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Consider a random process which samples uniformly with replacement from [n]. The expected time to find a duplicate is a constant factor times $\sqrt{n}$. This is a version of the famous Birthday Problem. How can one find the expected time to find the first copy of the first duplicate? This will obviously occur some time before. Also, what is the distribution of this time?

If you fix the position of the later copy of the duplicate then the earlier copy seems to occur uniformly before it but I am not sure how helpful that is to answer the questions.

By time I simply mean that each sample takes one unit of time so the time is just the number of samples at that point. This is not a question about algorithms or computation.

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Your comment at the end is quite helpful, at least for the mean. –  André Nicolas Nov 17 '12 at 22:10
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Call $2\leqslant D_n\leqslant n+1$ the time of the first duplicate and $1\leqslant C_n\leqslant D_n-1$ the time of the first copy of the first duplicate.

As you noted, for each $2\leqslant k\leqslant n+1$, conditionally on the event $[D_n=k]$, $C_n$ is uniformly distributed on $\{1,2,\ldots,k-1\}$ hence $\mathbb E(C_n\mid D_n=k)=\frac12k$. Thus, $\mathbb E(C_n)=\frac12\mathbb E(D_n)$ and each asymptotics on $\mathbb E(D_n)$ when $n\to\infty$ translates into an asymptotics on $\mathbb E(C_n)$.

Likewise, for every $1\leqslant i\leqslant n$, decomposing the event $[C_n=i]$ into its intersections with the events $[D_n=k]$ for $i+1\leqslant k\leqslant n+1$, one gets $$ \mathbb P(C_n=i)=\sum_{k=i+1}^{n+1}\frac{\mathbb P(D_n=k)}{k-1}. $$ Recall finally that, for every $2\leqslant k\leqslant n+1$, $$ \mathbb P(D_n\geqslant k)=\prod_{\ell=1}^{k-2}\frac{n-\ell}n, $$ hence, for every $1\leqslant i\leqslant n$, $$ \mathbb P(C_n=i)=\frac1n\sum_{k=i}^{n}\prod_{\ell=1}^{k-1}\frac{n-\ell}n=\frac{n!}{n^{n+1}}\sum_{k=0}^{n-i}\frac{n^k}{k!}. $$ Edit: Let $N_n$ denote a Poisson random variable with parameter $n$, then $$ \mathbb P(C_n=i)=\frac{n!\mathrm e^n}{n^{n+1}}\mathbb P(N_n\leqslant n-i). $$ Asymptotics follow from this identity since the prefactor $n!\mathrm e^n/n^{n+1}$ is equivalent to $\sqrt{2\pi/n}$ when $n\to\infty$, and $(N_n-n)/\sqrt{n}$ converges in distribution to a standard normal random variable. Thus, if $i_n/\sqrt{n}\to x$, $$ \mathbb P(C_n=i_n)\sim\sqrt{\frac{2\pi}n}\Phi(-x)=\frac1{\sqrt{n}}\int_x^{+\infty}\mathrm e^{-s^2/2}\mathrm ds. $$ Summing these yields $$ \mathbb P(C_n\geqslant i_n)\sim\int_x^{+\infty}(s-x)\mathrm e^{-s^2/2}\mathrm ds=\int_0^{+\infty}s\mathrm e^{-(s+x)^2/2}\mathrm ds. $$ In other words, $C_n/\sqrt{n}$ converges in distribution to a random variable $C$ whose probability density function $f_C$ is defined, for every $x\geqslant0$, by $$ f_C(x)=\int_x^{+\infty}\mathrm e^{-s^2/2}\mathrm ds. $$

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That last equality is not correct. In fact $\mathbb{E}(D_n) = 1 + \sum_{k=1}^n \tfrac{1}{k}$ if I recall correctly. –  WimC Nov 18 '12 at 9:33
    
@WimC Where do you see a formula for $\mathbb E(D_n)$ in my post? –  Did Nov 18 '12 at 9:35
    
Nowhere, I just put two remarks in a single comment. :-) –  WimC Nov 18 '12 at 9:35
    
@WimC Care to elaborate about the first part of your comment, then? –  Did Nov 18 '12 at 9:40
    
Take $k=3$. I get $\mathbb{P}(D_n \geq 3) = 1-1/n$ right? –  WimC Nov 18 '12 at 9:43
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