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Let $X_1,X_2, \dots, X_N $ be i.i.d sub-guaussian random variable ,with $E[X_i] = 0$ and $\operatorname{Var}[X_i] =1$ $a_1,a_2,\dots ,a_N \in R$

$$\left(\sum_{i=1}^N a_i^2\right)^{1/2} \leq \left( E \left| \sum_{i=1}^N a_iX_i\right|^p\right)^{1/p}$$ for all $p \geq2$ ?

is that possible to do that?

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  • $\begingroup$ If you add squares on the $a_i$'s, yes. It's from Jensen. $\endgroup$ – Clement C. Sep 18 '18 at 3:56
  • $\begingroup$ @ClementC. sorry not quite understand which mean , I tried Jensen from RHS to obtained lower bound.. $\endgroup$ – ShaoyuPei Sep 18 '18 at 5:34
  • $\begingroup$ The lower bound as stated is false. It's not even "homogeneous" is you say multiply all the $a_i$'s by the same factor. You need $$\left(\sum_{i=1}^n a_i^2\right)^{1/2}$$ for the lower bound to hold. $\endgroup$ – Clement C. Sep 18 '18 at 5:38
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Yes, it is possible. Actually, an upper bound holds as well -- namely, this is the Khintchine inequality for subgaussian random variables. See e.g. these lecture notes.

The lower bound you seek is the simple part, and id actually a simple consequence of Jensen's inequality: $$\begin{align*} \mathbb{E}\left[\left\lvert\sum_i a_i X_i\right\rvert^p\right]^{1/p} &= \mathbb{E}\left[\left(\left|\sum_i a_i X_i\right|^2\right)^{p/2}\right]^{1/p} \geq \left(\mathbb{E}\left[\left(\sum_i a_i X_i\right)^2\right]\right)^{1/2} = \left(\operatorname{Var}\left[\sum_i a_i X_i\right]\right)^{1/2}\\ &= \left(\sum_i a_i^2 \operatorname{Var}\left[X_i\right]\right)^{1/2} = \left(\sum_i a_i^2 \right)^{1/2} \end{align*}$$ using the fact that $p\geq 2$ (i.e., $p/2\geq 1$) for Jensen's inequality, then the fact that $\mathbb{E}\sum_i a_i X_i = 0$ to rewrite the expectation of the square as the variance, and finally independence of the $X_i$'s.

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  • $\begingroup$ thanks for helping me serval times , I should learn a lot from you , on both math and tcs $\endgroup$ – ShaoyuPei Sep 18 '18 at 5:47
  • $\begingroup$ @ShaoyuPei You're welcome -- glad to help. $\endgroup$ – Clement C. Sep 18 '18 at 5:48
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Consider a Random Variable $$Z=\sum_{i=1}^{N} a_iX_i$$

Then $Z$ is Gaussian with mean $E[Z]=0$ and variance $Var[Z] = \sigma^2=\sum_{i=1}^{N} a_i^2 $

$E[|Z|^p] = \sigma^p[(p-1)!!]\sqrt{\dfrac{2}{\pi}}$ if p is odd, and

$E[|Z|^p] = \sigma^p[(p-1)!!]$ if n is even. https://en.wikipedia.org/wiki/Normal_distribution#Moments

$\implies E[|Z|^p] = \sigma^p k$, let $k$ be a constant depending on whether n is even or odd. But note that $|k|\geq1$ for $p \geq 2$.

$\therefore$ $(E[|Z|^p])^{1/p} = (\sigma^p k)^{1/p} \geq \sigma$. But $\sigma = (\sum_{i=1}^{N}a_i^2)^{1/2}$

$\implies$ $(\sum_{i=1}^{N}a_i^2)^{1/2} \leq (E[|Z|^p])^{1/p}$

$$(\sum_{i=1}^{N}a_i^2)^{1/2} \leq (E[|\sum_{i=1}^{N} a_iX_i|^p])^{1/p} $$

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  • $\begingroup$ The $X_i$'s are sub-gaussian, not assumed actually Gaussian. $\endgroup$ – Clement C. Sep 18 '18 at 6:15

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