Minimum of Independent Exponentials¶

Theorem¶

Let $X_1,\ldots,X_n$ be independent exponential random variables with rates $\lambda_1,\ldots,\lambda_n$ . Then $\min(X_1,\ldots,X_n)$ is exponentially distributed with rate $\lambda_1+\cdots+\lambda_n$ $λ_{n}$ .

Interpretation¶

The minimum is the time at which the first of several independent clocks rings. Each running clock is another chance for an early ring, so the rates add up.

Survival curves e to the minus t, e to the minus 2t, and their product e to the minus 3t, which is the survival curve of the minimum.

The blue and red curves are the survival probabilities $\mathrm{e}^{-t}$ and $\mathrm{e}^{-2t}$ of two independent clocks with rates $1$ and $2$ . The green curve is their product $\mathrm{e}^{-3t}$ , the survival probability of the minimum. It lies below both, because the minimum survives past $t$ only if both clocks do: its mean is $\tfrac{1}{3}$ , against $1$ and $\tfrac{1}{2}$ .

Multiply the survival curves: the minimum survives only if both clocks do, which gives rate $\lambda_1+\lambda_2$ $λ_{2}$ . The simulation agrees with the exponential density.

For $t\ge0$ the proof is one line:

\mathbb{P}(\min(X_1,\ldots,X_n)>t) =\mathbb{P}(X_1>t)\cdots\mathbb{P}(X_n>t) =\mathrm{e}^{-(\lambda_1+\cdots+\lambda_n)t}.

The first step uses that the minimum exceeds $t$ if and only if every $X_i$ does, and independence to factor the probability. Without independence it can fail: if $X_2=X_1$ $X_{1}$ , the minimum is just $X_1$ with rate $\lambda_1$ . The second step uses the exponential form, since a product of exponential tails is again an exponential tail. For other distributions the type usually changes: for two independent $\operatorname{Unif}(0,1)$ variables, $\mathbb{P}(\min>t)=(1-t)^2$ on $[0,1]$ , which is not the tail of a uniform distribution.

Which clock rings first is also simple: clock $i$ wins with probability $\lambda_i/(\lambda_1+\cdots+\lambda_n)$ $λ_{n})$ , which is $\tfrac{1}{3}$ for the rate- $1$ clock in the figure. In two independent Poisson processes, the wait until the next arrival in either one is such a minimum, so the combined arrivals come at rate $\lambda_1+\lambda_2$ $λ_{2}$ .

Key Results¶

The minimum of independent exponentials is exponential with rate equal to the sum of the rates.
The probability that a particular clock is first is its rate divided by the total rate.
This result connects competing risks with Poisson process arrivals.

Examples¶

If two repair teams respond independently, the first response time can be modelled as the minimum of two exponential times.
In competing failure modes, the first failure time has rate equal to the sum of mode-specific rates.
If buses from two routes arrive independently, the next bus time can be the minimum of the two route waiting times.

Prerequisites¶

Sources¶

Probability Theory for EOR · Week 6

Aliases: minimum of exponentials Keywords: exponential race, competing exponential clocks Symbols: lambda

Last updated: 22 October 2026 19:04