Minimum of independent exponentially distributed random variables

Let $𝑇 𝑗 \sim E x p (𝜆 𝑗)$ be independently distributed for $𝑗 \in ℕ 𝑛$ and $𝑇 = m i n {𝐿 𝑗} 𝑛 𝑗 = 1$ . Then $𝑇 \sim E x p (\sum 𝑛 𝑗 = 1 𝑇 𝑗)$ . #m/thm/prob

Proof

Consider the Survival function of $𝑇$ :

ℙ (𝑇 > 𝑡) = ℙ (m i n {𝑇 𝑗} 𝑛 𝑗 = 1 > 𝑡) = ℙ (𝑛 ⋂ 𝑗 = 1 {𝑇 𝑗 > 𝑡}) = 𝑛 \prod 𝑗 = 1 ℙ (𝑇 𝑗 > 𝑡) = 𝑛 \prod 𝑗 = 1 e x p (- 𝜆 𝑗 𝑡) = e x p (- 𝑡 𝑛 \sum 𝑗 = 1 𝜆 𝑗)

as required.

Considering a Poisson process, this result is intuitive.

#state/tidy | #lang/en | #SemBr