Random numbers multiplied until the product drops below 1/e squared

Question

You draw independent numbers, each uniform on [0, 1], and keep a running product, stopping as soon as it first drops below 1/e^{2} \approx 0.1353. On average, how many numbers do you draw?

Accepted Answer

Take logs to convert the product to a sum. Since -\ln Ui is Exponential(1) for a uniform Ui on [0,1], the product drops below 1/e^2 exactly when the running sum of Exponential(1)s first exceeds -\ln(1/e^2) = 2. A running sum of Exponential(1)s is the arrival process of a rate-1 Poisson process, so the number of completed steps inside a window of length 	au = 2 is Poisson(2) with mean 2. The number of draws is one more than that: E[N] = 1 + 	au = 1 + 2 = 3 	ext{ draws.} In general, multiplying uniforms until the product falls below c averages 1 + \ln(1/c) draws; here \ln(e^2) = 2 gives 3. T…

Random numbers multiplied until the product drops below 1/e squared

Your answer

More Probability questions