Whichever fires first, the minimum of exponentials

Question

Three independent events are each "waiting to happen," with exponential timers of rates $\lambda_1, \lambda_2, \lambda_3$ (say, three resting orders that could each get hit). Let $T = \min(T_1, T_2, T_3)$ be the time until the first one occurs.

Find the distribution of $T$ and the probability that event $i$ is the one that happens first.

Accepted Answer

Distribution of the minimum. The minimum exceeds t only if all timers exceed t. By independence, P(T t) = \prod{i} P(Ti t) = e^{-\lambda1 t}\,e^{-\lambda2 t}\,e^{-\lambda3 t} = e^{-(\lambda1+\lambda2+\lambda3)t}. That is the survival function of an exponential with rate equal to the sum of the rates: T = \mini Ti \sim 	ext{Exponential}\!\left(	extstyle\sumi \lambdaiight). So the first event arrives at the combined rate, faster clocks together ring sooner. The expected time to the first event is 1/\sumi \lambdai. Which event wins. The probability that event i is the first to occur is its s…

Whichever fires first, the minimum of exponentials

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