As far as I understood, by saying a random variable/vector $X$ belongs to a space $S$ (or takes values in $S$), one means that the measurable function $X$ is $S$-valued: \begin{equation} X:(\Omega,\mathcal{F})\rightarrow (S,\mathcal{A}), \end{equation} e.g. $(S,\mathcal{A})=(\mathbb{R}^k,\mathcal{B}^k)$ so that for any $\omega\in\Omega$, we have $X(\omega)\in S$.

In a book I was reading, there was a theorem saying "any random vector $X$ that takes values in $\ell_p$ satisfies (something*)". Suppose now I could show that a random vector $Z=(Z_1,Z_2,...)^{T}\in\ell_p$, "almost surely". i.e. \begin{equation} \sum_{j=0}^{\infty}|Z_j|^p \end{equation} converges with probability 1. My questions are

  1. Will I be able to apply the theorem on $Z$, and argue that $Z$ satisfy the property (something*)?
  2. What if "almost sure" convergence is replaced by say "mean-squared" convergence?
  3. With reference to the definition far above, let me define \begin{equation} \mathbb{R}^{(\infty)}:=\Big\{x=(x_j)_{j=1}^{\infty}\in\mathbb{R}^{\mathbb{N}}\big|~\exists~i~~\text{with} ~x_j=0,~~ \forall j>i\Big\}. \end{equation} Then if I say a random vector $Y$ takes values in $\mathbb{R}^{(\infty)}$, does that mean for some $\omega_1\neq\omega_2\in\Omega$ the "number of non-zeros" of $X(\omega_1)$ and $X(\omega_2)$ can be different, is this correct?

Any short comments/partial answers would be greatly appreciated. Thanks very much.

  1. You can at best say Z satisfies the property almost surely.
  2. What do you actually mean by "mean-squared" convergence? We can talk about this in the comment. So, here what almost sure convergence means is that under some distribution (or measure) $\mathbb{P}$ you imbue on some more generic sequence space $\mathcal{F}$ containing $\mathcal{l}_p$, the following holds: $\mathbb{P}\left(Z\in \mathcal{l}_p\right)=1$. In other words, $\mathbb{P}\left(Z\in \mathcal{F}\backslash\mathcal{l}_p\right)=0$. So what I want to say here is that you actually need to define explicitly in which measure space $\mathcal{l}_p$ is lying in.
  3. Yes, you're right! So $Y$ can have different number of non-zeros at different realisations.
  • $\begingroup$ Thank you Seong! What I meant by mean squared convergence (to some limit say $S$) was simply $\lim_{n\rightarrow\infty}E|\sum_{j=0}^{n}|Z_j|^p-S|^2\rightarrow 0$. For example $p=1$ and $Z_j=\psi_jX_j$ where $\{X_j\}$ is any sequence of random variables, and $\psi_j$ is such that $\sum_{j=0}^{\infty}|\psi_j|<\infty$. Then we can easily show mean squared convergence provided $E|X_j|^2<\infty$ for all $j$. $\endgroup$
    – John
    Dec 11 '14 at 12:27
  • $\begingroup$ So here the normed space implicitly considered is $\mathcal{l}_1$, I guess, and $S$ is the $\mathcal{l}_1$ norm of $\bf{Z}$? Hmm, how about considering $S\in \mathcal{l}_2$? I think this would make more sense... i.e. in this case you say $\bf{Z}\rightarrow S$ in $\mathcal{l}_2$ if $E\left|\sum_{j=0}^n {\left(Z_j-S_j\right)^2} \right|\rightarrow 0$. Also, in using the expectation, I guess the expectation is with respect to the distribution of $\bf{Z}$? $\endgroup$ Dec 11 '14 at 12:35
  • $\begingroup$ Thanks once again. I think the choice of $p$ does not matter here because mean squared convergence should hold as long as $\sum_{j=0}^{\infty}|\psi_j|^{p}<\infty$ and $\sup_j E|X_j|^{2p}<\infty$. As you said, $S$ would be the $\ell_1$ norm of $Z$ in the example above. What did you mean by $S\in\ell_2$ (and $S_j$)? $\endgroup$
    – John
    Dec 11 '14 at 13:11

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