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It's a well known fact in probability theory that if $\lbrace X_i\rbrace$ is a family of random variables, $\phi : [0,\infty) \to [0,\infty)$ satisfies $$\lim_{x \to \infty} \frac{\phi(x)}{x} = \infty \quad (*) $$ and $\sup_i E\phi(|X_i|) < \infty$, then the family $\lbrace X_i\rbrace$ is uniformly integrable. I want to know whether there is a single function $\phi$ satisfying (*) such that $\sup_i E\phi(|X_i|) < \infty$ iff $\lbrace X_i\rbrace$ is uniformly integrable.

I suspect the answer is no, and that even a family of one absolutely continuous random variable should suffice. So I conjecture the following: given any $\phi$ satisfying (*), there exists a nonnegative function $f$ such that $$\int_0^\infty f(x)dx=1$$ $$\int_0^\infty x f(x)dx < \infty$$ $$\int_0^\infty \phi(x)f(x)dx = \infty.$$

But I can't see offhand what to use for $f$. Ideas?

Nb. This is not homework. It was going to be, but I thought I should be able to solve it myself before assigning it to my students :)

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By hypothesis, for any positive $k$, there is an $N_k$ such that $\phi(x) > kx$ if $x > N_k$. We can of course assume that $(N_k)$ is an increasing sequence, and so it splits the real line into intervals, which we furthermore can assume are of length at least one (say).

Now take $f$ to be piecewise constant: zero if $0 < x < N_1$, and chosen on the other intervals such that $\int_{N_k}^{N_{k+1}} x f(x) dx = \frac{c}{k^2}$, where $c$ is just a normalization constant (since you require $\int f(x) dx=1$).

(That is, choose $$ f(x) = \frac{c/k^2}{\int_{N_k}^{N_{k+1}} x dx} = \frac{2c}{k^2 (N_{k+1}^2-N_k^2)} $$ for $N_k \le x < N_{k+1}$. Then, because we arranged for the intervals to be sufficiently long, $\int_0^{\infty} f(x) dx$ reduces to a convergent sum.)

Then $$ \int_0^{\infty} \phi(x) f(x) dx \ge \sum_{k\ge 1} \int_{N_k}^{N_{k+1}} kx f(x) dx = \sum_{k\ge 1} k \frac{c}{k^2} = \infty.$$

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Very neat! Thanks. –  Nate Eldredge Oct 14 '10 at 12:52

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