Answer:
M₀ (t) = p / e^-t -q = p (e^-t -q) ^ -1
Step-by-step explanation:
Let the random variable Y have a geometric distribution g (y;p) = pq y-¹
The m.g.f of the geometric distribution is derived as below
By definition , M₀ (t) = E (e^ ty) = ∑ (e^ ty )( q ^ y-1)p ( for ∑ , y varies 1 to infinity)
= pe^t ∑(e^tq)^y-1
= pe^t/1- qe^t, where qe^t <1
In order to differentiate the m.g.f we write it as
M₀ (t) = p / e^-t -q = p (e^-t -q) ^ -1
M₀` (t) = pe^-t (e^-t -q) ^ -2 and
M₀^n(t) = 2pe^-2t (e^-t -q) ^ -3 - pe^-t (e^-t -q) ^ -2
Hence
E (y) = p (1-q)-² = 1/p
E (y²) =2 p (1-q)-³ - p (1-q)-²
= 2/p² - 1/p and
σ² = [E (y²) -E (y)]²
= 2/p² - 1/p - (1/p)²
= q/p²
