Memorylessness of the Geometric Distribution
Theorem
If $X \sim \text{Geo} ( m )$, then $P(X \ge s+ t ,|, X \ge s) = P(X \ge t)$
Explanation
The geometric distribution is a discrete probability distribution concerned with the number of times some event occurs. If we think of it in the sense that it is a discretization of the exponential distribution, this memorylessness of the geometric distribution can be said to be natural.
Here, the memoryless property is the property that what will happen in the future is not affected by the time that has already passed. For example, if a man in his 30s and a man in his 50s have all the same conditions regarding health, then we cannot know which of the two will die first. Even if one has lived 20 more years, if all of today’s health conditions are the same, then the timer of death also restarts from today. More extremely, we could say that for a newborn born today or tomorrow and an elderly person barely hanging on today or tomorrow, there is no order to their passing. The reason this does not hold in reality is that the assumption that 'all conditions regarding health are the same' is wrong.
Conversely, if we can show that all members belonging to some group satisfy the same assumption, we would be able to predict their lifespans. Promising a fixed sum of compensation when their lifespan ends, and collecting more money over a period shorter than the life expectancy, is precisely insurance.
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 $$ \begin{align*} P(X \ge s+ t ,|, X \ge s) =& {{P(X \ge s+ t)} \over {P(X \ge s)}} \\ =& { {(1-m)^{s+t}} \over {(1-m)^{s}} } \\ =& (1-m)^{t} \\ =& P(X \ge t) \end{align*} $$
■