You draw independent Uniform(0,1) values and stop at the first one below 0.2. What is the expected sum of all the values you drew, including the last one?

You draw independent Uniform(0,1) values and stop at the first one below 0.2. What is the expected sum of all the values you drew, including the last one?

Approach: The count is a stopping time that depends only on earlier draws, so Wald's identity applies. Then verify by splitting the sum into the non-terminal draws and the terminal one, each with its own conditional mean.

2.5. The number of draws N follows a geometric distribution with success probability 0.2, so E[N] = 5. The event that N is at least n depends only on the first n-1 draws, which is exactly the condition Wald's identity needs for a stopping time, so E[sum] = E[N] E[X] = 5 * 0.5 = 2.5. A direct check agrees: there are E[N] - 1 = 4 non-terminal draws with conditional mean E[X | X >= 0.2] = 0.6, plus one terminal draw with conditional mean E[X | X < 0.2] = 0.1, giving 4 * 0.6 + 0.1 = 2.5. Wald fails once the rule peeks ahead. Stopping one draw before the first value below 0.2 leaves every counted draw conditioned to be large, and the product formula no longer holds.

Follow-up: What is the variance of that sum?

Key concepts: wald's identity, stopping time, geometric distribution, conditional mean.