Does the does a random variable Y distributed as a negative binomial $Y \sim \text{NB}(r, p)$ possess the memorylessness property? By which I mean: $P(Y = a + b \mid Y > a) = P(Y > b)$
It's fairly simple to prove that a random variable $X$ with a geometric distribution, $X \sim \text{NB}(r=1, p)$ has this property:
$$
\begin{align}
P(X ≥ s + t \mid X ≥ t) &= \frac{P(X ≥ s + t, X ≥ t)}{P(X ≥ t)} \\
&= \frac{P(X ≥ s + t)}{P(X ≥ t)} \\
&= \frac{(1 − p)^{s+t}}{(1-p)^t} \\
&= (1 − p)^s \\
P(X ≥ s + t \mid X ≥ t) &= P(X ≥ s)
\end{align}
$$
Does this hold for the Negative Binomial in general?
Best Answer
As you've pointed out it does for a Geometric distribution, but not for a Negative Binomial distribution with $r>1$. This is pretty easy to rationalise as follows:
The condition $Y>a$ is fundamentally different in the special case of the Geomtetric distribution than in a Negative Binomial distribution with $r>1$. In a Geomtric distribution this condition immediately tells us there has been no successes by trial $a$ since $r=1$. While in the latter, it is possible that we have had some successes so long as there has been less than $r$ of them. The condition doesn't itself tell us how many successes we've had by stage $a$ so it's impossible to reduce it to $P(Y>b)$ since we cannot conclude that there's been no successes so far.
Since this is the case, of course the memoryless property does not hold as the condition clearly effects the probability of reaching $r$ successes by some number of trials if we've already had a few.
On the other hand, for the Geometric distribution if $Y>a$ then since $r=1$ this means there has been no successes, and the memoryless property follows from independence and constant probability of success.