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.

Gilles

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*}