# Proving that a positive-integer valued random variable has the lack of memory property iff it has a geometric distribution.

Suppose that $X$ is a positive-integer valued random variable with the lack of memory property which states:

Given that $X>n$, then $\mathbb{P}(X=n+k) = \mathbb{P}(X=k)$.

Consider the case where $p=\mathbb{P}(X=1)$ and let $q_n = \mathbb{P}(X>n)$ for each $n \in \Bbb N$ and $k=1$.

$$\mathbb{P}(X=n+1\mid X>n)=\dfrac {\mathbb{P} (X>n, X=n+1)}{\mathbb{P}(X>n)}$$

which by the conditional probability rule

$$= \frac {\mathbb{P} (X=n+1)}{\mathbb{P}(X>n)} = \frac {\mathbb{P} (X=n+1)}{q_n}$$

Here's where I get tripped up:

My textbook shows that

$$\frac {\mathbb{P} (X=n+1)}{q_n} = \frac {q_n-q_{n+1}}{q_n}.$$

How is that $q_n-q_{n+1}$ can be substituted for $\mathbb{P}(X=n+1)$?

I guess this is really a question of algebra, but I'm confused because my textbook skips over so many steps when it is showing how to derive certain equations.

$$\Pr(X>n) = \Pr(X=n+1 \text{ or }X>n+1) = \Pr(X=n+1) + \Pr(X>n+1).$$ So: $$\Pr(X>n) = \Pr(X=n+1) + \Pr(X>n+1).$$ Therefore $$\Pr(X>n) - \Pr(X>n+1) = \Pr(X=n+1).$$
• Just out of curiosity, did you arrive at this answer using the tail probabilities identity: $p_x$ = P(X>n-1) - P(x>n) ? – Nicholas Cousar Jun 18 '16 at 3:39
• @NicholasCousar : That identity is what this answer shows how to prove. $\qquad$ – Michael Hardy Jun 18 '16 at 17:15