probability distribution of coverage of a set after $X$ independently, randomly selected members of the set

I have a set of numbers where I am randomly and independently selecting elements within a set . After a number of these random element selections I want to know the coverage of the elements in the set. Coverage being how many elements from the set have been selected at least once divided by the total number of elements in the set.

To restate this: what is the probability distribution of the different coverage values on a set after $X$ randomly, independently selected elements of the set?

If there are $n$ elements of the set then the probability that $M=m$ have been selected after a sample of $x$ (with replacement) is

$$\frac{S_2(x,m) \; n!}{n^x \; (n-m)!}$$

where $S_2(x,m)$ is a Stirling number of the second kind.

The expected value of $M$ is: $n \left(1- \left(1-\dfrac{1}{n}\right)^x \right)$.

The variance is: $n\left(1-\dfrac{1}{n}\right)^x + n^2 \left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)^x - n^2\left(1-\dfrac{1}{n}\right)^{2x}.$

• I think you need to swap $m$ and $x$ in your expression for the probability. Does that affect the expected value and variance calculations, too? – Mike Spivey Apr 13 '11 at 18:31
• @Mike Spivey: you are right about the probability. $m$ should not appear in the expressions for mean and variance – Henry Apr 13 '11 at 19:32
• @Ross: The argument Henry is using (I think) is as follows: The number of ways to choose which $m$ elements are to be covered is $\binom{n}{m}$. Then the number of ways to have the $x$ elements in the sample chosen only from those $m$ elements is the same as the number of ways to distribute $x$ elements into $m$ distinguishable nonempty subsets; i.e., $m! S(x,m)$, which is a Stirling number of the second kind. Then the probability is obtained by dividing by the number of ways to choose $x$ elements with replacement from $n$ elements, which is $n^x$. The factor of $m!$ cancels. – Mike Spivey Apr 13 '11 at 22:36
• @Ross: There are a handful of us from the Pacific NW on this site. :) – Mike Spivey Apr 13 '11 at 22:36
• @ShelbyMooreIII If $x=m=n$ then you get $\frac{S_2(n,n) \; n!}{n^n \; (n-n)!}=\frac{n!}{n^n}$ as the probability that you select the $n$ different values in the first $n$ attempts. A direct calculation would give $\frac{n}{n} \times \frac{n-1}{n} \times \cdots \times \frac{2}{n} \times \frac{1}{n}$. This is not $1$ for $n\gt 1$: e.g. for $n=2$ it is $0.5$ and for $n=6$ it is about $0.0154321$ – Henry Jul 19 '18 at 7:50

The expected proportion of elements covered, $E\left(\frac{m}{n}\right)$, has a simple limiting form as $n \rightarrow \infty$ with the sampling rate $r / n$ fixed. Note that $\lim_{n \rightarrow \infty} \left(1-\frac{1}{n}\right)^n = e^{-1}$, and rewrite:

$$\lim_{n \rightarrow \infty} E\left(\frac{m}{n}\right) = 1 - e^{-\frac{r}{n}}$$

so that for example sampling $r=n$ times is expected to cover about 63% of the set. This is a reasonable approximation even for $n > 100$.

• Hi, if you have the time, how do you modify this to get coverage when you repeat the process? for example sampling 3000 out of 60000 for 100 times in a row – Austin Aug 5 '20 at 23:01

Derivation of $\operatorname E [M]$ with a classic use of indicator variables and total expectation:

We sample from set $X = \{1, \dots, n \}$. Let $X_i$ be $1$ if member $i$ is in our sample, $0$ otherwise.

For a sample of size $x$, the probability that none of the values are $i$ is $\left(\frac{n-1}{n}\right)^x$. Thus the probability that the sample includes $i$, and therefore $X_i = 1$, is $1-\left(\frac{n-1}{n}\right)^x$.

We have $$\operatorname E[M] = \operatorname E[X_1 + \dots + X_n] = \operatorname E[X_1] + \dots + \operatorname E[X_n] = n \left(1-\left(\frac{n-1}{n}\right)^x\right)$$

Variance is calculated similarly, but we have to consider separately $\operatorname E[X_i^2]$ and $\operatorname E[X_i X_j]$ for $i \ne j$.