# On a lower bound for the birthday problem

I'm now familiar with a lower bound for the birthday problem as exposed in the theorem A.16 of Katz and Lindell book (alternatively see this webpage).

If one denotes by $$C(q,N)$$ the probability of collision when taking elements independent and uniformly distributed from a set of size $$N$$ The bound is obtained by assuming $$q \le \sqrt{2N}$$:

$$C(q,N) \ge \frac{q(q-1)}{4N}$$

However, the bound that I has been hold in my class is (without the inequality assumption in $$q$$):

$$\forall N \in \mathbb{N}.C(q,N) \ge \frac{(q-1)^2}{2N}$$

How can I prove this bound correct? • Could you write the proof from your class? Dec 5, 2018 at 10:35
• @kelalaka unfortunately (to me) this was just stated and written to me by e-mail, i have no proof whatsoever, so what i'm expecting is to have a counterexample here. i will ask however more details myself Dec 5, 2018 at 10:37
• Note that the bound from class is weaker (that is the lower bound is higher) than the one from the book. Dec 5, 2018 at 10:44
• @SEJPM in that sense is a "better" lower bound right? that's why i ask Dec 5, 2018 at 10:49

The question's $$\displaystyle C(q,N) \ge \frac{(q-1)^2}{2N}$$ bound is wrong. That's incompatible with the well-known fact that for $$N=q^2$$ and large, $$C(q,N)\approx1-e^{-1/2}\approx39.3\%$$.

A correct bound is $$\displaystyle C(q,N) \le \frac{(q-1)\,q}{2N}$$, valid for all $$q$$ and $$N\ge1$$, and tight when $$q\ll\sqrt N$$.

Another correct bound is $$\displaystyle C(q,N) \ge \frac{(q-1)^2}{4N}$$ when $$1\le q\le\sqrt{2N}$$, which follows from the question's lemma.

See this answer for some derivations, but beware that the notation there is $$(n,k)$$ and $$p_n$$ where the question has $$(q,N)$$ and $$C(q,N)$$.

A larger lower bound is better. The Katz Lindell book bound gives the correct formula. It is $$(1-e^{-1})\frac{q(q-1)}{2N} \approx 0.316360 \frac{q(q-1)}{N},$$ which they weaken further to $$\frac{q(q-1)}{4N}$$ for simplicity.

The bound you ask about which is $$0.5 \frac{(q-1)^2}{N},$$ is actually not weaker but stronger than even the stronger bound of Katz Lindell with the $$(1-e^{-1})$$ factor, as $$N,q$$ grow, and I don't see how it can be correct, regardless of the value of $$q.$$

• for the part "you wrote it wrong" see the edit above, maybe we are using different versions Dec 5, 2018 at 11:12
• You’re right. Sorry. Dec 5, 2018 at 11:18