Relationship between binomial and hypergeometric distributions

If $𝑋 \sim B i n (𝑛, 𝑝)$ and $𝑌 \sim B i n (𝑚, 𝑝)$ are independently distributed then #m/thm/prob

𝑋 ∣ 𝑋 + 𝑌 = 𝑟 \sim H G e o m (𝑛, 𝑚, 𝑟)

Proof

Since

ℙ (𝑋 = 𝑘 ∣ 𝑋 + 𝑌 = 𝑟) = ℙ (𝑋 = 𝑘 \land 𝑋 + 𝑌 = 𝑟) ℙ (𝑋 + 𝑌 = 𝑟) = ℙ (𝑋 = 𝑘 \land 𝑌 = 𝑟 - 𝑘) ℙ (𝑋 + 𝑌 = 𝑟) = (𝑛 𝑘) 𝑝 𝑘 (1 - 𝑝) 𝑛 - 𝑘 (𝑚 𝑟 - 𝑘) 𝑝 𝑟 - 𝑘 (1 - 𝑝) 𝑚 - (𝑟 - 𝑘) (𝑛 + 𝑚 𝑟) 𝑝 𝑟 (1 - 𝑝) 𝑛 + 𝑚 - 𝑟 = (𝑛 𝑘) (𝑚 𝑟 - 𝑘) (𝑛 + 𝑚 𝑟)

as required.

If $𝑋 \sim H G e o m (𝑠, 𝑓, 𝑛)$ and $𝑁 = 𝑠 + 𝑓 \to \infty$ such that $𝑝 = 𝑠 𝑁$ remains fixed, then $𝑋 ⇝ B i n (𝑛, 𝑝)$ . #m/thm/prob

Proof

Since

ℙ (𝑋 = 𝑥) = (𝑠 𝑘) (𝑓 𝑛 - 𝑘) (𝑠 + 𝑓 𝑛) = (𝑛 𝑘) (𝑠 + 𝑓 - 𝑛 𝑠 - 𝑘) (𝑠 + 𝑓 𝑠) = (𝑛 𝑘) ((𝑠 + 𝑓 - 𝑛)! (𝑠 - 𝑘)! (𝑓 - 𝑛 + 𝑘)!) ((𝑠 + 𝑓)! 𝑠! 𝑓!) \to (𝑛 𝑘) 𝑝 𝑘 (1 - 𝑝) 𝑘

as required.

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