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I'm trying to research unexplored properties of the secretary problem, specifically the probability of choosing the k'th best candidate instead of the best one. Could be thought of as trying to calculate the PDF of the result of the optimal policy of skipping $n/e$.

I've come up with the following infinite sum which represents the probability of choosing $k=2$, in a method very similar to the original problem's proof. There is also a slightly longer and more algebra-heavy proof for a general k.

Using a Riemann approximation of an integral doesn't seem to work in this case, due to the $n-i$ expression stuck inside the sum. By running this calculation, I see that it does in fact converge to $e^{-2} \sim 0.135$, which does in fact match the empirical results. But how do I go about proving it?

$$r=n/e$$ $$\lim_{n\rightarrow\infty} \frac{r-1}{n(n-1)} \cdot \sum_{i=r}^{n-1}{\frac{n-i}{i-1}} \approx e^{-2}$$

EDIT: Maybe the following simplification can help? $$ = \lim_{n\rightarrow\infty} \frac{1}{e} ( \frac{1}{n} \cdot \sum_{i=r}^{n-1}{\frac{n-i}{i-1}}) \approx e^{-2} \Leftrightarrow \lim_{n\rightarrow\infty} \frac{1}{n} \cdot \sum_{i=r}^{n-1}{\frac{n-i}{i-1}} \approx e^{-1}$$

Screenshot of how the sum is arrived at

P.S. The formula for a general k, if anyone's interested: $$\lim_{n\rightarrow\infty} \frac{r-1}{n} \cdot \sum_{i=r}^{n-k+1}{\frac{1}{i-1} \cdot \prod_{j=0}^{k-2}{\frac{n-i-j}{n-j-1}}}$$ If anyone can help me find what the general k converges to, I will be much obliged. I suspect it converges to $e^{-k}$.

Tolstoyevsky
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I got to prove the case $k=2$, here it is: $$\lim_{n\to\infty} \frac{r-1}{n(n-1)}\sum_{i=r}^{n-1}\frac{n-i}{i-1} = \lim_{n\to\infty} \frac{r-1}{n(n-1)}\left(\frac{n-r}{r-1} + \frac{n-(r+1)}{r}+\frac{n-(r+2)}{r+1}+\dots\right) = \lim_{n\to\infty} \frac{r-1}{n}\left(\left(\frac{1}{r-1} - \frac{1}{n-1}\right) + \left(\frac{1}{r} - \frac{1}{n-1}\right) + \left(\frac{1}{r+1} - \frac{1}{n-1}\right) + \dots\right) = \lim_{n\to\infty} \frac{r-1}{n}\left(\sum_{k=r-1}^{n-2} \frac{1}{k} - \frac{n-r}{n-1}\right) \approx \lim_{n\to\infty} \frac{r-1}{n}\left(1 - \frac{n-r}{n-1}\right) = \lim_{n\to\infty} \frac{(r-1)^2}{n(n-1)} = \frac{1}{e^2}.$$ In the approximation step I used the following result: $$\sum_{k=a}^{b} \frac{1}{k} \approx \log\left(\frac{2b+1}{2a-1}\right),$$ which I found here: Summing Finitely Many Terms of Harmonic Series: $\sum_{k=a}^{b} \frac{1}{k}$

M4g1ch
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  • Thanks, clever simplification of the sum there. But isn't it a problem that you use a formula for summing finitely many terms of the harmonic series, while $n\rightarrow \infty$? – Tolstoyevsky Jul 30 '18 at 10:21
  • Or is it not a problem due to $a$ being a function of $b$ rather than being a constant? – Tolstoyevsky Jul 30 '18 at 10:25
  • You are right, I was kind of stuck in that step and I found that link which gave me some light, but now that you say it... I'm going to think about it and see if I come up with something else. If you figure it out (or anyone else), please tell me – M4g1ch Jul 31 '18 at 16:34