krackers · May 6, 2018 05:32
diff --git a/Proof of Reservoir Sampling b/Proof of Reservoir Sampling
 Say we want to generate a set of s elements and that we have already seen n>s elements.

 Let's assume that our current s elements have already each been chosen with probability s/n.

 By the definition of the algorithm, we choose element n+1 with probability s/(n+1).

 Each element already part of our result set has a probability 1/s of being replaced.

 The probability that an element from the n-seen result set is replaced in the n+1-seen result set is therefore (1/s)*s/(n+1)=1/(n+1). Conversely, the probability that an element is not replaced is 1-1/(n+1)=n/(n+1).

 Thus, the n+1-seen result set contains an element either if it was part of the n-seen result set and was not replaced---this probability is (s/n)*n/(n+1)=s/(n+1)---or if the element was chosen---with probability s/(n+1).
	Say we want to generate a set of s elements and that we have already seen n>s elements.

	Let's assume that our current s elements have already each been chosen with probability s/n.

	By the definition of the algorithm, we choose element n+1 with probability s/(n+1).

	Each element already part of our result set has a probability 1/s of being replaced.

	The probability that an element from the n-seen result set is replaced in the n+1-seen result set is therefore (1/s)*s/(n+1)=1/(n+1). Conversely, the probability that an element is not replaced is 1-1/(n+1)=n/(n+1).

	Thus, the n+1-seen result set contains an element either if it was part of the n-seen result set and was not replaced---this probability is (s/n)*n/(n+1)=s/(n+1)---or if the element was chosen---with probability s/(n+1).