# Explain Why a Geometric Mean Can Exceed Its Median Let `X` follow the geometric distribution counting the number of independent Bernoulli trials up to and including the first success, where each trial succeeds with probability `p` and `0 < p <= 1`. Derive the mean and a median of `X`, then explain intuitively how the mean can be greater than the median. Reconcile the two statistics rather than treating their difference as a contradiction. ### Constraints & Assumptions - Use the support `1, 2, 3, ...`; state how formulas shift under the alternative failures-before-success convention. - A median is the smallest integer `m` for which `P(X <= m) >= 1/2`. - Address the boundary case `p = 1`. ### Clarifying Questions to Ask - Which of the two common geometric-distribution parameterizations is intended? - Is the interviewer asking for an exact discrete median or a continuous approximation? - Should the explanation include a numerical example as well as the derivation? ```hint Use the survival probability The probability that success has not occurred after `m` trials is `(1-p)^m`. ``` ```hint Compare sensitivity to the tail The median depends on where cumulative probability crosses one half, while the mean weights every possible waiting time by its magnitude. ``` ### What a Strong Answer Covers - Correct PMF, CDF, mean, and discrete median under the stated support. - Care with the logarithm inequality and integer ceiling. - A right-tail explanation for why rare long waits pull the mean upward. - The equality boundary at `p = 1` and the parameterization shift. ### Follow-up Questions - Compute the mean and median when `p = 1/4`. - What is the memoryless property, and how does it relate to the tail? - How do the mean and median behave as `p` approaches zero?