Undergrad Poisson distribution with conditional probability

[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.

 

[Stephen Tashi]

I originally thought that P(X > x1 ∩ X > x2) = P(X > x1)

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

 

[Woolyabyss]

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

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?

 

[Stephen Tashi]

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?

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:

The only memoryless discrete probability distributions are the geometric distributions, which feature the number of independent Bernoulli trials needed to get one "success," with a fixed probability p of "success" on each trial. In other words those are the distributions of waiting time in a Bernoulli process.

- surprising (to me), if true.