# Question: 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)

(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)

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