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I frequently run into this question while modeling processes. I am wondering if there is a general solution or approximation.

The question I run into is: For a given distribution with a finite number of points, how much of the distribution have I explored after viewing $n$ samples with replacement? In other words, I have a box of balls with $n$ colors. The frequency of each color is given by a function $F(n)$. After sampling the distribution t times with replacement how many of the $n$ colors have been exposed in expectation?

There is an answer here. Using the given solution one could divide the distribution into $n$ discrete points to get the probability of each point and perform the given summation to arrive at the expected number of different colors seen.

Q: Is there a way to go directly from the continuous function $F(n)$ to an approximate or exact solution for the discrete case without the need for the summation given in the above solution?

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  • What is the function, F? –  Apr 21 '16 at 10:15
  • Is $F(n)$ the probability of drawing the $n$th colour in a single sample? Or is $F(n)$ the number of unique colours observed after $n$ samples with replacement? Or something else? – Henry Apr 21 '16 at 10:21
  • @Henry, Benedict $F(n)$ is the probability of drawing the $n$th color in a single sample; typically based on a chi-square distribution, power lognormal distribution etc. such that the area under the curve is approximately 1. – Toaster Apr 21 '16 at 12:08
  • If $F(n)$ is the probability of drawing the $n$th colour in a single sample, with $\sum_n F(n)=1$ then the expected number of unique colours after $m$ draws is $ \sum_n \left(1-\left(1-F(n)\right)^m \right)$ as the answer to your linked question suggests. You will not be able to avoid the sum or some equivalent way of combining the $F(n)$ except in special cases, such as when they are all equal – Henry Apr 21 '16 at 12:15
  • @Henry Thank you! You have lifted the nagging feeling that there was a better way. – Toaster Apr 21 '16 at 15:59

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