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(1) Simple bounds:

For $m=1$ (see here (or here and here)) the inequalities

\begin{align*} \sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}&\leq \mathbb{E}(M_1)\leq \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}+ {\max_i( p_i)\over \lVert p\rVert_2^2}\mbox{ and }\\ \sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}&\leq \mathbb{E}(M_1)\leq 2 \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}\end{align*} hold. Since $M_m$ is obviously stochastically smaller than the sum of $m$ independent copies of $M_1$

$$\sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}+1 \leq \mathbb{E}(M_m)\leq 2m \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}$$

(2) Precise asymptotics:

Assume in the sequel the conditions and notation of Thm. 4 in [CP00]. Then $$\lVert p(n)\rVert_2\, M_m(n)=s_nR_{nm}\mbox{ in [CP00]})$$ converges in distribution to a random variable $\eta_m$, where the joint distribution of $(\eta_1,\eta_2,\ldots,\eta_m)$ is given in Thm. 6. It is not hard to see that the random variables $\lVert p(n)\rVert_2\, M_m(n)$ are uniformly integrable. Thus $\eta_m$ is integrable and $$\mathbb{E}\left(\lVert p(n)\rVert_2 M_m(n)\right)\longrightarrow \mathbb{E}(\eta_m)$$ Thus asymptotically $$\mathbb{E}(M_m(n))\sim \frac{C_m}{\lVert p(n)\rVert_2}$$ where $C_m=\mathbb{E}(\eta_m)$. The distribution of $\eta_m$ depends on all the "large cell" parameters $\theta_1,\theta_2,...$ and $C_m$ will in general not be of a convenient explicit form. But the case $\theta_1=0$ is simple. Here we get from Thm. 6 that $\mathbb{P}(\eta_m>x)=\mathbb{P}(\mathrm{Poiss}(\tfrac{1}{2}x^2)\leq m-1)$ and integrating this gives \begin{align*} C_m&={m-\tfrac{1}{2} \choose m-1}\sqrt{{\pi \over 2}} &\mbox{ for } \theta_1=0 \\ %\mbox{ and } C_m&=m+1 &\mbox{ for } \theta_1=1 \end{align*} In particular, for the uniform distribution on $\{1,\ldots,n\}$ $$E(M_2)\sim \tfrac{3}{2}\sqrt{\tfrac{1}{2}\pi n},\; \mathbb{E}(M_3)\sim \tfrac{15}{8} \sqrt{\tfrac{1}{2}\pi n},\ldots\;\;.$$ (This certainly looks "classical", but I haven't found anything in the literature.)

(1) Simple bounds:

For $m=1$ (see here (or here and here)) the inequalities

\begin{align*} \sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}&\leq \mathbb{E}(M_1)\leq \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}+ {\max_i( p_i)\over \lVert p\rVert_2^2}\mbox{ and }\\ \sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}&\leq \mathbb{E}(M_1)\leq 2 \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}\end{align*} hold. Since $M_m$ is obviously stochastically smaller than the sum of $m$ independent copies of $M_1$

$$\sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}+1 \leq \mathbb{E}(M_m)\leq 2m \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}$$

(2) Precise asymptotics:

Assume in the sequel the conditions and notation of Thm. 4 in [CP00]. Then $$\lVert p(n)\rVert_2\, M_m(n)=s_nR_{nm}\mbox{ in [CP00]})$$ converges in distribution to a random variable $\eta_m$, where the joint distribution of $(\eta_1,\eta_2,\ldots,\eta_m)$ is given in Thm. 6. It is not hard to see that the random variables $\lVert p(n)\rVert_2\, M_m(n)$ are uniformly integrable. Thus $\eta_m$ is integrable and $$\mathbb{E}\left(\lVert p(n)\rVert_2 M_m(n)\right)\longrightarrow \mathbb{E}(\eta_m)$$ Thus asymptotically $$\mathbb{E}(M_m(n))\sim \frac{C_m}{\lVert p(n)\rVert_2}$$ where $C_m=\mathbb{E}(\eta_m)$. The distribution of $\eta_m$ depends on all the "large cell" parameters $\theta_1,\theta_2,...$ and $C_m$ will in general not be of a convenient explicit form. But the case $\theta_1=0$ is simple. Here we get from Thm. 6 that $\mathbb{P}(\eta_m>x)=\mathbb{P}(\mathrm{Poiss}(\tfrac{1}{2}x^2)\leq m-1)$ and integrating this gives \begin{align*} C_m&={m-\tfrac{1}{2} \choose m-1}\sqrt{{\pi \over 2}} &\mbox{ for } \theta_1=0 \\ %\mbox{ and } C_m&=m+1 &\mbox{ for } \theta_1=1 \end{align*} In particular, for the uniform distribution on $\{1,\ldots,n\}$ $$E(M_2)\sim \tfrac{3}{2}\sqrt{\tfrac{1}{2}\pi n},\; \mathbb{E}(M_3)\sim \tfrac{15}{8} \sqrt{\tfrac{1}{2}\pi n},\ldots\;\;.$$ (This certainly looks "classical", but I haven't found anything in the literature.)

(1) Simple bounds:

For $m=1$ (see here (or here and here)) the inequalities

\begin{align*} \sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}&\leq \mathbb{E}(M_1)\leq \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}+ {\max_i( p_i)\over \lVert p\rVert_2^2}\mbox{ and }\\ \sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}&\leq \mathbb{E}(M_1)\leq 2 \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}\end{align*} hold. Since $M_m$ is obviously stochastically smaller than the sum of $m$ independent copies of $M_1$

$$\sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}+1 \leq \mathbb{E}(M_m)\leq 2m \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}$$

(2) Precise asymptotics:

Assume in the sequel the conditions and notation of Thm. 4 in CM[00]. Then $$\lVert p(n)\rVert_2\, M_m(n)(=s_nR_{nm}\mbox{ in CM[00]})$$ converges in distribution to a random variable $\eta_m$, where the joint distribution of $(\eta_1,\eta_2,\ldots,\eta_m)$ is given in Thm. 6. It is not hard to see that the random variables $\lVert p(n)\rVert_2\, M_m(n)$ are uniformly integrable. Thus $\eta_m$ is integrable and $$\mathbb{E}\left(\lVert p(n)\rVert_2 M_m(n)\right)\longrightarrow \mathbb{E}(\eta_m)$$ Thus asymptotically $$\mathbb{E}(M_m(n))\sim \frac{C_m}{\lVert p(n)\rVert_2}$$ where $C_m=\mathbb{E}(\eta_m)$. The distribution of $\eta_m$ depends on all the "large cell" parameters $\theta_1,\theta_2,...$ and $C_m$ will in general not be of a convenient explicit form. But the case $\theta_1=0$ is simple. Here we get from Thm. 6 that $\mathbb{P}(\eta_m>x)=\mathbb{P}(\mathrm{Poiss}(\tfrac{1}{2}x^2)\leq m-1)$ and integrating this gives \begin{align*} C_m&={m-\tfrac{1}{2} \choose m-1}\sqrt{{\pi \over 2}} &\mbox{ for } \theta_1=0 \\ %\mbox{ and } C_m&=m+1 &\mbox{ for } \theta_1=1 \end{align*} In particular, for the uniform distribution on $\{1,\ldots,n\}$ $$E(M_2)\sim \tfrac{3}{2}\sqrt{\tfrac{1}{2}\pi n},\; \mathbb{E}(M_3)\sim \tfrac{15}{8} \sqrt{\tfrac{1}{2}\pi n},\ldots\;\;.$$ (This certainly looks "classical", but I haven't found anything in the literature.)

