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Suppose that $X$ is a random element with values in a separable Hilbert space $\mathbb H$ such that $\operatorname EX=0$ and $\operatorname E\|X\|^2<\infty$. Suppose that $f_1,f_2,\ldots$ form an orthonormal basis of $\mathbb H$ (these might be eigenvectors of the covariance operator of $X$). Then, by Parseval's identity, we have that $$ \sum_{n=1}^\infty\operatorname E|\langle X,f_n\rangle|^2<\infty. $$ In particular, $\operatorname E|\langle X,f_n\rangle|^2\to0$ as $n\to\infty$.

I am interested in $\operatorname E|\langle X,f_n\rangle|^p$ with $p>2$. If $\operatorname E\|X\|^{2(p-1)}<\infty$, using the fact that $\|f_j\|=1$ for $j\ge1$ and the Cauchy-Schwarz inequality twice, we obtain \begin{align*} \operatorname E|\langle X,f_n\rangle|^p &=\operatorname E[|\langle X,f_n\rangle||\langle X,f_n\rangle|^{p-1}]\\ &\le\operatorname E[|\langle X,f_n\rangle|\|X\|^{p-1}]\\ &\le(\operatorname E|\langle X,f_n\rangle|^2)^{1/2}(\operatorname E\|X\|^{2(p-1)})^{1/2}. \end{align*} This shows that $\operatorname E|\langle X,f_n\rangle|^p\to0$ as $n\to\infty$ as well, but at a slower rate than $\operatorname E|\langle X,f_n\rangle|^2$. The rate of convergence does not depend on $p$, which I find counterintuitive. Also, we need to assume that $\operatorname E\|X\|^{2(p-1)}<\infty$.

At the moment, I have that $\operatorname E|\langle X,f_n\rangle|^p=O((\operatorname E|\langle X,f_n\rangle|^2)^{1/2})$ as $n\to\infty$. I suspect that $\operatorname E|\langle X,f_n\rangle|^p$ goes to $0$ faster than that. Is it possible to show that $\operatorname E|\langle X,f_n\rangle|^p\to0$ as $n\to\infty$ faster than $(\operatorname E|\langle X,f_n\rangle|^2)^{1/2}$? Perhaps even $\operatorname E|\langle X,f_n\rangle|^p=O((\operatorname E|\langle X,f_n\rangle|^2)^{p/2})$ as $n\to\infty$?

Any help is much appreciated!

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I don't think so. Take X to be the random element of $l^2 = \sum a_i Z_i$ where, say, $\sum n^2a_n^2 < \infty$ and the $Z_i$ are independent, mean 0 variance 1, but not iid. $f_1 = (1, 0, 0, ...) $ etc. so that $\langle f_n ,X \rangle = a_i Z_i$. In order to get $||X|| \in L^4$ you only need $$\sum a_i^4 E(Z_i^4) < \infty $$ but if your last condition were true it would entail $$E(|Z_i|^4) \approx (E(|Z_i|^2))^2$$ I built in a bit of room so I think I can make the lhs much bigger than the right without altering the fact of $||Z||$ being $L^4$.

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  • $\begingroup$ Thanks (+1)! I’m sorry, but your notation is confusing. So we take $X=\{a_nZ_n\}$ with values in $\ell^2$, where $\sum a_n^2<\infty$, $Z_n$ are independent rv's with $EZ_n=0$, $EZ_n^2=1$ and $E|Z_n|^p<\infty$. We have that $\langle f_n,X\rangle=a_nZ_n$. Hence, $E|\langle f_n,X\rangle|^2=|a_n|^2$ and $E|\langle f_n,X\rangle|^p=|a_n|^pE|Z_n|^p$. If $E|\langle f_n,X\rangle|^p=O((E|\langle f_n,X\rangle|^2)^{p/2})$, then $|a_n|^pE|Z_n|^p\le C (|a_n|^2)^{p/2}$ for large $n$ and this would imply that $\sup_nE|Z_n|^p<\infty$, which is not necessarily true. Do I understand correctly? $\endgroup$ – Cm7F7Bb May 8 '18 at 12:30
  • $\begingroup$ Yes. It would imply that because the ambitious condition ($\operatorname E|\langle X,f_n\rangle|^p=O((\operatorname E|\langle X,f_n\rangle|^2)^{p/2})$ makes the p norm the same order as the 2 norm, which is 1, and I claimed without proof that it can be made to become infinite slowly enough so that $\sum a_i^4 E(Z_i^4) < \infty$. $\endgroup$ – user83457 May 8 '18 at 12:46
  • $\begingroup$ Would you think that it is possible to show that $\operatorname E|\langle X,f_n\rangle|^p\to0$ faster than $(\operatorname E|\langle X,f_n\rangle|^2)^{1/2}$? $\endgroup$ – Cm7F7Bb May 8 '18 at 12:50
  • $\begingroup$ If p < 2 yes but if p > 2 no, by Jensen the rhs must be bigger than the lhs^p ( it makes no difference that you dropped the p form the rhs, but it makes this a little awkward). This is just a question of constructing distributions that are bounded in $L^2$ but converge to 0 or $\infty$ in $L^p$ $\endgroup$ – user83457 May 9 '18 at 8:06
  • $\begingroup$ If $\xi_1,\xi_2,\ldots$ are rv's such that $E|\xi_n|^p<\infty$ with $p>2$ and $E|\xi_n|^2\to0$, it does not follow that $E|\xi_n|^p\to0$. In general, $(E|\xi_n|^q)^{1/q}\le(E|\xi_n|^p)^{1/p}$ for $p\ge q$. However, I have a particular situation and I can show that in this particular situation $E|\langle X,f_n\rangle|^p\le C(E|\langle X,f_n\rangle|^2)^{1/2}$ for $p>2$ provided that $\operatorname E\|X\|^{2(p-1)}<\infty$, where $C=(\operatorname E\|X\|^{2(p-1)})^{1/2}$. The bound that I obtain is very rough so I suspect that $E|\langle X,f_n\rangle|^p=o((E|\langle X,f_n\rangle|^2)^{1/2})$. $\endgroup$ – Cm7F7Bb May 9 '18 at 8:56

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