Uniform Lp bound implies finite Lp norm with random index?

Question

Suppose that $\{X_N:N\in\mathbb N\}$ is a sequence of random variables uniformly bounded in $L^p$ norm for all $p>1$: $$\sup_N\|X_N\|_{L^p}<C(p).$$

Let $\mathcal N$ be some random variable taking values in $\mathbb N$ that is independent of the sequence $\{X_N\}$, and define $X_{\mathcal N}$ as the random variable $$\omega\to X_{\mathcal N(\omega)}(\omega).$$ Is it then possible to prove that $\|X_{\mathcal N}\|_{L^{p}}<\infty$?

It is easy to see that $$\|X_{\mathcal N}\|_{L^{p}}=\left\|\sum_{N}X_N\cdot1_{\{N=\mathcal N\}}\right\|_{L^{p}},$$ and thus we get the result by Minkowski's and Hölder's inequalities if $$\sum_N\|1_{\{N=\mathcal N\}}\|_{L^{2p}}<\infty.$$ However, it is possible to prove this with weaker (or ideally no) conditions on the distribution of $\mathcal N$?

Answer 1

Notice that for all $\omega$ $$ X_{\mathcal N}\left(\omega\right)=\sum_{n=1}^{+\infty}X_{n}(\omega)\mathbf 1\left\{\mathcal N(\omega)=n\right\} $$ and using pairwise disjointness of the events $\left\{\omega\mid \mathcal N(\omega)=n\right\}$, we get $$ \left\lvert X_{\mathcal N}\left(\omega\right)\right\rvert^p=\sum_{n=1}^{+\infty}\left\lvert X_{n}(\omega)\right\rvert^p\mathbf 1\left\{\mathcal N(\omega)=n\right\}. $$ Integrating and using independence between $\mathcal N$ and the sequence $\left(X_n\right)_{n\geqslant 1}$, we get $$ \mathbb E\left\lvert X_{\mathcal N}\left(\omega\right)\right\rvert^p=\sum_{n=1}^{+\infty}\mathbb E\left\lvert X_{n}(\omega)\right\rvert^p\mathbb P\left\{\mathcal N(\omega)=n\right\}\leqslant c(p)\sum_{n=1}^{+\infty} \mathbb P\left\{\mathcal N(\omega)=n\right\}=c(p). $$

Answer 2

My guess is that no this isn't true without an assumption on $\mathcal{N}$. Constructing an explicit counterexample may require more time than I have right now, but more or less the only thing you have for free is that

$\sum_{n=1}^{\infty} \mathbb{P}(\mathcal{N} = n) = 1$.

However this series could go so slowly that $\sum_{n=1}^{\infty} \mathbb{P}(\mathcal{N} = n)^{\alpha} = \infty$ for every $\alpha < 1$. Which seems to rule out any Holder inequality-type bound working (unless you have a uniform $L^{\infty}$ bound on $|X_n|$).

And if an easy Holder argument doesn't work then my guess is there is a genuine counterexample that exploits this. The tricky thing in constructing one is the independence. Admittedly this is on your side, but I don't think Holder can be effectively improved, even when the RVs are independent. I think that basically you can still come very close to being equality which is bad in this case.