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If $X\sim\mathrm{Poisson}(\lambda)$, it is known that $$ \mathbb{E}[|X-\lambda|] = \frac{2e^{-\lambda} \lambda^{\lfloor \lambda\rfloor +1}}{\lfloor \lambda\rfloor!} \operatorname*{\sim}_{\lambda\to\infty} \sqrt{\frac{2}{\pi}\lambda} $$

Based on this, it is not hard to show, via Jensen's inequality, that if $X,Y\sim\mathrm{Poisson}(\lambda)$ are independent, $$ \mathbb{E}[|X-\lambda|] \leq \mathbb{E}[|X-Y|] \leq 2\cdot \mathbb{E}[|X-\lambda|] $$ which gives that $\mathbb{E}[|X-Y|] = \Theta(\sqrt{\lambda})$ as $\lambda\to\infty$. Moreover, the exact leading constant in the $\Theta(\cdot)$ is between $\sqrt{\frac{2}{\pi}}$ and $2\sqrt{\frac{2}{\pi}}$. One can even refine the upper bound by Cauchy—Schwarz to get $$ \mathbb{E}[|X-\lambda|] \leq \mathbb{E}[|X-Y|] \leq \sqrt{\mathbb{E}[(X-Y)^2]} = \sqrt{2\lambda} $$ so we know that the right asymptotic constant is between $\sqrt{\frac{2}{\pi}}$ and $\sqrt{2}$.

Based on numerical evidence (see below), it looks like the right answer should be $$ \mathbb{E}[|X-Y|] \operatorname*{\sim}_{\lambda\to\infty} \frac{2}{\sqrt{\pi}}\sqrt{\lambda} $$

Is this result known? Is there a proof available somewhere?

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Clement C.
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1 Answers1

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$Z=X-Y$ follows the Skellam distribution which has probability mass function $$ p(Z=k) = e^{-2\lambda}I_{|k|}(2\lambda), $$ where $I_{\alpha}$ is the modified Bessel function of the first kind. Thus we need to compute the infinite sum $2e^{-x}\sum_{k=1}^{\infty} kI_k(x)$ where $x=2\lambda$. Since we have the following recurrence relation: $$ kI_k(x) = \frac{x}{2}(I_{k-1}(x)-I_{k+1}(x)), $$ the partial sum of the series can be simplified as: $$ 2\sum_{k=1}^{N} kI_k(x) = x(I_0(x)+I_1(x))-x(I_N(x)+I_{N+1}(x)).$$ We have an asymptotic formula for $I_N(x)$ as $N\to\infty$: $$ I_N(x) \sim \frac{1}{\sqrt{2\pi N}}\left(\frac{ex}{2N}\right)^N \underset{N\to\infty}{\to} 0. $$ Finally, we use the asymptotic formula $I_k(x)\sim \frac{e^x}{\sqrt{2\pi x}}$ as $x\to\infty$: \begin{align*}2e^{-x}\sum_{k=1}^{\infty} kI_k(x) = xe^{-x}(I_0(x)+I_1(x)) &\underset{x\to\infty}{\sim} x\cdot\frac{2}{\sqrt{2\pi x}} \\ &= \sqrt{\frac{2x}{\pi}} \\ &= \frac{2}{\sqrt{\pi}}\sqrt{\lambda}. \end{align*}

DP_
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