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 $$ \Pr(X>n) - \Pr(X>n+1) = \Pr(X=n+1). $$

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

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