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Memoryless

A variable x is memoryless with respect to t if, for all s with t!=0,

P(x>s+t|x>t)=P(x>s).
(1)

Equivalently,

(P(x>s+t,x>t))/(P(x>t))=P(x>s)
(2)
P(x>s+t)=P(x>s)P(x>t).
(3)

The exponential distribution satisfies

P(x>t)=e^(-lambdat)
(4)
P(x>s+t)=e^(-lambda(s+t)),
(5)

and therefore

P(x>s+t)=P(x>s)P(x>t)
(6)
=e^(-lambdas)e^(-lambdat)
(7)
=e^(-lambda(s+t)),
(8)

is the only memoryless random distribution.

If s and t are integers, then the geometric distribution is memoryless. However, since there are two types of geometric distribution (one starting at 0 and the other at 1), two types of definition for memoryless are needed in the integer case. If the definition is as above,

P(x>s+t|x>t)=P(x>s),
(9)

then the geometric distribution that starts at 1 is memoryless. If the definition becomes

P(x>s+t|x>=t)=P(x>s),
(10)

then the geometric distribution that starts at 0 is memoryless. Note that these two cases are equivalent in the continuous case.

A useful consequence of the memoryless property is

<x-t|x>t>=<x>,
(11)

where <x> indicates an expectation value.

See also

Exponential Distribution, Geometric Distribution

Portions of this entry contributed by Andrew M. Ross

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Cite this as:

Ross, Andrew M. and Weisstein, Eric W. "Memoryless." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/Memoryless.html

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