Memoryless property
Theorem taught in Random variables and distributions, part of Probability
Memoryless property
| Type | Theorem |
| Taught in | Random variables and distributions |
| Part | Probability |
| Worked examples | 2 |
| Questions using it | 1 |
A positive random variable X is memoryless when \mathbb{P}(X>s+t\mid X>s)=\mathbb{P}(X>t) for all s,t>0. The exponential distributions are the continuous memoryless laws, and the geometric distributions are the discrete ones.
Statement
For the exponential law with rate \lambda, the survival probability is e^{-\lambda x}, so
\mathbb{P}(X>s+t\mid X>s)=\frac{e^{-\lambda(s+t)}}{e^{-\lambda s}}=e^{-\lambda t}.
The geometric calculation is the same with (1-p)^{s+t} in place of e^{-\lambda(s+t)}.
Examples
Two components with the same mean life
Question page: Two components with the same mean life
Two components each have expected lifetime 3 years. The first lifetime is exponential. The second is normal with standard deviation 1/2 year.
Find the probability that the first still works after 4.5 years, given that it works after 4 years.
Find the probability that the second still works after 4.5 years, given that it works after 4 years.
- The first lifetime is exponential with rate 1/3. By the memoryless property, surviving 4 years makes no difference:
\mathbb{P}(T>4.5\mid T>4)=\mathbb{P}(T>0.5)=e^{-0.5/3}=0.846.
- The second lifetime is \operatorname{N}(3,0.25). Standardising, 4 and 4.5 years are 2 and 3 standard deviations above the mean, so
\mathbb{P}(T>4.5\mid T>4)=\frac{1-\Phi(3)}{1-\Phi(2)}=\frac{0.00135}{0.02275}=0.059.
The normal component is worn out: having already lasted far longer than usual, it is very unlikely to last another half year. The exponential component does not age at all.
A memoryless component
Question page: A memoryless component
A component has an exponential lifetime with expectation 2 years.
Find the probability that it still works after 2 years.
Find the probability that it still works after 10 years, given that it still works after 7 years.
The mean is 2, so the rate is \lambda=1/2.
\mathbb{P}(X>2)=e^{-1}\approx 0.368.
Memorylessness gives \mathbb{P}(X>10\mid X>7)=\mathbb{P}(X>3)=e^{-3/2}\approx 0.223.
Used in
- Taught in Random variables and distributions, where the derivation and examples are.
- Also used in Stochastic processes · Discrete-time Markov chains · Poisson processes
- Questions that use it (1) — Two components with the same mean life (answer)
- All 27 questions on Random variables and distributions
See also
Random variable · Exponential distribution · Geometric distribution · Cumulative distribution function · Discrete distribution · Continuous distribution · Expectation · Expectation of a function