Statistics Question

Question

Suppose that mutations in a virus occur in the time interval [0, t] according to the following assumptions:

i) Divide the interval [0, t] into n equal intervals, each of width t/n

ii) Let n be very large

iii) Let Y_i = 1 (if 1 mutation occurs in the interval i, i = 1, 2, . . . , n) or 0 (if 0 mutations occur in the interval i, i = 1, 2, . . . , n)

iv) Let P(Y_i = 1) ≈ λt/n, for large n and for all i = 1, 2, . . . , n

v) Let P(Y_i > 1) ≈ 0, for large n, for all i = 1, 2, . . . , n

vi) Let the random variables Y_i be independent for i = 1, 2, . . . , n for all large n

Now let X be the number of mutations that occur in the interval [0, t].

Prove that:

P(X = k) = [(λt)^ke^-λt] / [k!], for k = 0, 1, 2, . . .

[Hint: Begin by finding a “starting” probability model for X that uses the given assumptions. Then let n → ∞]

Answer 1

P(X = k)

← P(Y₁ + ... + Y_n = k) (as n → ∞)

= P(Y₁ = 1, ..., Y_k = 1, Y_k+1 = 0, ..., Y_n = 0)n!/k!(n – k)! (since all choices have same P)

= (λt/n)^k(1 – λt/n)^n–kn!/k!(n – k)! (by independence)

= (λt)^k(n – λt)^–k(1 – λt/n)ⁿn!/k!(n – k)!. (algebra)

Since

n!/(n – k)! = n(n – 1)...(n – k + 1),

we can factor

(n – λt)^–kn!/(n – k)! = (n/(n – λt))((n – 1)/(n – λt))...((n – k + 1)/(n – λt)) → 1.

Also (1 – λt/n)ⁿ → exp(λt), (not easy to prove)

so finally

P(X = k) = (λt)^kexp(λt)/k!.