Exponential

The continuous, memoryless distribution of waiting times between events in a Poisson process.

1. The idea

An Exponential $(λ)$ random variable models the waiting time until the next event in a process with constant rate $λ$ , with density $f (x) = λ e^{- λ x}$ for $x \geq 0$ and $P (X > x) = e^{- λ x}$ .

It is the continuous analogue of the Geometric distribution and shares its memoryless property: $P (X > s + t ∣ X > s) = P (X > t)$ . Having waited $s$ time units with no event tells you nothing about how much longer you'll wait.

When racing several independent exponentials, the minimum is itself exponential with the summed rate, and each competitor wins with probability proportional to its own rate — the exponential race trick.

2. Why it matters

Waiting-time and "first of several events" questions collapse instantly once you recognize the exponential race structure, avoiding messy integrals over joint densities.

3. Core mathematics

Parameter

λ > 0

rate; mean waiting time is $1/ λ$

Support

x \geq 0

Expectation

E [X] = \frac{1}{λ}

Variance

Var (X) = \frac{1}{λ ^{2}}

Key property

P (X > s + t ∣ X > s) = P (X > t)

memorylessness

Exponential race

X_{i} \sim Exp (λ_{i}) indep. \Rightarrow i min X_{i} \sim Exp (i \sum λ_{i}), P (X_{j} wins) = \frac{λ _{j}}{\sum _{i} λ _{i}}

4. How to recognize it

Think of this when…

"Time until the next arrival/failure/trade" with a constant rate
Continuous waiting-time questions phrased with "on average, every $x$ minutes"
Multiple independent processes competing to happen first — "which event happens first"
Conditioning phrases like "given no event has happened yet"

5. Worked example

Worked example

Two independent servers process requests with exponential service times of rate $λ_{1} = 2$ and $λ_{2} = 3$ per minute. Find the expected time until the first one finishes and the probability server 1 finishes first.

1
Let $T_{1} \sim Exp (2)$ , $T_{2} \sim Exp (3)$ , independent.
2
The time until the first finish is $min (T_{1}, T_{2}) \sim Exp (λ_{1} + λ_{2}) = Exp (5)$ .
3
Its expectation follows directly.
$E [min (T_{1}, T_{2})] = \frac{1}{5} = 0.2 minutes$
4
Server 1 wins with probability proportional to its rate.
$P (T_{1} < T_{2}) = \frac{λ _{1}}{λ _{1} + λ _{2}} = \frac{2}{5} = 0.4$
Answer: Expected time to first finish is $0.2$ minutes; server 1 finishes first with probability $0.4$ .

6. Common mistakes

Treating the sum of independent exponentials as exponential (it is Gamma/Erlang, not exponential)
Confusing the minimum's rate (sum of rates) with the maximum's, which has no simple exponential form
Forgetting units: mean waiting time is $1/ λ$ , not $λ$
Applying memorylessness to a distribution that isn't Exponential or Geometric

7. Quick check

If $X \sim Exp (4)$ (rate per hour), what is $E [X]$ ?

Three independent alarms ring after Exp $(1)$ , Exp $(2)$ , Exp $(3)$ times. What is the probability the Exp $(2)$ alarm rings first?

8. Practice

16 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Memorylessness Exponential Race

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