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Let $$ X_0 , X_1 , ...$$ be independent random variables, all have the same uniform distribution over the interval [0, 1]. How to find the distribution for $$ Y = min\{n : \prod_{k=0}^n X_k < e^{-\lambda}\}$$

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We can rewrite $$\prod_{k=1}^n X_k < e^{-\lambda}$$ as $$\sum_{k=1}^n-\log(X_k) > \lambda. $$

Since when $X_k$ is $U(0,1)$, $-\log(X_k)$ is a standard exponential, the problem becomes computing the distribution of $$ Y =\min\{n\mid \sum_{k=1}^nZ_k > \lambda\}$$ where $Z_k$ are independent exponentials.

If we remember some facts about Poisson processes, we recall that $\sum_{k=1}^nZ_k$ is the wait time until the $n$-th jump of the Poisson process (with rate $1$ since the exponentials are standard).

So $Y $ is the first value of $n$ so that the wait time is longer than $\lambda.$ In other words, $n-1$ is the value of the Poisson process (with unit rate) at time $\lambda.$ So $n-1$ is distributed like Poisson($\lambda$).

  • 0
    When $F_{X_k}(x)=x$, why $-log(X_k)$ is a standard exponential? – 2017-01-20
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    Let $Y= -\log(X).$ Then $P(Y e^{-y}) =1-F_X(e^{-y})= 1-e^{-y}. $ This is the CDF of a standard exponential. – 2017-01-20