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I'm having a difficult time constructing a counter example to this. My intuition (sloppily) is to construct a family of functions {$X_n$} that have Dirac pulses at $n$ and $-n$. Such that $\sup_n \Bbb E[|X_n|\Bbb 1_{X≥n}]=1$

However, I'm not sure if this is correct.

user237393
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2 Answers2

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The thing which breaks this (and other things like martingale convergence) is the fact that you can have things be very large on small sets. If you have $L^p$ bounds for $p > 1$, this just won't happen. But for $p = 1$, it can! Neal provided a good answer, but allow me to add one that won't require much thinking for you!

$X_n = 2^{n} 1_{[0,2^{-n}]}$

Notice that $||X_n||_{L^1}$ = 1, but no matter how small a set you take, you can find an $n$ large enough where all of your mass will still be there.

Rikimaru
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Hint:

Your intuition is good, but you can make it even simpler. To kill uniform integrability, you need your family to "pile up" on small sets. Dirac deltas are not actually functions in $L^1$, so you can't use them explicitly, but you can probably get the same behavior using approximations ...

Solution: The example I have in mind is a family of normal distributions with standard deviation tending to zero. Explicitly, $$\{X_n\} = \mathcal{N}_{0,1/n} = \frac{n}{\sqrt{2\pi}}e^{-(nx)^2/2}.$$ Each of these has $\|X_n\|_{L^1} = 1$ for all $n$, so the family is bounded in $L^1$, and you can use standard facts about the normal distribution (found in the back of any intro stats textbook) to show that for $n$ large, most of the mass of $X_n$ is concentrated on a very small neighborhood of $0$. Parse the symbols in the definition of "uniform integrability" and you'll see that this gives you a counterexample.

Mittens
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Neal
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  • I have seen it mentioned a few times that to kill uniform integrability you need the family to "pile up" on small sets. But the sample mean is uniformly integrable yet it piles up on a small set since it piles up around the mean. Actually the sample mean corresponds closely to your example. So why is your example not uniformly integrable yet the sample mean is uniformly integrable? – sonicboom Mar 16 '21 at 10:49
  • Hmmm..maybe its because of the $\frac{1}{n}$ in $\overline X_n = \frac{1}{n} \sum_{i=1}^n X_i$ that the sample mean is uniformly integrable but your example isnt? – sonicboom Mar 16 '21 at 10:51
  • Your example does not work for if $X_n\sim N(0,1/n)$ then $\sup_nE[|X_n|^2]\leq1$ and so, $\lim_{a\rightarrow\infty}\sup_nE[\mathbf{1}_{{|X_n|>a}}|X_n|]=0$, which means that $(X_n)$ is uniformly integrable (for the underlaying space is fine measure). – Mittens Jan 14 '25 at 20:06