# I Poisson distribution with conditional probability

1. Oct 19, 2016

### Woolyabyss

Hi guys,
I have a question about computing conditional probabilities of a Poisson distribution.
Say we have a Poisson distribution P(X = x) = e^(−λ)(λx)/(x!) where X is some event.
My question is how would we compute P(X > x1 | X > x2), or more specifically P(X> x1 ∩ X > x2) with x1 > x2?
I originally thought that P(X > x1 ∩ X > x2) = P(X > x1) but recently read about the memorylessness property of exponential distributions and I'm not sure if it applies to Poisson distributions.

2. Oct 19, 2016

### Stephen Tashi

If x1 > x2 then $\{X: X > x1, X > x2\}$ is the same event as $\{X:X>x1\}$ , isn't it?

3. Oct 20, 2016

### Woolyabyss

Yes. Also I know that the events X = x of a Poisson distribution are independent of one another but surely P(X >= 70) and P(X >= 80) for example can't be, because given at least 70 events happen, the probability that at least 80 events happen would be 10 no?

4. Oct 20, 2016

### Stephen Tashi

But my remark wasn't about the independence of events. If $A \subset B$ then $Pr(A \cap B) = Pr(A)$.

As far as independence goes, in most cases if $A \subset B$ then $A$ and $B$ are not independent events. Exceptions would be cases like $Pr(A) = Pr(B) = 0$ or $Pr(A) = Pr(B) = 1$.

To find $Pr(X > 80 | X > 70)$, what does Bayes theorem tell you ?

In the current Wikipedia article on "Memorylessness" https://en.wikipedia.org/wiki/Memorylessness there is the interesting claim:

- surprising (to me), if true.