Suppose that X has a geometric distribution with parameter p ∈(0,1) (see formula (3.17) on page 129 for the pmf). For any natural number n show that P(X ≤n) = 1 −(1 −p)n+1. (b) Suppose now that Xn has a geometric distribution with parameter λ/n, where λ ∈(0,1). Show that the distribution function of Xn/n converges to the distribution function of an exponential random variable X with parameter λ, that is, for every x > 0, P(Xn n ≤x) −→P(X ≤x). You can pretend as if nx is always a natural number. (c) What does the geometric distribution model in a binomial process? From here, explain what the exponential distribution models in a Poisson process.
Suppose that X has a geometric distribution with parameter p ∈(0,1) (see formula (3.17) on
page 129 for the pmf). For any natural number n show that P(X ≤n) = 1 −(1 −p)n+1.
(b) Suppose now that Xn has a geometric distribution with parameter λ/n, where λ ∈(0,1). Show
that the distribution
random variable X with parameter λ, that is, for every x > 0,
P(Xn
n ≤x) −→P(X ≤x).
You can pretend as if nx is always a natural number.
(c) What does the geometric distribution model in a binomial process? From here, explain what
the exponential distribution models in a Poisson process.
Sol:-
(a) The probability that X is less than or equal to n is given by:
P(X ≤ n) = P(X = k) = (1-p)(k-1)p
Using the formula for the sum of a geometric series, we have:
P(X ≤ n) = 1 - (1-p)n
Multiplying both sides by (1-p), we obtain:
P(X ≤ n) = 1 - (1-p)(n+1)
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