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Let $x_k$ be a random vector such that its expectation $$ E[\Vert x_k \Vert]<a $$ for some $a>0$. Then can we say that $$ E[\Vert x_k \Vert^2]<a^2 ? $$

Ron
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  • Actually the opposite always holds, that is, $$E[|x_k|^2]\geqslant E[|x_k|]^2.$$ – Did Jul 24 '14 at 16:08
  • Thanks. In fact its true from Jensen's inequality. But I needed a bound on $E[\Vert x_k \vert^2]$ given a bound on $E[\vert x_k \Vert]$, which was the reason I asked this question. – Ron Jul 24 '14 at 18:10
  • Cauchy-Schwarz, no need to require Jensen actually. – Did Jul 24 '14 at 18:13

2 Answers2

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$\mathbb{E}[\lVert X\rVert^2]$ need not even be defined in general. Consider the random variable on $\mathbb{N}^\ast$ (vector of dimension 1) with probability mass function $$ \mathbb{P}\{ X = n\} = \frac{1}{\zeta(3)}\cdot\frac{1}{n^3}. $$ (for which you do have $\mathbb{E}[X] = \sum_{n=1}^\infty \frac{1}{\zeta(3)}\cdot\frac{1}{n^2} = \frac{\pi^2}{6\zeta(3)}$)

Clement C.
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  • Assuming the existence of both expectations, is it possible to obtain a bound on $E[\Vert x_k \Vert^2]$ by tweaking Jensens's inequality of some other way? – Ron May 14 '14 at 09:11
  • If $X$ is supported on a bounded domain (i.e., $\lVert X\rVert \leq M$ a.s.), you can get a very loose bound $\mathbb{E}[\lVert X\rVert^2]\leq \sqrt{M}\cdot \sqrt{\mathbb{E}[\lVert X\rVert]}$, but I'm not sure that's the type of things you are looking for... – Clement C. May 14 '14 at 10:17
  • Another comment: you cannot get any bound depending only on $\mathbb{E}\lVert X\rvert$ (without assumptions like e.g. the support being bounded) in general. For instance, take the $X$ defined in my answer, and tweak it a bit to have the support being ${,\dots, N}$ for an arbitrary $N$ (this will change the normalization constant $\zeta(3)$, but the sums will now only be $\sum_{n=1}^N$). Both expectations will be defined, $\mathbb{E}[\lVert X\rVert]$ will depend on $N$ but be bounded by a constant (independent of $N$) yet $\mathbb{E}[\lVert X\rVert^2]\propto \ln N$ (thus arbitrarily large). – Clement C. May 14 '14 at 10:39
  • Sorry but I fail to understand the "very loose bound" in your comment on May 14 at 10:17. – Did Jul 24 '14 at 16:07
  • Looked back, I failed to see it too I think I missed a $4$ in the exponent of $\lVert X\rVert$: $\sqrt{\mathbb{E}[\lVert X\rVert^4]}$, if what I was thinking was a Cauchy-Schwarz application. – Clement C. Aug 05 '14 at 11:01
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No, not in general, but you can say that $\left[\mathbb{E}\left(\biggr|\biggr|\vec{X_k}\biggr|\biggr|\right)\right]^2 <a^2$ .

afedder
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