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

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.

$$\newcommand{\la}{\lambda}$$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 $$c=1$$).

• That's great, thanks! From numerical experiments (in particular, for the case $\lambda=\mu$), it looks like the lower bound on that term could be improved to some value $\approx 0.67$, and the upper bound (for the constant in front of the $\lambda+\mu$ term, i.e., large values) as well... any idea how achievable that could be? Jun 30 '21 at 8:10
• @ClementC. : Now we have the optimal constant, $\approx0.67$. Jul 1 '21 at 9:54
• Wonderful! ${}{}{}{}{}$ Jul 1 '21 at 12:18