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Suppose we have a family of uniformly integrable random variables $\{X_n:n\in\mathbb{N}\}$. Suppose also we have a sequence of continuous functions $f_n:\mathbb{R}\to\mathbb{R}$ converging to $f$ and that $\{f(X_n):n\in\mathbb{N}\}$ is uniformly integrable. Also assume that $f_n(X_n)$ is integrable for all $n$.

If $f_n$ converges to $f$ uniformly, then we know that $\{f_n(X_n):n\in\mathbb{N}\}$ is uniformly integrable.

If we replace uniform convergence with compact convergence (uniform convegergence on compact sets), is $\{f_n(X_n):n\in\mathbb{N}\}$ also uniformly integrable?

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1 Answer 1

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Let $f_n\colon x\mapsto x/n$. Then $f_n$ is continuous for all $n$ and the sequence $\left(f_n\right)_{n\geqslant 1}$ converges to $0$ uniformly on compact sets. It thus suffices to find a sequence of random variables $(X_n)_{n\geqslant 1}$ such that for each $n$, $X_n/n$ is integrable but the family $\left\{X_n/n,n\geqslant 1\right\}$ is not uniformly integrable. By letting $Y_n:=X_n/n$, it suffices to find a sequence of integrable functions which is not uniformly integrable.

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