Sampling with replacement
Some time ago I considered the problem of sampling with replacement exactly n-elements from n-element set. As a result of such an operation output set may contains duplicates – let us here assume that we received exactly k unique elements (of course 1 ≤ k ≤ n). Then the question came:
Number of ways with k-unique outputs
What is the number of ways to obtain exactly k unique elements sampling with replacement n-elements from n-element set?
Above question relates to bootstrap estimation, as the answer gives distribution of unique elements (its number) in bootstrap samples.
Let B(n,k) be the function representing the number of such ways.
1st step – partial permutation (k-permutation of n aka variation without repetition)
Pretty standard – the number of ways to choose k different elements of n paying attention to the order – this is a variation without repetition V(n,k).
2nd step – Stirling numbers of the second kind
Let us think form the other perspective, forgetting for a moment about just sampled k unique elements and the need of additional n-k (though it’s true). Instead let’s imagine that we have n items, including k unique. The trick is now to understand that having k different elements in a set of n elements is generating division of original set into k non-empty disjunctive subsets. How many ways to divide n-element set into k-subsets? This is the Stirling number of the second kind marked S2 (n, k). Finally
mXparser – Math Parser – definition
import org.mariuszgromada.math.mxparser.*; ... Function V = new Function("V(n,k) = n! / (n-k)!"); Function B = new Function("B(n,k) = V(n,k) * Stirl2(n,k)", V); int n = 5; for (int k = 0; k <= n; k++) mXparser.consolePrintln("B(" + n + "," + k + ") = " + B.calculate(n,k) );
[mXparser-v.2.3.1] B(5,0) = 0.0 [mXparser-v.2.3.1] B(5,1) = 5.0 [mXparser-v.2.3.1] B(5,2) = 300.0 [mXparser-v.2.3.1] B(5,3) = 1500.0 [mXparser-v.2.3.1] B(5,4) = 1200.0 [mXparser-v.2.3.1] B(5,5) = 120.0