An alternative proof of Theorem 5 2 may be based on
An alternative proof of Theorem 5.2 may be based on the fact that if X1, X2, . . ., and Xn are independent random variables having the same Bernoulli distribution with the parameter ., then Y = X1 + X2 + · · · + Xn is a random variable having the binomial distribution with the parameters n and ..
(a) Verify directly (that is, without making use of the fact that the Bernoulli distribution is a special case of the binomial distribution) that the mean and the variance of the Bernoulli distribution are µ = θ and σ2 = θ(1 – θ).
(b) Based on Theorem 4.14 and its corollary on pages 135 and 136, show that if X1, X2, . . ., and Xn are independent random variables having the same Bernoulli distribution with the parameter θ and Y = X1 + X2 + · · · + Xn, then
E(Y) = nθ and var(Y) = nθ(1 – θ)
Membership TRY NOW
  • Access to 800,000+ Textbook Solutions
  • Ask any question from 24/7 available
    Tutors
  • Live Video Consultation with Tutors
  • 50,000+ Answers by Tutors
OR
Relevant Tutors available to help