I came across an inequality in the paper of 'Estimation of a function with discontinuities ...' (AoS, 1998, p.1374) and tried to prove it, but could not get to the result. Some simulations also seemed to confirm that the inequality is not always true; I however don't know if computer precision could not interfere. Hereafter is what I have got so far. The element I particularly question is the last part about the value of C(q). Does anyone has ever encountered this inequality ? Is there a better way to get to the result ? Is my 'proof' even correct ?

Thank you very much for your help.


Proposition: $E\vert x+ \xi\vert^q \le (\xi+c(q))^q$, where $\xi$ is a positive constant, $C(q)$ is a quantity depending on q only and $x \sim \mathcal{N}(0,1).$ $C(q)\le 2$.

Proof(?): \begin{align*} l.h.s & \le \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty (\lvert x \rvert +\lvert\xi\rvert)^q \ e^{-x^2/2 } dx\\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \sum_{k=0}^q {{q} \choose {k}} \lvert x \rvert^k \ \xi^{q-k} \ e^{-x^2/2 }dx \\ &= \frac{1}{\sqrt{2\pi}} \sum_{k=0}^q {{q} \choose {k}} \xi^{q-k} \int_{-\infty}^\infty \lvert x \rvert^k \ \ e^{-x^2/2 }dx \\ \end{align*} Recalling that $\int_0^\infty x^k \ \ e^{-x^2/2 }dx = \mathcal{O}((k-1)!!)$ [where !! denotes a pattern alike double factorial], because $\mathbb{E}\lvert x\rvert^k = (k-1)\mathbb{E}\lvert x\rvert^{k-2}$. We further would like to bound the term $([(k-1)!!]^{1/k})$ to use the binomial formula. Replacing k by q in the formula, we get an upper bound and finally obtain: \begin{align*} E\vert x+ \xi\vert^q & \le \sum_{k=0}^q {{q} \choose {k}} \xi^{q-k} \mathcal{O}([(q-1)!!]^{1/q})^k \\ & \le (\xi + C(q))^q \end{align*}

  • $\begingroup$ Isn't the right hand side of the inequality $(\xi+C(q))^q$? $\endgroup$ Jun 5, 2018 at 7:18
  • $\begingroup$ Sorry for the typo. Thank you for finding it $\endgroup$ Jun 5, 2018 at 7:21
  • $\begingroup$ You're claiming $\mathbb E(|N+\alpha|^q)\le (\alpha+2)^q$ for each $\alpha\ge 0$ and each $q$. I don't think this can possibly be true. As $q$ gets large, the left side approaches the $q$th power of the $L^\infty$ norm of $N+\alpha$. This approaches $\infty$. $\endgroup$ Jun 5, 2018 at 7:22
  • $\begingroup$ I do not believe this to be true. It is however at the heart of a theorem published in the paper in the Annals of Statistics. For this reason, I wanted to try to get the best bound and see how far it is from what is claimed. $\endgroup$ Jun 5, 2018 at 7:28
  • $\begingroup$ So doesn't my answer prove that $C(q)$ (assuming it exists, which I'm sure it does) is not less than 2 for sufficiently large $q$? $\endgroup$ Jun 5, 2018 at 7:57

1 Answer 1


If the claim is that $C(q)\le 2$ for all $q$, then the proposition is false: Let $\mathbb P(N\ge 3)=p$. Now for all $q$, you have $\mathbb E(|N|^q)\ge 3^qp$. For sufficiently large $q$, you have $2^q/3^q<p$, so that $\mathbb E(|N|^q)> 2^q$. (The claimed inequality fails for $\xi=0$).

If there's no claim about the value of $C(q)$, you argument shows the proposition should be true: since $((q-1)!!)^{1/q}$ is an increasing sequence, the upper bound you use is correct. Now $((q-1)!!)^{1/q}\sim \sqrt{q/e}$ by Stirling's approximation, so that the best $C(q)$ is something like $\sqrt{q/e}$. (you even see this with $\xi=0$).

  • $\begingroup$ Thank you very much for the swift reply. Have a good day. $\endgroup$ Jun 5, 2018 at 7:39

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