Memorylessness

In probability and statistics, memorylessness is a property of probability distributions. It describes situations where previous failures or elapsed time does not affect future trials or further wait time. Only the geometric and exponential distributions are memoryless.

Definition

A random variable <math>X</math> is memoryless if <math display="block">\Pr(X>t+s \mid X>s)=\Pr(X>t)</math>where <math>\Pr</math> is its probability mass function or probability density function when <math>X</math> is discrete or continuous respectively and <math>t</math> and <math>s</math> are nonnegative numbers.<ref name=":1">Template:Cite book</ref><ref>Template:Cite book</ref> In discrete cases, the definition describes the first success in an infinite sequence of independent and identically distributed Bernoulli trials, like the number of coin flips until landing heads.<ref>Template:Cite book</ref> In continuous situations, memorylessness models random phenomena, like the time between two earthquakes.<ref>Template:Cite book</ref> The memorylessness property asserts that the number of previously failed trials or the elapsed time is independent, or has no effect, on the future trials or lead time.

The equality characterizes the geometric and exponential distributions in discrete and continuous contexts respectively.<ref name=":1" /><ref name=":2">Template:Cite book</ref> In other words, the geometric random variable is the only discrete memoryless distribution and the exponential random variable is the only continuous memoryless distribution.

In discrete contexts, the definition is altered to <math display="inline">\Pr(X>t+s \mid X \geq s)=\Pr(X>t)</math> when the geometric distribution starts at <math>0</math> instead of <math>1</math> so the equality is still satisfied.<ref>Template:Cite book</ref><ref name=":0">{{#invoke:citation/CS1|citation |CitationClass=web }}</ref>

Characterization of exponential distribution

If a continuous probability distribution is memoryless, then it must be the exponential distribution.

From the memorylessness property,<math display="block">\Pr(X>t+s \mid X>s)=\Pr(X>t).</math>The definition of conditional probability reveals that<math display="block">\frac{\Pr(X > t + s)}{\Pr(X > s)} = \Pr(X > t).</math>Rearranging the equality with the survival function, <math>S(t) = \Pr(X > t)</math>, gives<math display="block">S(t + s) = S(t) S(s).</math>This implies that for any natural number <math>k</math><math display="block">S(kt) = S(t)^k.</math>Similarly, by dividing the input of the survival function and taking the <math>k</math>-th root,<math display="block">S\left(\frac{t}{k}\right) = S(t)^{\frac{1}{k}}.</math>In general, the equality is true for any rational number in place of <math>k</math>. Since the survival function is continuous and rational numbers are dense in the real numbers (in other words, there is always a rational number arbitrarily close to any real number), the equality also holds for the reals. As a result,<math display="block">S(t) = S(1)^t = e^{t \ln S(1)} = e^{-\lambda t}</math>where <math>\lambda = -\ln S(1) \geq 0</math>. This is the survival function of the exponential distribution.<ref name=":2" />

Characterization of geometric distribution

If a discrete probability distribution is memoryless, then it must be the geometric distribution.

From the memorylessness property,<math display="block">\Pr(X>t+s \mid X\geq s)=\Pr(X>t)</math>The definition of conditional probability reveals that<math display="block">\frac{\Pr(X > t + s)}{\Pr(X \geq s)} = \Pr(X > t)</math>From this it can be proven by induction that <math display="block">\Pr(X > kt) = \Pr(X > 1)^k</math>Then it follows that<math display="block">f_X(x)=Pr(X\leq x)=1-Pr(X>x)=1-Pr(X>1)^x</math> and if we let <math display="block">Pr(X>1)=1-p</math>for some <math>0\leq p \leq 1</math>. we can easily see that X is geometrically distributed with some parameter p. in other words <math display="block">X\sim Geo(p)</math>