## Refer to Exercise 3.145. Use the uniqueness of moment-generating functions to give the distribution of a random variable with moment-generat

Question

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## Answers ( )

Answer: it follows a binomial random variable with parameters p= 0.6 and n=3

Step-by-step explanation:

Let’s look at the moment -generating function of a binomial distribution which is given as,

m(t) = [pet+q]n. If we look closely, these moment – generating functions are exactly the same or we can say they are identical to each other, where

p= 0.6, q= 0.4 and n= 3.Thus, the random variable with moment-generating function m(t) = (0.6et+ 0.4)3 follows binomial random variable with parameters p= 0.6 and n= 3

Answer:

The random variable with m(t) = (0.6e^t+ 0.4)^3

follows binomial random variable with parameters p= 0.6 and n= 3

Step-by-step explanation:

Given that p + q = 1

q = 1 – p

The general form of moment generating function, MGF, m(t) is given as

m(t) = [pe^t + q]^n for a binomial distribution

Comparing this to the moment-generating function to [0.6e^t + 0.4]^3

These m(t) functions are exactly the same with

p= 0.6,

q = 1 – 0.6

q= 0.4

n= 3.

Thus, the random variable with m(t) = (0.6e^t+ 0.4)^3

follows binomial random variable with parameters p= 0.6 and n= 3