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Suppose we have b bins and each bin could have max k balls. We have sufficient balls and what is the expected number if we need to full all bins? I know the answer when k = 1, the expected number is b * (lnb -1). But what the expected number when k>1 ? I am thinking this question for a long time but still can't figure out. Anyone would help me out? Thank you in advanced.

Li

DingLi
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1 Answers1

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Consider two modifications of the model which make it simpler to analyze:

  • Assume the balls arrive at the times of a Poisson process of rate $b$ and that each one is allocated randomly to a bin chosen uniformly randomly. Equivalently each bin receives balls at rate $1$.
  • Continue to add the balls to the bins they land in even when the bin is full.

The first time when there is at least $k$ balls in each of the $b$ bins is $T=\max\{T_1,\ldots,T_b\}$ when $T_i$ is the arrival time of the $k$th ball in bin $i$. Since balls arrive at rate $b$, the mean number of balls used at time $T$, which is also the number of steps of the process in discrete time, is $bE_k^b[T]$. Furthermore, $$ E_k^b[T]=\int_0^\infty P_k^b[T\geqslant t]\,\mathrm dt,\qquad[T\lt t]=\bigcap_{i=1}^b[T_i\lt t]. $$ For each $i$, $T_i$ is the sum of $k$ i.i.d. standard exponential times hence $T_i$ has density $f_k$ where $$ f_k(t)=\frac1{(k-1)!}t^{k-1}\mathrm e^{-t}\,\mathbf 1_{t\geqslant0}. $$ Thus, $$ E_k^b[T]=\int_0^\infty \left(1-F_k(t)^b\right)\,\mathrm dt,\qquad F_k(t)=\int_0^tf_k. $$ For example, if $k=1$, $F_1(t)=1-\mathrm e^{-t}$ hence $$ E_1^b[T]=\int_0^\infty \left(1-(1-\mathrm e^{-t})^b\right)\,\mathrm dt=\int_0^1 \left(1-s^b\right)\,\frac{\mathrm ds}{1-s}=\int_0^1\sum_{n=1}^{b}s^{n-1}\mathrm ds=\sum_{n=1}^{b}\frac1n. $$ Thus, $bE_1^b[T]=bH_b$ where $H_b$ is the $b$th harmonic number $H_b=\sum_{n=1}^{b}\frac1n$, and $$ bE_1^b[T]\sim b\log b. $$ Likewise, for every $k\geqslant1$, $$ E_k^b[T]=\int_0^1 \left(1-s^b\right)\,\frac{\mathrm ds}{f_k(t)}, $$ where $s=F_k(t)$, that is, $t=F_k^{-1}(s)$. Note that $$ 1-F_k(t)=\sum_{n=0}^kf_n(t), $$ hence $1-F_k(t)\sim f_k(t)$ when $t\to\infty$, that is, $t=F_k^{-1}(s)$ implies that $f_k(t)\sim1-s$ when $s\to1$. For example, $$ E_2^b[T]=\int_0^1 \left(1-s^b\right)\,\frac{\mathrm ds}{t\mathrm e^{-t}},\qquad 1-s=(t+1)\mathrm e^{-t}. $$ An obvious bound is $E_k^b[T]\leqslant kE_1^b[T]$. Since $E_k^b[T]\geqslant E_1^b[T]$, this shows that, for every fixed $k$, when $b\to\infty$, $$ bE_k^b[T]=\Theta(b\log b), $$ and suggests that, for some constant $c_k$ depending on $k$ such that $1\leqslant c_k\leqslant k$, $$ bE_k^b[T]\sim c_kb\log b. $$

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