If Y has a geometric distribution with probability of success p, show that the moment-generating function for Y is

  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

50x5=$250

step-by-step explanation: