of a bunch of random variables is random), which are further correlated with the various $X_n$s. Thus $\mathbb{E}[X_{a_j}]$ in fact need not equal $\mathbb{E}[X_n]$ for any $n$ (and certainly the whole group of these $ \mathbb{E}[X_{a_j}]$s is very different). $\endgroup$ Jul 7 '20 at 16:59
  • $\begingroup$ but $X^{\ast} = \sup_n X_n$ regardless of the index, so $EX^{\ast} = \sup E X_n$ since $E$ is order-preserving. Why is this wrong? $\endgroup$
    – Alex
    Jul 7 '20 at 19:28
  • $\begingroup$ This perhaps got a little muddled. The point I was trying to make was that in general, the multisets of the values of $\mathbb{E}[X_{a_j}]$ and $\mathbb{E}[X_n]$ are not the same. I.e., $$ \{ \mathbb{E}[X_{a_j}]; j \in \mathbb{N}\} \neq \{ \mathbb{E}[X_n]: n \in \mathbb{N}\}.$$ It is true that $\mathbb{E}[X_{a_j}] \le \mathbb{E}[X_{a_k}]$ for $k \ge j$ but this only shows that $\mathbb{E}[\sup X_n] = \sup \mathbb{E}[X_{a_j}]$. This does not further tell you that $\sup \mathbb{E}[X_{a_j}] = \sup \mathbb{E}[X_n]$. The inequality of the multisets is a way to point this out. (contd) $\endgroup$ Jul 7 '20 at 19:52
  • $\begingroup$ Of course this is not a proof that your approach doesn't work, since you're interested in only the suprema of these multisets, but it should at least demonstrate that you're skipping some steps in the argument presented. It so happens that in this case the supremum also doesn't work in general. Both grand_chat and Oliver give simple demonstrations of this. $\endgroup$ Jul 7 '20 at 19:52
  • $\begingroup$ Thanks. In Rosenthal's textbook only $\sup_n EX_n$ is used in the proof of implicationof DCT, I decided to try proving it in a different way. $\endgroup$
    – Alex
    Jul 7 '20 at 20:48

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.

  • $\begingroup$ In you example $X \equiv 1$ $\endgroup$
    – Alex
    Jul 7 '20 at 19:34
  • $\begingroup$ @Alex Yes, if you mean $X^*$ is identically $1$ $\endgroup$
    – grand_chat
    Jul 7 '20 at 19:47

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.

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

Are there cases where the equality hold?

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

  • $\begingroup$ $\int \sup_n X_n$ doesn't exist even for a finite collection of $X_n$? I understand it diverges as $n \to \infty$ $\endgroup$
    – Alex
    Jul 7 '20 at 20:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.