Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
If an event is Poisson distributed, does that mean it can occur more than once? –  Henry Oct 23 '11 at 8:13
    
Yes, all events may occur more than once –  fghftd Oct 23 '11 at 9:17
    
I 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. –  joriki Oct 23 '11 at 9:23
    
Yes 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<T|exactly\: one\: event\: occurs) butI do not have the derivation for this –  fghftd Oct 23 '11 at 9:41

1 Answer 1

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 $e^{-\lambda(1-F(t))} (1 + \lambda (1 - F(t)))$.

share|improve this answer
    
" 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? –  fghftd Oct 23 '11 at 9:51
    
You are following a course presenting Poisson processes and you are puzzled by this? What is in your lecture notes then? –  Did Oct 23 '11 at 10:15
    
"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? –  fghftd Oct 23 '11 at 10:23
    
Hence 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. –  Did Oct 23 '11 at 13:21
1  
Wrong. 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). –  Did Oct 23 '11 at 18:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.