0
$\begingroup$

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.

  • 0
    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\: severity2011-10-23

1 Answers 1

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

  • 0
    ... you should get $e^{-\lambda p} \frac{(\lambda p)^k}{k!}$2011-10-23