Random numbers multiplied until the product drops below a threshold

Question

You draw independent numbers, each uniform on [0, 1], and keep a running product. You stop as soon as the product first drops below 1/e \approx 0.3679. On average, how many numbers do you draw?

Accepted Answer

Turn the product into a sum by taking logarithms. If the running product is U1 U2 \cdots Un, then -\ln(U1 \cdots Un) = \sum{i=1}^n (-\ln Ui). Each term -\ln Ui is an Exponential(1) random variable (a standard fact for a uniform on [0,1]). The product drops below 1/e exactly when this growing sum of Exponential(1)s first exceeds -\ln(1/e) = 1. But a running sum of Exponential(1) variables is just the arrival times of a rate-1 Poisson process. The number of arrivals within a time window of length 	au = 1 is Poisson(1), with mean 1. The number of draws N is one more than the number of completed…

Random numbers multiplied until the product drops below a threshold

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