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Setup

Let $X \sim \text{Binomial}(p, n)$, and $r \ge 1$.

Question

What is a good upper-bound for $\mathbb E[|X-np|^r]$ ?

Solution for small $r$

  • If $r=2$, then $\mathbb E[|X-np|^2] = np(1-p)$.

  • If $1 \le r < 2$, then Jensen's inequality gives $$ \mathbb E[|X-np|^r] =(\mathbb E[|X-np|^r])^{2/r})^{(r/2)} \le (\mathbb E[|X-np|^2])^{r/2} = (np(1-p))^{(r/2)}. $$

Notes

I'm ultimately interested in bounding (via McDiamid's inequality) the $\ell_r$-distance between a distribution $P$ and it's empirical version $\hat{P}_n$.

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1 Answer 1

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By the main result of the paper Exact Rosenthal-type bounds, we have $$E|X-np|^r\le c^r E|\Pi_\lambda-\lambda|^r $$ for real $r\in(2,\infty)\setminus(3,4)\setminus(4,5)$, where real $c>0$ and $\lambda>0$ are defined by the conditions $$c^r\lambda=n(q^rp+p^rq)\quad \text{and}\quad c^2\lambda=npq; $$ $q:=1-p$; and $\Pi_\lambda$ is a Poisson random variable with parameter $\lambda$.

Other results on Rosenthal-type bounds can be found e.g. in this paper or its arXiv version, and in references therein.

Added: In a comment, the OP stated that it may be assumed that $1\le r\le 2$. This simplifies the matter a great deal. Indeed, in this case we have \begin{equation} E|X-np|^r\le\min((npq)^{r/2},2npq(q^{r-1}+p^{r-1}))\ll s^r\wedge s^2,\tag{1} \end{equation} where $(npq)^{r/2}$ is the bound the OP obtained by using Jensen's inequality; and $2npq(q^{r-1}+p^{r-1})$ is a bound immediately obtained by using the von Bahr--Esseen inequality -- see e.g. this paper or its arXiv version; $s:=\sqrt{npq}$. For positive expressions $e_1$ and $e_2$, we write $e_1\ll e_2$ or, equivalently, $e_2\gg e_1$ if $e_1\le C e_2$ for some universal positive real constant $C$.

The upper bound on $E|X-np|^r$ in (1) is optimal up to a universal constant factor: for $r\in[1,2)$ \begin{equation} E|X-np|^r\gg s^r\wedge s^2.\tag{2} \end{equation} This lower bound on $E|X-np|^r$ is obtained by using the log-convexity of $m_t:=E|X-np|^t$ in $t>0$ and an upper Rosenthal-type bound -- as follows: By (say) Theorem 1.5 in the already cited paper Exact Rosenthal-type bounds, we have $m_3\ll s^3\wedge s^2$. Now using the log-convexity of $m_t:=E|X-np|^t$ in $t>0$ or, equivalently, the Hölder inequality, we have \begin{equation*} m_2\le m_r^{1/a}m_3^{1-1/a}, \end{equation*} where $a:=3-r$. Hence, \begin{equation*} E|X-np|^r=m_r\ge m_2^{3-r}m_3^{r-2}\gg s^{6-2r}(s^{3r-6}\wedge s^{2r-4}) =s^r\wedge s^2. \end{equation*} Thus, (2) indeed holds.

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    $\begingroup$ Thanks for the references. Can the bound be made more explicit ? I'm not away that the problem is simpler for Poisson variables. Thanks in advance. $\endgroup$
    – dohmatob
    Commented May 28, 2019 at 2:54
  • $\begingroup$ @dohmatob : Yes, in [2] the bounds are quite explicit. However, they are much more general and contain a number of parameters, in which it should be straightforward to minimize, depending on how $n$ and $p$ are related in your setting. $\endgroup$ Commented May 28, 2019 at 3:20
  • $\begingroup$ Hum, upvoted, but I'm afraid this is not readily exploitable. I'd expect there exists explicit bounds for such an explicit problem. [2] appears to be rather too ambitious, and I'm willing to strike a compromise. I only require a bound with a "decent" dependence on the parameters $n$, $p$, and $r$ (it may be assumed that $1 \le r \le 2$, BTW). $\endgroup$
    – dohmatob
    Commented May 28, 2019 at 3:51
  • $\begingroup$ Thanks for the nice addendum. It's really helpful. Also Bahr-Esseen is fresh news for me. Thanks. $\endgroup$
    – dohmatob
    Commented May 29, 2019 at 12:31
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    $\begingroup$ I have now shown that the bound is the best possible. $\endgroup$ Commented May 29, 2019 at 14:27

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