Timeline for The conditions in the definition of Poisson process (and a Lévy process generalization)
Current License: CC BY-SA 2.5
16 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Oct 26, 2010 at 11:50 | comment | added | Shai Covo | Yes, it works (just note the typo in "$X_3=0$ if ..."). | |
| Oct 26, 2010 at 1:32 | comment | added | Louigi Addario-Berry | I put in a simpler "example" than the one I initially found but it was too simple. The modified one should work. | |
| Oct 26, 2010 at 1:31 | history | edited | Louigi Addario-Berry | CC BY-SA 2.5 |
Fixed my Bernoulli example.
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| Oct 25, 2010 at 23:19 | comment | added | Shai Covo | In the binomial example given above, $X_2 + X_3$ does not have Binomial(2,1/2) distribution (for example, $P(X_2+X_3=0)=1/8$). | |
| Oct 25, 2010 at 20:23 | history | edited | Louigi Addario-Berry | CC BY-SA 2.5 |
Responded to comments which showed my answer was incomplete, added a bit of a new answer.
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| Oct 25, 2010 at 19:42 | comment | added | Louigi Addario-Berry | Yes I see now. Sorry for being confused! Actually this is already an interesting question in the discrete case. Does there exist a sequence $(X_n)_{n \in \mathbb{N}}$ of random variables with $X_{j+1} + \ldots +X_{j+n}$ having Binomial$(n,1/2)$ distribution for all $j$ and $n$, which is not simply a sequence of independent binomial random variables? | |
| Oct 25, 2010 at 19:26 | comment | added | Shai Covo | My first comment corresponds exactly to George's comment. | |
| Oct 25, 2010 at 19:10 | comment | added | George Lowther | Hi. I think this answer is still wrong. We don't know the probability of there being no jump times in sets which are not intervals. So the result from the other thread on point processes does not apply. In fact, I have a counterexample in mind, which I'll post later. | |
| Oct 25, 2010 at 18:37 | comment | added | Louigi Addario-Berry | I've elaborated on my answer. If I'm misunderstanding something please let me know. | |
| Oct 25, 2010 at 18:36 | history | edited | Louigi Addario-Berry | CC BY-SA 2.5 |
Elaborated my answer.
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| Oct 25, 2010 at 18:04 | comment | added | Shai Covo | Of course, a trivial mistake of mine. It would have been correct if $X$ was only assumed cadlag with $X(0)=0$ and $X(t) \sim Poi(t)$. However, the question remains unanswered. | |
| Oct 25, 2010 at 17:42 | comment | added | Louigi Addario-Berry | With probability one, for every pair $0 < p < q$, $p,q$ rationals, $X(q)-X(p)$ is a non-negative integer. Since $X$ is cadlag the same property must hold for every real pair $0 < s < t$, i.e. $X$ is increasing and integer-valued, so it is a point process. | |
| Oct 25, 2010 at 16:34 | comment | added | Shai Covo | My last comment is incorrect, but the situation is only more complicated, as the process $X$ is not even known to be a point process (for example, the function $t \mapsto X(t)$ might not be monotone). | |
| Oct 25, 2010 at 15:48 | comment | added | Shai Covo | In our situation, we only know $P(X(A)=0)$ for sets $A$ of the form $A=(s,t]$, which cannot characterize the law of $X$. Hence, the question is still unanswered. | |
| Oct 25, 2010 at 14:56 | history | answered | Louigi Addario-Berry | CC BY-SA 2.5 |