What are the mean and variance of a geometric distribution with parameter p?

What are the mean and variance of a geometric distribution with parameter p?

• A. Mean p; Variance p(1-p)
• B. Mean np; Variance np(1-p)
• C. Mean 0; Variance 1
• D. Mean 1/p; Variance (1-p)/p^2

Answer: (D)

When a geometric distribution counts the number of trials until the first success, the mean and variance follow specific forms because each trial is an independent chance p of succeeding, and we’re waiting for the first success. The average number of trials you expect to perform is 1/p. This comes from the fact that the waiting time until success has a geometric structure: on each trial you have probability p of stopping, so the expected waiting time is the reciprocal of p. The variance, which measures the spread of the waiting time, is (1−p)/p^2. This comes from computing E[N^2] for the geometric PMF P(N=k) = (1−p)^{k−1}p and then using Var(N) = E[N^2] − (E[N])^2, yielding (1−p)/p^2. If you instead defined the geometric distribution as the number of failures before the first success, the mean would be (1−p)/p and the variance would still be (1−p)/p^2. The values 1/p and (1−p)/p^2 specifically correspond to counting the total number of trials up to and including the first success.

When a geometric distribution counts the number of trials until the first success, the mean and variance follow specific forms because each trial is an independent chance p of succeeding, and we’re waiting for the first success.

The average number of trials you expect to perform is 1/p. This comes from the fact that the waiting time until success has a geometric structure: on each trial you have probability p of stopping, so the expected waiting time is the reciprocal of p. The variance, which measures the spread of the waiting time, is (1−p)/p^2. This comes from computing E[N^2] for the geometric PMF P(N=k) = (1−p)^{k−1}p and then using Var(N) = E[N^2] − (E[N])^2, yielding (1−p)/p^2.

If you instead defined the geometric distribution as the number of failures before the first success, the mean would be (1−p)/p and the variance would still be (1−p)/p^2. The values 1/p and (1−p)/p^2 specifically correspond to counting the total number of trials up to and including the first success.