# Birthday problem extension to unequal probabilities and multiple collisions

Let $$p_1, ... ,p_k$$ denote the probabilities of drawing bin $$1, .. ,k$$, where $$\sum_{i = 1}^{k} p_i= 1$$. My question is if we draw $$n$$ times, how can I show that the probability that all bins are drawn less than $$C$$ times is maximized by $$p_1 = p_2 = ... = p_k = 1/k$$ ?

I've tried using conditional probability to solve inductively such as assuming bin $$i$$ is drawn $$ times and trying to reason about the remaining $$k-1$$, but no luck so far. Computing explicitly seems to be really difficult.

Here is a proof using generating functions. Let $$c\geq 2$$ and $$k\geq 2$$ be fixed.

Let $$X=(X_1(n),\ldots,X_k(n))$$ be the $$k$$-tuple of occupancy numbers at time $$n$$, i.e. $$X_i(n)$$ = number of bins of type $$i$$ drawn at time $$n$$.
Clearly $$X$$ has the multinomial distribution with parameters $$n$$ and $$p=(p_1,\ldots,p_k)$$. Therefore
$$\mathbb{P}( \max X_i(n)\leq c-1)=n!\,[t^n] \prod _{i=1}^n q_c(p_it)$$

where $$q_c(t):=\sum_{j=0}^{c-1}\frac{t^j}{j!}$$ is the $$c$$th partial sum of $$\exp(t)$$.

The following lemma is the key.

Lemma
When $$p_i,p_j$$ are each replaced by their arithmetic mean $$a:=(p_i+p_j)/2$$
$$\mathbb{P}(\max X_i(n) \leq c-1 )$$ will not decrease.
If $$c\leq n \leq k(c-1)$$ and all other $$p_\ell$$ are positive $$\mathbb{P}(\max X_i(n) \leq c-1 )$$ will strictly increase if $$p_i\not = p_j$$.

Proof Consider first the case of two factors. Let $$x,y\in\mathbb{R}_+,x. We show that $$[t^m] q_c(xt)q_c(yt)\leq [t^m] \left(q_c(((x+y)/2)t\right)^2\;\;.$$It is easy to see that equality holds for $$m\leq c-1$$ and $$m>2c-2$$. Let $$c\leq m \leq 2c-2$$. We have $$m!\,[t^m] q_{c}(xt)q_c(yt)=(x+y)^m-\sum_{i=0}^{m-c} {m \choose i}\left(x^i y^{m-i}+y^i x^{m-i}\right)$$ For fixed sum $$s=x+y$$ the function $$x \mapsto f(x):=\sum_{i=0}^{m-c} {m \choose i}\left(x^i y^{m-i}+y^i x^{m-i}\right)$$ has the derivative $$f^\prime(x)=(m-c+1){m \choose c} \left(x^{m-c+1}(s-x)^c-(s-x)^{m-c+1}x^c\right)$$. Thus $$x\mapsto f(x)$$ is strictly decreasing (resp. increasing) on $$[0,s/2]$$ (resp. $$[s/2,s]$$), attaining its minimum at $$x=s/2$$. The rest is easy. End proof.

For any distribution $$p=(p_1,\ldots,p_k)$$ we have that $$\mathbb{P}( \max X_i(n)\leq c-1)=1$$ for $$n\leq c-1$$, and $$\mathbb{P}( \max X_i(n)\leq c-1)=0$$ for $$n > k(c-1)$$, so that only the cases $$c\leq n \leq k(c-1)$$ are of interest.

Corollary
For $$c\leq n \leq k(c-1)$$ the uniform distribution $$p_1=\ldots=p_k=\tfrac{1}{k}$$ uniquely maximises $$\mathbb{P}(\max X_i(n) \leq c-1 )$$.

Proof: Since we are maximising a continuous function (multivariate polynomial) on a compact set (simplex) the maximum is attained. Let $$p$$ be a maximising distribution. No $$p_i$$ can be $$0$$ (else replacing the last $$0$$ strictly increases the probability). But then $$p$$ can only be maximising if $$p$$ is the uniform distribution. End proof.

Remark Lemma 2 shows that the probability $$\mathbb{P}(\max X_i(n) \leq c-1 )$$ is a Schur-concave function on the simplex $$p_i\geq 0, \sum_{i=1}^k p_i=1$$. Possibly the result can be found under this heading in the literature (but a brief Google-search didn't reveal anything).

Experts may have a better answer.

This is I believe very hard if not impossible to do exactly, but if you use poissonization, i.e., model $$X_i$$ as independent poisson arrivals into bin $$i$$ with rate $$np_i$$ then the new process $$(X_1,\ldots, X_k)$$ equals the original bin arrival process $$(Y_1,\ldots,Y_k)$$ in expectation.

There is also a property that the probability of a monotone event can only be at most a factor of $$2$$ different in the bin arrival model and this model (I saw this in Mitzenmacher and Upfal's Probability and Computing textbook). Your event

$$A=\{ \omega: \textrm{Max } X_i \leq C\}$$ is monotone. Maybe you can show strong concentration with upper and lower bounds on $$A$$ in the poisson model.
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