Let $(\Omega,\mathcal F, P)$ be a probability space and let $L_1$ be the collection of all integrable (finite expectation) real-valued random variables defined on this space. Assume that for $X_k\in L_1,k=1,2,\dots n$ it holds that $\mathbb E \lvert \sum_k a_k^m X_k \rvert \leq 1,\forall m$, where $(a_1^m,a_2^m,\dots,a_n^m)=:a_m$ is a sequence in $\mathbb R^n$ such that $\max_k \lvert a_k^m \rvert\to \infty, m \to \infty.$ Prove that $\exists i$ and real numbers $b_k$ such that $\mathbb E \lvert X_i - \sum_{k\neq i} b_k X_k \rvert=0$.

Note that this is related to homework.

I tried a number of things but can't pin down where I fail. I tried using contradiction and Fatou to show that there is a set of non zero measure where the $\lim \inf$ is infinite but did not manage to convince myself. I also tried arguing by standardizing all coefficients by the $\max$, leading to $\mathbb E \lvert \sum_k \alpha_k^m X_k \rvert\to0$ where at least one $\alpha_k^m \in \{ -1,1\},\forall m$ but could not convince myself that this is helpful. From here I also tried using the fact that there is a subsequence converging almost surely to zero.

In short, I am at a loss where to even start. I feel like I've tried all the standard tools from a graduate level probability course but am not making progress. I am looking for a pointer in the right direction. The overall goal is to show that the span of finite number of integrable r.v.s is a closed set, but this is a step on the way.

  • $\begingroup$ Thank you for expanding the question! I suspect you might be able to offer a constructive demonstration by considering the sequence $\mathrm{a}/||\mathrm{a}||$ of normalized vectors: show that it must have an accumulation point on the sphere $S^{n-1}\subset \mathbb{R}^n$ and calculate the $b_k$ from the coefficients of such an accumulation point. $\endgroup$
    – whuber
    Mar 19, 2015 at 14:53
  • $\begingroup$ @whuber thanks for the suggestion! I'll go and work on that and see where it gets me. $\endgroup$
    – KOE
    Mar 19, 2015 at 14:54

1 Answer 1


Based on @whuber's hint, I came up with the following. Please do let me know if I made a mistake somewhere.

Let $\alpha^{m}:=\left(a_{1}^{m},\dots, a_{n}^{m}\right)/\left\Vert \left(a_{1}^{m},\dots\,a_{n}^{m}\right)\right\Vert$. Since every bounded sequence in $\mathbb{R}^{n}$ has a convergent sub-sequence (Bolzano-Weierstrass), we may pick $m_{l}:\alpha^{m_{l}}\to b$ for some $b\in\mathbb{R}^{n}$. Then $\sum_{k}\alpha_{k}^{m_{l}}X_{k}\to\sum_{k}b_{k}X_{k}$ pointwise. By assumption we also have $$ \mathbb{E}\left|\sum_{k}\alpha_{k}^{m_{l}}X_{k}\right|\leq1/\left\Vert a^{m_{l}}\right\Vert \to0 $$

Since $\sum_{k}\mathbb{E}\left|X_{k}\right|<\infty$ and $\left|\sum_{k} \alpha_{k}^{m_{l}}X_{k}\right|\leq\sum_{k}\left| \alpha_{k}^{m_{l}}X_{k}\right|\leq\sum_{k}\left|X_{k}\right|$ dominated convergence gives: $$0=\lim_{l\to\infty}\mathbb{E}\left|\sum_{k} \alpha_{k}^{m_{l}}X_{k}\right|=\mathbb{E}\left|\sum_{k}b_{k}X_{k}\right|,$$ which is what we wanted to prove (upon standardizing $b$ so that $b_k=1$ for at least one $k$).


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