For a distribution with probability p0 of zero, the zero-truncated probability for k > 0 is Pn_truncated(k) = Pn(k) / (1 - p0). Which describes this relationship for k > 0?

• A. Pn_truncated(k) = Pn(k) / (1 - p0) for k > 0
• B. Pn_truncated(k) = Pn(k) * (1 - p0)
• C. Pn_truncated(k) = Pn(k) / p0
• D. Pn_truncated(k) = (1 - p0) * Pn(k)

Answer: (A)

Zero-truncated probabilities come from conditioning on nonzero outcomes. If the original distribution assigns p0 to zero and Pn(k) to k > 0, then the probability that X equals k given X > 0 is P(X = k | X > 0) = Pn(k) / P(X > 0). Since P(X > 0) = 1 − p0, the zero-truncated probability for k > 0 is Pn_truncated(k) = Pn(k) / (1 − p0). This renormalizes the mass on positive values so the probabilities sum to 1 over k > 0. For example, with p0 = 0.2 and Pn(1) = 0.3, Pn(2) = 0.5, the truncated values are 0.3/0.8 = 0.375 and 0.5/0.8 = 0.625, which sum to 1. Other forms would not correctly normalize to a proper distribution over positive outcomes.