# Question

Prove Theorem 5.6 by first determining E(X) and E[X(X + 1)].

Theorem 5.6

The mean and the variance of the negative binomial distribution are

µ = k/θ and σ2 = k/θ(1/θ – 1)

Theorem 5.6

The mean and the variance of the negative binomial distribution are

µ = k/θ and σ2 = k/θ(1/θ – 1)

## Answer to relevant Questions

Show that the moment-generating function of the geometric distribution is given by A variation of the binomial distribution arises when the n trials are all independent, but the probability of a success on the ith trial is θi, and these probabilities are not all equal. If X is the number of successes ...Suppose that f(x, t) is the probability of getting x successes during a time interval of length t when (i) The probability of a success during a very small time interval from t to t + △t is α ∙ △t, (ii) The ...Use the result of Exercise 5.3 to show that for the Bernoulli distribution, (a) α3 = 1 – 2θ / √(1 - θ), where α3 is the measure of skewness defined in Exercise 4.26 on page 129; (b) α4 = 1 – 3θ(1 - θ) / √(1 - ...Verify that (a) b(x; n,θ) = b(n - x; n, 1 - θ). Also show that if B(x; n,θ) For x = 0,1,2,…,n, then (b) b(x; n,θ) = B(x; n,θ)- B(x- 1; n,θ); (c) b(x; n,θ) = B(n- x; n, 1-θ)- B(n- x- 1; n,1- θ); (d) B(x; n,θ) ...Post your question

0