Bounds on the expectation of $|X-Y|$ for $X,Y$ Poisson

Question

I would have a proof of the following fact; but it's a bit clunky, and am wondering if one can get a more elegant one (and/or improve the constants). I couldn't find this anywhere, and searching properties of the Skellam distribution didn't help much either.

Let $X\sim\operatorname{Poi}(\lambda)$ and $Y\sim\operatorname{Poi}(\mu)$ be two independent random variables. Then $$ \max(|\lambda-\mu|, \frac{1}{40}\min(\sqrt{\lambda+\mu}, \lambda+\mu)) \leq \mathbb E[|X-Y|] \leq\min(|\lambda-\mu|+\sqrt{\lambda+\mu}, \lambda+\mu) $$

Here is the current proof I came up with:

The upper bound follows from Cauchy–Schwarz, as $\mathbb E[(X-Y)^2] = (\lambda-\mu)^2 + \lambda+\mu$; and by the triangle inequality, as $\mathbb{E}[|X-Y|]\leq \mathbb{E}[X+Y]$. The lower bound $|\lambda-\mu|$ is a direct consequence of Jensen's inequality; we now proceed to establish the second term.

• Consider first the case $\lambda+\mu\geq 1$. We will use Paley–Zygmund, noting that for any $c\in(0,1)$ we have $$ \mathbb{E}[|X-Y|] \geq c\sqrt{\mu+\lambda}\cdot\mathbb{P}\{|X-Y|\geq c\sqrt{\mu+\lambda} \} $$ and $$ \begin{align*} \mathbb{P}\{|X-Y|&\geq c\sqrt{\mu+\lambda}\} \\ &= \mathbb{P}\{(X-Y)^2\geq c^2(\mu+\lambda)\} \\ &\geq \mathbb{P}\{(X-Y)^2\geq c^2\mathbb{E}[(X-Y)^2]\}\\ &\geq (1-c^2)^2\frac{\mathbb{E}[(X-Y)^2]^2}{\mathbb{E}[(X-Y)^4]}\\ &= \frac{ (1-c^2)^2(\lambda+\mu+(\lambda-\mu)^2)^2}{\lambda+\mu+6(\lambda^2+\mu^2)+6(\lambda-\mu)^2(\lambda+\mu)+(\lambda-\mu)^2+(\lambda-\mu)^4}\\ &\overset{(\dagger)}{\geq} (1-c^2)^2 \frac{(\lambda+\mu)^2+(\lambda-\mu)^4}{(\lambda+\mu)+10(\lambda+\mu)^2+3(\lambda-\mu)^4}\\ &\overset{(\ddagger)}{\geq} \frac{(1-c^2)^2}{11} \end{align*} $$ where $(\dagger)$ is using the AM-GM inequality and $(a-b)^2\leq (a+b)^2$; and $(\ddagger)$ relies on $\lambda+\mu \leq (\lambda+\mu)^2$, which holds since $\lambda+\mu\geq 1$. Taking $c=1/\sqrt{5}$ concludes the proof of this case, as then $\frac{c(1-c^2)^2}{11} > \frac{1}{40}$.

• If $\lambda+\mu \leq 1$, we will use properties of the modified Bessel function of the first kind $I_0$ (namely, that it is nondecreasing on $[0,\infty)$, with $e^{-x}I_0(x) \geq 1-\frac{x}{2}$ for $x\in[0,1]$) to conclude, as $$ \begin{align} \mathbb{E}[|X-Y|] &\geq 1 - \mathbb{P}\{|X-Y|=0\} \\ &= 1- e^{-(\lambda+\mu)} I_0(2\sqrt{\lambda\mu}) \\ &\geq 1- e^{-(\lambda+\mu)} I_0(\lambda+\mu) \geq \frac{\lambda+\mu}{2} \end{align}$$ where the first inequality is the AM-GM inequality.

Answer 1

$\newcommand{\la}{\lambda}$Let us show that \begin{equation*} E|Z|\ge J(1)\min[c,\sqrt c\,], \tag{1} \end{equation*} where \begin{equation*} Z:=X-Y,\quad c:=\la+\mu, \end{equation*} \begin{equation*} J(x):=\frac2\pi\,\int_0^\infty\frac{1-\exp\{-x\, (1-\cos t)\}}{t^2}\,dt. \end{equation*} Mathematica's command NIntegrate[] produces \begin{equation*} J(1)=0.67390\dots. \end{equation*} So, the constant factor $J(1)$ is an about 27-times improvement of the corresponding coefficient $\frac1{40}$ in the OP. Moreover, the constant factor $J(1)$ is the best possible.

Inequality (1) is based on the Zolotarev identity \begin{equation*} E|Z|=\frac2\pi\,\int_0^\infty\frac{1-\Re Ee^{itZ}}{t^2}\,dt; \end{equation*} see e.g. formula (3.26) in this paper or formula (48) in the corresponding arXiv preprint. In our case, \begin{equation*} \Re Ee^{itZ}=\exp\{-c(1-\cos t)\}\cos((\la-\mu)\sin t)\le\exp\{-c(1-\cos t)\}; \end{equation*} Moreover, the latter inequality will turn into the equality when $\la=\mu$.

So, \begin{equation*} E|Z|\ge J(c), \tag{2} \end{equation*} and the latter inequality will turn into the equality when $\la=\mu$.

Note that, for each real $z>0$, \begin{equation*} \frac{1-e^{-zx}}x=\int_0^z e^{-xu}\,du \end{equation*} is decreasing in $x$. So, $J(x)/x$ is decreasing in real $x>0$. Moreover, as shown in this answer, $J(x)/\sqrt x$ is increasing in real $x>0$. So, $J(c)\ge cJ(1)$ if $c\in(0,1]$ and $J(c)\ge\sqrt c\,J(1)$ if $c\in[1,\infty)$.

Thus, (1) follows from (2). Moreover, (1) turns into the equality when $\la=\mu=1/2$ (so that $c=1$).