Geometric Distribution's Memorylessness
Theorem
$X \sim \text{Geo} ( m )$ implies $P(X \ge s+ t ,|, X \ge s) = P(X \ge t)$
Explanation
The geometric distribution concerns the number of trials needed for an event to occur and is a discrete probability distribution. Thinking of it as a discretization of the exponential distribution, it’s natural to consider the memorylessness of the geometric distribution.
Here, the memoryless property refers to the characteristic where past occurrences do not influence future outcomes. For example, if all health conditions are equal, it is impossible to determine whether a man in his 30s or a man in his 50s will die first. Even if one has lived 20 years longer, if the health conditions today are the same, the countdown to death restarts from today. More extremely, there is no order to who passes away first between a baby born today or tomorrow and a senior citizen passing today or tomorrow. The practical reason this seems incorrect is due to the false premise that ‘all health conditions are the same’.
Conversely, if it can be demonstrated that all members of a certain group satisfy the same assumptions, it would be possible to predict their lifespan. The idea of insurance revolves around promising a certain payout at the end of these lifespans, earning more money in a shorter period than the expected lifespan.
Proof
$$ \begin{align*} P( X \ge a) =& 1- \sum_{i=0}^{a-1} m ( 1- m)^{i} \\ =& 1 - m{{1- (1-m)^a } \over {1 - (1-m) }} \\ =& (1-m)^{a} \end{align*} $$ therefore $$ \begin{align*} P(X \ge s+ t ,|, X \ge s) =& {{P(X \ge s+ t)} \over {P(X \ge s)}} \\ =& { {(1-m)^{s+t}} \over {(1-m)^{s}} } \\ =& (1-m)^{t} \\ =& P(X \ge t) \end{align*} $$
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