I've seen two different definitions for the memoryless property of discrete random variable. Why they are equal (if they are) ?
$P\left(X=n+k\mid X>n\right)=P\left(X=k\right)$
$P\left(X>n+k\mid X>n\right)=P\left(X>k\right)$
Equal memoryless defenition ?
0
$\begingroup$
probability
random-variables
1 Answers
1
Assume that $P(X=n+k|X>n)=P(X=k)$ $\forall n,k$. Then
\begin{align*} P(X>n+k|X>n)&=\sum_{x>n+k}P(X=x|X>n)\\ &=\sum_{k'>k}P(X=n+k'|X>n)\\ &=\sum_{k'>k}P(X=k')\\ &=P(X>k) \end{align*} where I used the assumption to go from the second to the third line. This shows that the second equation follows from the first.
Now vice-versa assume that $P(X>n+k|X>n)=P(X>k)$ $\forall n,k$. Then
\begin{align*} P(X=n+k|X>n)&=P(X>n+k-1|X>n)-P(X>n+k|X>n)\\ &=P(X>k-1)-P(X>k)\\ &=P(X=k) \end{align*} where the assumption was used to go from the first to the second line.
Thus the first condition follows from the second and the second from the first and they are therefore equivalent.