(1) Simple bounds:

For $m=1$ (see here (or here and here)) the inequalities

\begin{align*} \sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}&\leq \mathbb{E}(M_1)\leq \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}+ {\max_i( p_i)\over \lVert p\rVert_2^2}\mbox{ and }\\ \sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}&\leq \mathbb{E}(M_1)\leq 2 \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}\end{align*} hold. Since $M_m$ is obviously stochastically smaller than the sum of $m$ independent copies of $M_1$

$$\sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}+1 \leq \mathbb{E}(M_m)\leq 2m \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}$$

(2) Precise asymptotics:

Assume in the sequel the conditions and notation of Thm. 4 in [CP00]. Then $$\lVert p(n)\rVert_2\, M_m(n)=s_nR_{nm}\mbox{ in [CP00]})$$ converges in distribution to a random variable $\eta_m$, where the joint distribution of $(\eta_1,\eta_2,\ldots,\eta_m)$ is given in Thm. 6. It is not hard to see that the random variables $\lVert p(n)\rVert_2\, M_m(n)$ are uniformly integrable. Thus $\eta_m$ is integrable and $$\mathbb{E}\left(\lVert p(n)\rVert_2 M_m(n)\right)\longrightarrow \mathbb{E}(\eta_m)$$ Thus asymptotically $$\mathbb{E}(M_m(n))\sim \frac{C_m}{\lVert p(n)\rVert_2}$$ where $C_m=\mathbb{E}(\eta_m)$. The distribution of $\eta_m$ depends on all the "large cell" parameters $\theta_1,\theta_2,...$ and $C_m$ will in general not be of a convenient explicit form. But the case $\theta_1=0$ is simple. Here we get from Thm. 6 that $\mathbb{P}(\eta_m>x)=\mathbb{P}(\mathrm{Poiss}(\tfrac{1}{2}x^2)\leq m-1)$ and integrating this gives \begin{align*} C_m&={m-\tfrac{1}{2} \choose m-1}\sqrt{{\pi \over 2}} &\mbox{ for } \theta_1=0 \\ %\mbox{ and } C_m&=m+1 &\mbox{ for } \theta_1=1 \end{align*} In particular, for the uniform distribution on $\{1,\ldots,n\}$ $$E(M_2)\sim \tfrac{3}{2}\sqrt{\tfrac{1}{2}\pi n},\; \mathbb{E}(M_3)\sim \tfrac{15}{8} \sqrt{\tfrac{1}{2}\pi n},\ldots\;\;.$$ (This certainly looks "classical", but I haven't found anything in the literature.)

(1) Simple bounds:

For $m=1$ (see here (or here and here)) the inequalities

\begin{align*} \sqrt{\frac{\pi}{2}}{1\over ||p||_2}&\leq \mathbb{E}(M_1)\leq \sqrt{\frac{\pi}{2}}{1 \over ||p||_2}+ {\max_i( p_i)\over ||p||_2^2}\mbox{ and }\\ \sqrt{\frac{\pi}{2}}{1\over ||p||_2}&\leq \mathbb{E}(M_1)\leq 2 \sqrt{\frac{\pi}{2}}{1 \over ||p||_2}\end{align*} hold. Since $M_m$ is obviously stochastically smaller than the sum of $m$ independent copies of $M_1$

$$\sqrt{\frac{\pi}{2}}{1\over ||p||_2}+1 \leq \mathbb{E}(M_m)\leq 2m \sqrt{\frac{\pi}{2}}{1 \over ||p||_2}$$

(2) Precise asymptotics:

