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I derived a function in the form: $$ f(x)={\left[ 1 - \frac{N-m \choose x}{ N \choose x } \right]}^{\frac{N}{x}} $$ $m$ can be treated as a constant and $m=\alpha N$, where $ 0 < \alpha < 1$. $N$ also ban be regarded as a constant. I want to analyse how $f(x)$ changes with respect to $x$, i.e. how to minimize $f(x)$ with respect to $x$. I guess there could be a closed-form expression for $f(x)$. But I cannot figure it out. Therefore my question is:

Is it possible to derive a closed-form expression for $f(x)$? If not, is there any approximation method to analyse how $f(x)$ changes in terms of $x$.

Thank you!

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2 Answers 2

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Observe that $$ g(x) = \frac{\binom{N - m}{x}}{\binom{N}{x}} = \frac{N-m}{N} \cdot \frac{N-m - 1}{N-1} \cdot \cdots \cdot \frac{N - m - x + 1}{N - x + 1} $$ is a decreasing function on $x$; that is, for $y > x$, we have $g(y) < g(x)$. Also note $0 \leq g(x) \leq 1$. We have then $$ f(x) = (1 - g(x))^{N / x} < \left(1 - g(y)\right)^{N/x} < (1 - g(y))^{N / y} = f(y) $$ for $y > x$.

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Note the Stirling's approximation $$ \ln n! \approx n\ln n - n .$$

Also note $$ \frac{\binom{N-m}{x} }{ \binom{N}{x} } = \frac{(N-m)!(N-x)!}{(N-m-x)!} .$$

This yields $$ f(x) \approx \left[ 1 - e^{g(x)} \right]^{\frac{N}{x}} $$ where $$ g(x) = (N-m)\ln(N-m) + (N-x)\ln(N-x) - (N-m-x)\ln(N-m-x) - N. $$

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  • $\begingroup$ You answer is problematic, where the second equation is greater than one, which should not be possible considering it is a probability. By the way, the first equation for Stirling's approximation should be $n\ln{n}-n$. $\endgroup$
    – yusixie
    Apr 5, 2017 at 21:31

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