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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

cakkby
April 26, 2022

 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

cakkby

April 26, 2022
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  1. View full-size slide

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

    View full-size slide

  3. 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

    View full-size slide

  4. 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

    View full-size slide

  5. 𝜆
    𝜆
    𝜆1
    𝜆2

    View full-size slide

  6. 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

    View full-size slide

  7. 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 = ∞

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  8. 𝜙
    𝜎2 = 𝜇
    𝜎2 = 𝜇 +
    𝜇2
    𝜙

    View full-size slide

  9. 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

    View full-size slide

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

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  11. 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

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  12. 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

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  13. View full-size slide