Assume in the sequel the conditions and notation of Thm. 4 in CM[00]. Then $$||p(n)||_2\, M_m(n)(=s_nR_{nm}\mbox{ in CM[00]})$$ converges in distribution to a random variable $\eta_m$, where the joint distribution of $(\eta_1,\eta_2,\ldots,\eta_m)$ is given in Thm. 6. It is not hard to see that the random variables $||p(n)||_2\, M_m(n)$ are uniformly integrable. Thus $\eta_m$ is integrable and $$\mathbb{E}\left(||p(n)||_2 M_m(n)\right)\longrightarrow \mathbb{E}(\eta_m)$$ Thus asymptotically $$\mathbb{E}(M_m(n))\sim \frac{C_m}{||p(n)||_2}$$ where $C_m=\mathbb{E}(\eta_m)$. The distribution of $\eta_m$ depends on all the "large cell" parameters $\theta_1,\theta_2,...$ and $C_m$ will in general not be of a convenient explicit form. But the case $\theta_1=0$ is simple. Here we get from Thm. 6 that $\mathbb{P}(\eta_m>x)=\mathbb{P}(\mathrm{Poiss}(\tfrac{1}{2}x^2)\leq m-1)$ and integrating this gives \begin{align*} C_m&={m-\tfrac{1}{2} \choose m-1}\sqrt{{\pi \over 2}} &\mbox{ for } \theta_1=0 \\ %\mbox{ and } C_m&=m+1 &\mbox{ for } \theta_1=1 \end{align*} In particular, for the uniform distribution on $\{1,\ldots,n\}$ $$E(M_2)\sim \tfrac{3}{2}\sqrt{\tfrac{1}{2}\pi n},\; \mathbb{E}(M_3)\sim \tfrac{15}{8} \sqrt{\tfrac{1}{2}\pi n},\ldots\;\;.$$ (This certainly looks "classical", but I haven't found anything in the literature.)

(1) Simple bounds:

For $m=1$ (see here (or here and here)) the inequalities

\begin{align*} \sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}&\leq \mathbb{E}(M_1)\leq \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}+ {\max_i( p_i)\over \lVert p\rVert_2^2}\mbox{ and }\\ \sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}&\leq \mathbb{E}(M_1)\leq 2 \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}\end{align*} hold. Since $M_m$ is obviously stochastically smaller than the sum of $m$ independent copies of $M_1$

$$\sqrt{\frac{\pi}{2}}{1\over \lVert p\rVert_2}+1 \leq \mathbb{E}(M_m)\leq 2m \sqrt{\frac{\pi}{2}}{1 \over \lVert p\rVert_2}$$

(2) Precise asymptotics:

Assume in the sequel the conditions and notation of Thm. 4 in CM[00]. Then $$\lVert p(n)\rVert_2\, M_m(n)(=s_nR_{nm}\mbox{ in CM[00]})$$ converges in distribution to a random variable $\eta_m$, where the joint distribution of $(\eta_1,\eta_2,\ldots,\eta_m)$ is given in Thm. 6. It is not hard to see that the random variables $\lVert p(n)\rVert_2\, M_m(n)$ are uniformly integrable. Thus $\eta_m$ is integrable and $$\mathbb{E}\left(\lVert p(n)\rVert_2 M_m(n)\right)\longrightarrow \mathbb{E}(\eta_m)$$ Thus asymptotically $$\mathbb{E}(M_m(n))\sim \frac{C_m}{\lVert p(n)\rVert_2}$$ where $C_m=\mathbb{E}(\eta_m)$. The distribution of $\eta_m$ depends on all the "large cell" parameters $\theta_1,\theta_2,...$ and $C_m$ will in general not be of a convenient explicit form. But the case $\theta_1=0$ is simple. Here we get from Thm. 6 that $\mathbb{P}(\eta_m>x)=\mathbb{P}(\mathrm{Poiss}(\tfrac{1}{2}x^2)\leq m-1)$ and integrating this gives \begin{align*} C_m&={m-\tfrac{1}{2} \choose m-1}\sqrt{{\pi \over 2}} &\mbox{ for } \theta_1=0 \\ %\mbox{ and } C_m&=m+1 &\mbox{ for } \theta_1=1 \end{align*} In particular, for the uniform distribution on $\{1,\ldots,n\}$ $$E(M_2)\sim \tfrac{3}{2}\sqrt{\tfrac{1}{2}\pi n},\; \mathbb{E}(M_3)\sim \tfrac{15}{8} \sqrt{\tfrac{1}{2}\pi n},\ldots\;\;.$$ (This certainly looks "classical", but I haven't found anything in the literature.)