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

Question

Refer to Exercise 3.145. Use the uniqueness of moment-generating functions to give the distribution of a random variable with moment-generating function m(t) = (.6et + .4)3.

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1 week 2021-10-08T08:29:12+00:00 2 Answers 0

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    0
    2021-10-08T08:30:58+00:00

    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

    0
    2021-10-08T08:31:03+00:00

    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

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