Proving the lack of memory property of the Geometric distribution Can someone help me prove this for Geometric distribution?:
For any positive integer $k$ and $x$:
Memorylessness is defined as:
$P(X \geq k+x|X>x)=P(X \geq k)$
 A: I don't have enough reputation to comment, but for those confused as to why $ P(X \geq t) = (1-p)^t $, here's a quick derivation:
$$P(X \geq t) = \sum_{x = t}^\infty p(1-p)^x = p(1-p)^t[1+(1-p)+ (1-p)^2+ \ldots]\\
=p(1-p)^t\sum_{x = 0}^\infty (1-p)^x = (1-p)^t\sum_{x = 0}^\infty p(1-p)^x = (1-p)^t$$
Where the last step follows from the fact that an infinite sum of a pmf equals 1. 
Thus we have that $$P(X \geq s+t |X \geq t) = \frac{P(X \geq s + t)} {P(X \geq t)} = \frac{(1-p)^{s+t}}{(1-p)^t} = (1-p)^s $$ as needed.
A: A geometric random variable X has the memoryless property if for all nonnegative
integers s and t , the following relation holds . 
$  P(X>s+t| X>t) = P(X>s)$ or $ \frac{ P(X>s+t \text{ and } X>t)}{P(X>t)} = P(X>s)$
or $  P(X>s+t) = P(X>t)P(X>s)$
The probability mass function for a geometric random variable X is $f(x)=p(1-p)^x $
The probability that X is greater than or equal to x is $ P(X ≥ x) = (1 − p)^x $ .
So the conditional probability of interest is
$P(X ≥ s + t| X ≥ t) = \frac{P(X ≥ s + t, X ≥ t)}{P(X ≥ t)}
                     = \frac{P(X ≥ s + t)}{P(X ≥ t)}
                     =\frac{(1 − p)^{(s+t)}}{(1 − p)^t}
                     = (1 − p)^s
                     = P(X ≥ s), $
which proves the memoryless property
