I'm trying to develop some intuition for the concepts of tightness and uniform integrability, in a probabilistic context.

(i) Does uniform integrability imply tightness?

(ii) If not, is $X_n(x)=I_{[n, n+1]}$ uniformly integrable? If we take the measure space composed of Borel $\sigma$-algebra on $\mathbb{R}$ with the Lebesgue measure, $(\mathbb{R}, \mathcal{B}, \lambda)$, then is $f_n(x)=I_{[n, n+1]}$ uniformly integrable? It is certainly not tight, but $\mathbb{E}[|f_n|I_{[|f_n| \geq K]}] \leq \varepsilon$ seems to satisfy the uniform integrability condition.


(i) In fact, even a sequence $(X_n)_{n\geqslant 1} $ which is bounded in $\mathbb L^1$ (that is, $\sup_n\mathbb E|X_n| <\infty$) is tight. This can be seen from the inequality $$\mathbb P\{|X_n|\gt R  \}\leqslant \frac 1R\mathbb E[|X_n|]\leqslant \frac 1R\sup_j\mathbb E[|X_j|].$$

(ii) When the underlying measure space is not finite, we rather use the following definition of uniform integrability: the family $\left(f_i\right)_{i\in I}$ is uniformly integrable if for each positive $\varepsilon$, there exists an integrable function $g$ such that $\sup_{i\in I}\int_{\{|f_i|\geqslant g }|f_i|<\varepsilon$. With the classical definition in the context of probability space, we would find that the sequence $ (\mathbf 1_{(0,n)})_{n\geqslant 1}$ is uniformly integrable, which is not acceptable because it is not even bounded in $\mathbb L^1$.


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.