Lecture 11: The Poisson distribution

피망이·2023년 12월 4일
0
post-thumbnail

Sympathetic magic

  • Don't confuse a r.v. with its distribution

    • Sum of r.v. vs Sum of PMF is different!
  • "Word is not the thing, the map is not the territory."

    • random variable : random house, distribution : blue print(청사진)

Poisson distribution X ~ Pois(λ)

  • P(X=k)=eλλk/k!P(X=k) = e^{-\lambda} \lambda^{k} / k!, k{0,1,2,...}k \in \{0, 1, 2, ... \}

    • λ\lambda is a "rate(속도)" parameter(모수), λ>0\lambda > 0
  • Valid : k=0eλλkk!=eλeλ=1\displaystyle \sum_{k=0}^{∞} \frac{e^{- \lambda} \lambda^{k}}{k!} = e^{- \lambda} e^{\lambda} = 1 (Sum of prob. = 1)

    • cf. ex=n=0xnn!e^{x} = \displaystyle \sum_{n=0}^{∞} \frac{x^n}{n!} (by Taylor series)
  • E(X)=eλk=0kλk/k!E(X) = e^{- \lambda} \displaystyle \sum_{k=0}^{∞} k \lambda^{k} / k!

    =eλk=1λk(k1)!=λeλk=1λk1(k1)!= e^{- \lambda} \displaystyle \sum_{k=1}^{∞} \frac{\lambda^{k}}{(k-1)!} = \lambda e^{- \lambda} \displaystyle \sum_{k=1}^{∞} \frac{\lambda^{k-1}}{(k-1)!}

    =λeλeλ=λ= \lambda e^{- \lambda} e^{\lambda} = \lambda

  • V(X)=λV(X) = \lambda, Variance is also λ\lambda!

  • Often used for application where counting # of "successes" where there are a large # trials each with small prob. of success.

    • 여러 번의 시행을 하지만, 성공 가능성이 거의 없는 확률을 찾고자 할 때 쓰임

    (1) # emails in an hour.
    (2) # chips in chocolate chip cookie.
    (3) # earthquakes in a year in same region.

Pois Paradigm

  • Events A1,A2,...,AnA_1, A_2, ... , A_n, P(Aj)=pjP(A_j) = p_j, n large, pjp_j is small

    • Events indep. or "weakly dependent" then # of AjA_j's that occur is approximate

    • Pois(λ)Pois(\lambda), λ=j=1npj\lambda = \displaystyle \sum_{j=1}^{n} p_j

  • XBin(n,p)X \sim Bin(n, p), let n → ∞, p → 0, λ=np\lambda = np is held constant.

    • Find what happens to P(X=k)=(nk)pk(1p)nkP(X=k) = \begin{pmatrix}n\\k\\ \end{pmatrix} p^{k} (1-p)^{n-k}, k fixed.

    • p=λ/np = \lambda / n,

      P(X=k)=n(n1)...(nk+1)k!(λn)k(1λn)nkP(X=k) = \frac{n(n-1)...(n-k+1)}{k!} (\frac{\lambda}{n})^k (1-\frac{\lambda}{n})^{n-k}

      =n(n1)...(nk+1)λkk!nk(1λn)n(1λn)k= \frac{n(n-1)...(n-k+1) \lambda^{k}}{k! n^{k}} (1-\frac{\lambda}{n})^{n} (1-\frac{\lambda}{n})^{-k}

      • n → ∞,

        n(n1)...(nk+1)λkk!1\frac{n(n-1)...(n-k+1) \lambda^{k}}{k!} → 1, (1λn)k1(1-\frac{\lambda}{n})^{-k} → 1

      • (1+xn)nex(1+\frac{x}{n})^{n} → e^{x} as n → ∞

        : 복리 계산 식이라 생각하면 쉬움

      P(X=k)P(X=k)λkk!eλ\frac{\lambda^k}{k!} e^{- \lambda}, Pois PMF at k.

    • 이항분포가 n → ∞, p → 0 조건을 가지면, 포아송 분포로 수렴함!

  • ex.

    길바닥에 빗방울이 떨어지는 횟수 또한 포아송 근사로 설명할 수 있다.
    각 사각형에 빗방울이 떨어지는 사건은 이항분포이지만, 그 사건은 서로 독립이다.
    빗방울은 많이 떨어지지만 한 사각형 안에 떨어질 확률은 작기 때문에, 포아송 분포로도 볼 수 있다.

Birthday problem

  • Ex. Have n people, find approximate prob. that there are 3 people with same birthday.

    • (n3)\begin{pmatrix}n\\3\\ \end{pmatrix} triplets of people, indicator r.v. for each, FijkF_{ijk}, i<j<ki < j < k
  • E(# triple matches) = (n3)13652\begin{pmatrix}n\\3\\ \end{pmatrix} \frac{1}{365^2}

    • 한 명의 생일은 정해져 있어야 하므로, 나머지 둘이 이와 같을 때의 확률로 계산!
  • XX = # of triple matches. Approx Pois(λ\lambda)

    • λ=E(X)=(n3)13652\lambda = E(X) = \begin{pmatrix}n\\3\\ \end{pmatrix} \frac{1}{365^2}

    • I123=I124I_{123} = I_{124} are not indep.

  • P(X1)=1P(X=0)1eλλ0/0!=1eλP(X \ge 1) = 1- P(X=0) \approx 1-e^{- \lambda} \lambda^0 / 0! = 1-e^{- \lambda}


profile
물리학 전공자의 프로그래밍 도전기

0개의 댓글