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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?

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    $\begingroup$ No, it does not. In fact, you can prove that the geometric distribution is the ONLY discrete distribution which is memoryless. That is, a discrete random variable $X$ is memoryless if, and only if, it has a geometric distribution. For continuous variables, the same property holds for the exponential distribution. If you want, I can provide the proof. $\endgroup$ Sep 10, 2016 at 19:03
  • $\begingroup$ By the way, there is an error in the second line of your question. That should be $P(y>a+b|y>a)=P(y>a)$. $\endgroup$ Sep 10, 2016 at 19:06
  • $\begingroup$ See this MSE post: math.stackexchange.com/questions/923009/… $\endgroup$
    – Math1000
    Sep 10, 2016 at 19:27
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    $\begingroup$ @StijnDeVuyst I presume you mean it should be changed to "$=P(y>b)$" - but yes, I'd suggest editing that. $\endgroup$
    – Silverfish
    Sep 10, 2016 at 23:21
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    $\begingroup$ @StijnDeVuyst: This is a snarky comment but: it does for $r=1$, in which case it reduces to the geometric distribution. $\endgroup$
    – Alex R.
    Sep 11, 2016 at 17:43

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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.

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  • $\begingroup$ Yes, of course (duh). I've derived the proof – also very straight forward. Thanks! $\endgroup$ Sep 14, 2016 at 0:32

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