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$\newcommand{\N}{\mathbb{N}}\newcommand{\P}{\mathrm{P}}\newcommand{\esssup}{\mathrm{esssup}}$Let $(Z_j)_{j\in\N}$ be a nonnegative stochastic process ($Z_j\geq 0$ $\P$-a.s) for which the following holds: $$ \esssup\left(\sum_{j=0}^{\infty}Z_j\right)\leq M\tag{1}\label{1}, $$ for some $M\geq 0$.

Using the definition of essential supremum we have that the sequence is almost surely summable: $$ \begin{align} &\P\left[ \{\omega\in\Omega: \sum_{j=0}^{\infty}Z_j > M\}\right] = 0 \\ \Leftrightarrow\ &\P\left[ \{\omega\in\Omega: \sum_{j=0}^{\infty}Z_j \leq M\}\right] = 1 \end{align}$$ and since $\{\omega\in\Omega: \sum_{j=0}^{\infty}Z_j(\omega) \leq M\}\subseteq \{\omega\in\Omega: Z_j(\omega)\to 0\}$, we have that \eqref{1} implies $\P[\{\omega: Z_j(\omega)\to 0\}]=1$, that is, $Z_j\to 0$ a.s., and, a fortiori, $Z_j\to 0$ in probability.

I wonder whether we can prove that $\esssup(Z_j)\to 0$.

If not, there will be a $\delta>0$ and an increasing sequence of integers $(j_k)_k$ so that $\esssup(Z_{j_k})>\delta$ for all $k\in\N$. Since $Z_j$ are nonnegative, I guess (but I am not sure) that we can show that $\esssup(\sum_{k\in\N}Z_{j_k})>\delta$ which is a contradiction. Is this correct? If not, is there a counterexample where \eqref{1} holds but $\esssup(Z_j)$ does not converge to $0$?

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No, this need not be the case. For instance, it is possible that with probability $1$, there exists $j$ such that $Z_j=M$ and $Z_k=0$ for all $k\neq j$. For an explicit example, let $X$ be a Poisson random variable and define $Z_j:=\mathbf1_{\{X=j\}}$ for each $j$. Then $\sum_{j=0}^\infty Z_j=1$ almost surely, but $\operatorname{esssup}(Z_j)=1$ for all $j$, so in particular it does not converge to zero.

Note that the conclusion you are trying to draw - that $Z_j\to0$ in $L^\infty$ - is extremely strong. This is rarely something we would try to prove in probability theory.

EDIT: As we have discussed in the comments, it is true that $Z_j\to0$ in $L^p$ for any $1\le p<\infty$. To see this, it is sufficient to show that there is a constant $C$ such that $\sum_jZ_j^p\le C$ almost surely, since this would imply $\sum_j\mathbb E[Z_j^p]=\mathbb E\left[\sum_jZ_j^p\right]\le C$ by the monotone convergence theorem, which of course shows that $\mathbb E[Z_j^p]\to0$.

Let $A=\{j\in\mathbb N:Z_j\le1\}$ and $B=\mathbb N\setminus A$. Notice that $Z_j^p\le Z_j$ for all $j\in A$, $Z_j\le M$ for all $j\in B$ almost surely, and $|B|\le\sum_{j\in B}Z_j\le M$. This implies

$$\sum_jZ_j^p=\sum_{j\in A}Z_j^p+\sum_{j\in B}Z_j^p\le\sum_{j\in A}Z_j+M\cdot M^p\le M+M^{p+1}$$

almost surely, completing the proof.

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  • $\begingroup$ Thanks a lot for the answer. Is it true, however, that $Z_j\to 0$ a.s.? Given (1), can we deduce anything additional about $(Z_j)_j$? (perhaps under additional assumptions?) $\endgroup$ – Pantelis Sopasakis Aug 8 '17 at 21:23
  • $\begingroup$ Yes, certainly - $\sum_jZ_j\le M$ almost surely, so clearly $Z_j\to0$ almost surely. As far as other things $(1)$ implies, you can for instance show that $Z_j\to0$ in $L^p$ for all $1\le p<\infty$. Try to think about how you might show this for $p=1$ (this is the easiest case). $\endgroup$ – Jason Aug 8 '17 at 21:31
  • $\begingroup$ For $p=1$ I would start by $\mathbb{E}(\sum_j Z_j) \leq \mathrm{esssup}(\sum_j Z_j) \leq M$ and use the linearity of $\mathbb{E}$. $\endgroup$ – Pantelis Sopasakis Aug 8 '17 at 21:40
  • $\begingroup$ Right. You technically also need the monotone convergence theorem since it's an infinite series, but you get the idea. For $p>1$, it's the same principle - we just need to find an (almost sure) upper bound for $\sum_jZ_j^p$, which i'll leave you to think a little bit about. $\endgroup$ – Jason Aug 8 '17 at 21:45
  • $\begingroup$ For $p=1$ we also have the Beppo-Levi Theorem which applies directly. Let me think about the $L^p$ convergence case... $\endgroup$ – Pantelis Sopasakis Aug 8 '17 at 21:52

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