Hypergeometric distribution

The hypergeometric distribution $𝑋 \sim H G e o m (𝑠, 𝑓, 𝑛)$ describes the probability of a sample of size $𝑛$ containing $𝑥$ successes, drawn from a pool consisting of $𝑠$ successes and $𝑓$ failures, without replacement. prob It has the probability mass function

ℙ (𝑋 = 𝑥) = (𝑠 𝑥) (𝑓 𝑛 - 𝑥) (𝑠 + 𝑓 𝑛)

Proof

We draw $𝑛$ times from a pool of size $𝑠 + 𝑓$ , so the total number of outcomes is $(𝑠 + 𝑓 𝑛)$ . Using the naïve definition of probability, the number of outcomes with $𝑋 = 𝑥$ will be equal to the number of ways of choosing $𝑥$ of $𝑠$ successes and $𝑛 - 𝑥$ of $𝑓$ failures, giving $(𝑠 𝑥) (𝑓 𝑛 - 𝑥)$ .

Properties

Let $𝑋 \sim H G e o m (𝑠, 𝑓, 𝑛)$ . Let $𝑁 = 𝑠 + 𝑓$ , $𝑝 = 𝑠 / 𝑁$ , and $𝑞 = 𝑓 / 𝑁$ .

Expectation: $𝔼 [𝑋] = 𝑛 𝑠 𝑠 + 𝑓 = 𝑛 𝑝$
Variance: $V a r [𝑋] = 𝑁 - 𝑛 𝑁 - 1 𝑛 𝑝 𝑞$

Proof of 1–2

Let $𝐼 𝑗$ be the indicator random variable for the $𝑗$ th draw being a success, so that $𝑋 = \sum 𝑛 𝑗 = 1 𝐼 𝑗$ . It follows $𝔼 [𝐼 𝑗] = 𝑝$ and hence $𝔼 [𝑋] = \sum 𝑛 𝑗 = 1 𝔼 [𝐼 𝑗] = 𝑛 𝑝$ , proving ^P1. We also have $V a r [𝐼 𝑗] = 𝑝 𝑞$ . Notice that by symmetry, $C o v [𝐼 𝑖, 𝐼 𝑗] = C o v [𝐼 1, 𝐼 2]$ for $𝑖 \neq 𝑗$ . Now
$V a r [𝑋] = V a r [𝑛 \sum 𝑗 = 1 𝐼 𝑗] = 𝑛 \sum 𝑗 = 1 V a r [𝐼 𝑗] + 𝑛 \sum 𝑖 = 1 𝑛 \sum 𝑗 = 1 (1 - 𝛿 𝑖 𝑗) C o v [𝐼 𝑖, 𝐼 𝑗] = 𝑛 𝑝 𝑞 + 𝑛 (𝑛 - 1) C o v [𝐼 1, 𝐼 2]$
where
$C o v [𝐼 1, 𝐼 2] = 𝔼 [𝐼 1 𝐼 2] - 𝔼 [𝐼 1] 𝔼 [𝐼 2] = 𝑠 𝑁 𝑠 - 1 𝑁 - 1 - 𝑝 2 = 𝑝 (𝑠 - 1 𝑁 - 1 - 𝑝)$
so
$V a r [𝑋] = 𝑛 𝑝 𝑞 + 𝑛 (𝑛 - 1) 𝑝 (𝑠 - 1 𝑁 - 1 - 𝑝) = 𝑁 - 𝑛 𝑁 - 1 𝑛 𝑝 𝑞$
proving ^P2.

Furthermore

$H G e o m (𝑠, 𝑓, 𝑛) \sim H G e o m (𝑛, 𝑠 + 𝑓 - 𝑛, 𝑓)$

Table of Contents

Hypergeometric distribution

Properties

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