cakkby
April 26, 2022
740

# なぜ人はRNA-Seqのリードカウントを負の二項分布に従うと考えるのか

なぜ人はRNA-Seqのリードカウントを負の二項分布に従うと考えるのかについて解説してみました

April 26, 2022

## Transcript

1. 𝑖 𝑙𝑖
𝑒𝑖
𝑁 𝑖 𝑁𝑖
𝑁𝑖
𝑁
=
𝑙𝑖
𝑒𝑖
σ
𝑗
𝑙𝑗
𝑒𝑗
𝑒𝑖
= 1
𝑒𝑖
𝑞𝑖
= 𝑙𝑖
𝑒𝑖
𝑞𝑖

2. 𝑛 𝑔 = σ𝑗
𝑞𝑗
𝑖 𝑞𝑖
𝑁 𝑝 = 𝑞𝑖
/𝑔 𝑖
Pr 𝑁 = 𝑘 = 𝐵𝑖𝑛𝑜𝑚 𝑘|𝑛, 𝑝 =
𝑛
𝑘
𝑝𝑘 1 − 𝑝 𝑛−𝑘

3. 𝑛 𝑔
𝜆 = 𝑛𝑝 = 𝑛𝑞𝑖
𝑔
𝑛, 𝑔
lim
𝜆=𝑛𝑝: fix
𝑛,𝑔→∞
𝑛
𝑘
𝑝𝑘 1 − 𝑝 𝑛−𝑘 =
𝜆𝑘𝑒−𝜆
𝑘!
= 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝑘|𝜆
𝜆 = 𝑛𝑝
𝑖 𝜆
𝜆 = 1
𝜆 = 2
𝜆 = 3

4. 𝜆
𝜆
𝜆1
𝜆2

5. 𝜆 𝑖
𝑗 𝜆𝑖𝑗
𝜑 𝜃 𝜃 𝑃 𝑥|𝜃
𝜃
𝑃 𝑥 =
1
𝑀

𝑗=1
𝑀
𝑃 𝑥|𝜃𝑗
≃ න𝜑 𝜃 𝑃 𝑥|𝜃 𝑑𝜃

6. 𝐺𝑎𝑚𝑚𝑎 𝜆|𝜇, 𝜙 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝑥|𝜆

0

𝐺𝑎𝑚𝑚𝑎 𝜆|𝜙,
𝜇
𝜙
𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝑥|𝜆 𝑑𝜆 =
Γ 𝑥 + 𝜙
Γ 𝑥 + 1 Γ 𝜙
𝜙
𝜇 + 𝜙
𝜙
𝜇
𝜇 + 𝜙
𝑥
= 𝑁𝑒𝑔𝐵𝑖𝑛𝑜𝑚 𝑥|𝜇, 𝜙
𝜇, 𝜙
𝑉 𝑥 = 𝜇 +
𝜇2
𝜙
> 𝜇
𝜙 = ∞

7. 𝜙
𝜎2 = 𝜇
𝜎2 = 𝜇 +
𝜇2
𝜙

8. 𝑒𝑖
𝑞𝑖
𝑥𝑖
′ = 𝑥𝑖
𝑙𝑖
𝐸 𝑥′ =
𝐸 𝑥
𝑙𝑖
=
𝜇
𝑙𝑖
= ෤
𝜇, 𝑉 𝑥′ =
𝑉 𝑥
𝑙𝑖
2
=
𝜇
𝑙𝑖
2
+
𝜇2
𝑙𝑖
2𝜙
=

𝜇
𝑙𝑖
+

𝜇2
𝜙
≠ ෤
𝜇 +

𝜇2
𝜙

9. log2
𝑦 = 𝑋𝑓𝑢𝑙𝑙
𝛽
log2
𝑦 = 𝑋𝑟𝑒𝑑𝑢𝑐𝑒𝑑
𝛽
𝑃 𝑌|𝑋𝑓𝑢𝑙𝑙
𝑃 𝑌|𝑋𝑟𝑒𝑑𝑢𝑐𝑒𝑑

10. 𝑃 𝑘 =
𝑛 𝑛 − 1 ⋯ 𝑛 − 𝑘
𝑘!
𝑝𝑘 1 − 𝑝 𝑛−𝑘
𝜆 = 𝑛𝑝 𝑛 → ∞ 𝑝 → 0
𝑃 𝑘 ≈
𝑛𝑘
𝑘!
𝑝𝑘 1 − 𝑝 𝑛 =
𝜆𝑘
𝑘!
1 −
𝜆
𝑛
𝑛
𝑒 1 − 𝜆
𝑛
𝑛
→ 𝑒−𝜆
𝑃 𝑘 ≈
𝜆𝑘
𝑘!
𝑒−𝜆

11. 0

𝐺𝑎𝑚𝑚𝑎 𝜆|𝜙,
𝜇
𝜙
𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝑥|𝜆 𝑑𝜆 = න
0
∞ 𝜆𝜙−1𝑒−
𝜆𝜇
𝜙
𝜙
𝜇
𝜙
Γ 𝜙
𝜆𝑥
𝑥!
𝑒−𝜆𝑑𝜆
=
𝜇
𝜙
𝜙
Γ 𝑥 + 1 Γ 𝜙

0

𝜆𝜙+𝑥−1𝑒−𝜆
𝜇+𝜙
𝜙 𝑑𝜆
=
𝜙
𝜇 + 𝜙
−𝜙 𝜇
𝜇 + 𝜙
𝜙
Γ 𝑥 + 1 Γ 𝜙
𝜙
𝜇 + 𝜙
𝜙+𝑥
Γ 𝜙 + 𝑥
=
Γ 𝜙 + 𝑥
Γ 𝑥 + 1 Γ 𝜙
𝜙
𝜇 + 𝜙
𝑥
𝜇
𝜇 + 𝜙
𝜙
=
Γ 𝑥 + 𝜙
Γ 𝑥 + 1 Γ 𝜙
𝜙
𝜇 + 𝜙
𝜙
𝜇
𝜇 + 𝜙
𝑥