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Memoryless Property

Definition

A nonnegative waiting time is memoryless if, once it has lasted a certain time, the time still to wait has the same distribution as the original waiting time. For a nonnegative integer-valued GG this means that for all nonnegative integers m,nm,n

P(G≥m+n∣G≥m)=P(G≥n).\mathbb{P}(G\ge m+n\mid G\ge m)=\mathbb{P}(G\ge n).

For an absolutely continuous nonnegative TT it means that for all s,t≥0s,t\ge0

P(T>s+t∣T>s)=P(T>t).\mathbb{P}(T>s+t\mid T>s)=\mathbb{P}(T>t).

The conditioning events must have positive probability. The property also works in reverse. Suppose the waiting time is finite with probability 11 and all its tail probabilities are positive. Then a memoryless GG has a Geo⁡(p)\operatorname{Geo}(p) distribution for some 0<p<10<p<1 (counting the failures before the first success), and a memoryless TT has an Exp⁡(λ)\operatorname{Exp}(\lambda) distribution for some λ>0\lambda>0. For the first-success distribution, which counts trials and takes the values 1,2,…1,2,\dots, the same condition is written with strict tails: P(X>m+n∣X>m)=P(X>n)\mathbb{P}(X>m+n\mid X>m)=\mathbb{P}(X>n) for all nonnegative integers m,nm,n.

Interpretation

A memoryless waiting time does not age: having waited for a while says nothing about how much longer the wait will be. A working component with a memoryless lifetime is as good as new.

An exponential survival curve, and a second copy of it restarted at time s; at s + t the restarted curve has the same value as the original curve at t.

The blue curve is the survival probability P(T>u)=e−λu\mathbb{P}(T>u)=\mathrm{e}^{-\lambda u} of an exponential waiting time. The red dashed curve is the conditional probability of survival given T>sT>s: it starts at 11 at time ss and is an exact copy of the blue curve shifted to the right by ss. Its height at s+ts+t equals the height of the blue curve at tt:

P(T>s+t∣T>s)=e−λ(s+t)e−λs=e−λt=P(T>t).\mathbb{P}(T>s+t\mid T>s) =\frac{\mathrm{e}^{-\lambda(s+t)}}{\mathrm{e}^{-\lambda s}}=\mathrm{e}^{-\lambda t}=\mathbb{P}(T>t).

Since {T>s+t}\{T>s+t\} is contained in {T>s}\{T>s\}, the condition can be written without conditioning as P(T>s+t)=P(T>s) P(T>t)\mathbb{P}(T>s+t)=\mathbb{P}(T>s)\,\mathbb{P}(T>t): the survival probability turns sums of times into products, and that is what forces the exponential form. The discrete version reads the same way: {G≥m}\{G\ge m\} means that the first mm trials all failed, and given this the number of further failures is again Geo⁡(p)\operatorname{Geo}(p), since the trials are independent.

Most lifetimes with wear do not have it: for T∼Unif⁡(0,1)T\sim\operatorname{Unif}(0,1), P(T>0.75∣T>0.5)=0.5\mathbb{P}(T>0.75\mid T>0.5)=0.5 while P(T>0.25)=0.75\mathbb{P}(T>0.25)=0.75. The property also does not say that a long wait makes an end "due": after ten failures in independent trials, a success is no more likely than before. It is what makes the Poisson process start afresh at every moment: from any fixed time, the wait for the next arrival is again Exp⁡(λ)\operatorname{Exp}(\lambda).

Key Results

  • For G∼Geo⁡(p)G\sim\operatorname{Geo}(p) counting failures, P(G≥m+n∣G≥m)=(1−p)n\mathbb{P}(G\ge m+n\mid G\ge m)=(1-p)^n for admissible nonnegative integers m,nm,n.
  • For T∼Exp⁡(λ)T\sim\operatorname{Exp}(\lambda), P(T>s+t∣T>s)=e−λt\mathbb{P}(T>s+t\mid T>s)=\mathrm{e}^{-\lambda t} for s,t≥0s,t\ge0.

Examples

  • For G∼Geo⁡(1/4)G\sim\operatorname{Geo}(1/4), P(G≥5∣G≥2)=(3/4)3=27/64\mathbb{P}(G\ge5\mid G\ge2)=(3/4)^3=27/64.
  • For T∼Exp⁡(2)T\sim\operatorname{Exp}(2), P(T>4∣T>3)=e−2=P(T>1)\mathbb{P}(T>4\mid T>3)=\mathrm{e}^{-2}=\mathbb{P}(T>1).

Prerequisites

Sources

Aliases: memorylessness, lack of memory, lack-of-memory property Keywords: Memoryless Symbols: lambda

Last updated: 22 October 2026 19:04