Question

Let N have a Geometric (p) distribution on 1,2,3… Suppose that given N=n, random variable X follows a Binomial(n,p) distribution. E(X) is:

a. none of the choices given
b. 1/p
c.​ p
d. 1

Answer 1

The expected value E(X) of a binomial distribution, given that N follows a geometric distribution and given N=n, X follows a Binomial(n, p) distribution, is 1.

Step-by-step explanation:

We are given that the random variable N follows a Geometric (p) distribution and, given N=n, the random variable X follows a Binomial(n, p) distribution. To find the expected value E(X), we can use the fact that the expected value of X in a binomial distribution is given by the formula μ = np.

However, since N is geometrically distributed, its expected value is E(N) = 1/p. By using the law of total expectation, we can express E(X) as E(N)*p, which simplifies to (1/p)*p, yielding an expected value of 1 for X.