Consider M events that are all independent and poisson distributed in occurence with individual frequencies $\{\lambda_{k}\}_{k=1}^{M}$. Once they occur, they occur with a certain severity, event $k$ has severity distribution $F_{k}(T)$. I would like to know the distribution of the second largest event.
Distribution of the second largest event
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0If an event is Poisson distributed, does that mean it can occur more than once? – 2011-10-23
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0Yes, all events may occur more than once – 2011-10-23
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0I presume that by the second largest event you mean the second most severe event. Since the events all occur arbitrarily often over time, with varying severities, it doesn't make sense to speak of "the" second most severe event without saying something about the time span over which the events are compared. By the distribution of the second largest event, do you mean its severity distribution? I'd expect that you'd have to say something about $F_k(T)$ to say anything useful about that. – 2011-10-23
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0Yes by second largest event i mean the second most severe outcome of any event. I actually know the distribution so it is given by D(T)=e^{\lambda(1-F(T))}(1+\lambda(1-F(T)) where F(T)=P(event\: severity
– 2011-10-23
1 Answers
I suspect what is meant is something like this. Each event that occurs has a type (1 to $M$) and a severity $S$, such that the probability of any given occurrence being of type $k$ is $\lambda_k/\lambda$, and the cumulative distribution function of severity for events of type $k$ is $F_k(t)$, these being independent of whatever other events occur. Thus the combined cumulative distribution function of severity for each occurrence is $F(t) = \sum_{k=1}^M \frac{\lambda_k}{\lambda} F_k(t)$. The number of events that occur is a Poisson random variable with parameter $\lambda$, so the number of events of severity $> t$ that occur is a Poisson random variable with parameter $\lambda (1 - F(t))$. The probability that at most one event of severity $> t$ occurs (i.e. that the severity of the second most severe occurrence, if any, is at most $t$), is $e^{-\lambda(1-F(t))} (1 + \lambda (1 - F(t)))$.
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0" so the number of events of severity >t that occur is a Poisson random variable with parameter λ(1−F(t))" Im sorry this probably trivial but I dont see how you reach this conclusion, what steps are taken? – 2011-10-23
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0You are following a course presenting Poisson processes and you are puzzled by this? What **is** in your lecture notes then? – 2011-10-23
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0"You are following a course presenting Poisson processes and you are puzzled by this" no I am not, perhaps you could be slightly more constructive and point me to what property that is being used? – 2011-10-23
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0Hence what is the context which you meet this question in? And what do you know? And what did you try? You indicate none of these although to do so is suggested in the FAQ and would help potential answerers to be *more constructive* (in your lingo), or, more accurately, to be able to choose the appropriate level of sophistication for their answers. At present, if I wanted to answer your question I could not, because you left people completely in the dark in this respect. – 2011-10-23
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0Robert got the whole context right, all I would like to see is the explanation for the missing step I quoted. – 2011-10-23
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1Wrong. In fact, @Robert (very politely) signaled exactly that this kind of information is missing in your post, in the first sentence of his answer. If you *ask* for something here, you might consider caring to *provide* what you are supposed to (and worse, what you flatly refuse to provide, even once the fact that you did not is mentioned). – 2011-10-23
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0This is usually discussed under the heading of "compound Poisson processes". Suppose the number of events that occur is a Poisson random variable $X$ with parameter $\lambda$ and each occurrence is "special" with probability $p$, independent of all other occurrences. Let $Y$ be the number of "special" events. The conditional distribution of $Y$ given $X=x$ is binomial with parameters $x$ and $p$. So $$P(Y = k) = \sum_{x=k}^\infty P(Y = k | X = x) P(X = x) = \sum_{x=k}^\infty {x \choose k} p^k (1-p)^{x-k} e^{-\lambda} \frac{\lambda^x}{x!}$$ After some manipulation... – 2011-10-23
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0... you should get $e^{-\lambda p} \frac{(\lambda p)^k}{k!}$ – 2011-10-23