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Perhaps my question is very simple, however I could not get a reliable solution for my problem.

Here is my problem: Consider something as 'event' if this causes me a monetary loss equal to or above $x$ amount (say in USD). The number of such events (in a particular time interval, like say 10 years) follows a Poisson distribution with rate $\alpha$. Now my question is: Under this framework, what will be the distribution of the number of 'another event' which causes me $\$\ x_1$ loss, where $x_1 > x$.

I believe that distribution will be Poisson with rate $\alpha\frac{x}{x_1}$? Is it correct? How can I prove this mathematically?

Any suggestion will be highly appreciated.

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    $\begingroup$ There does not appear to be enough information to answer this question, but it's not clear. Are you saying that there is a stochastic process $Y(t)$ of losses over time? And if we define a new process $x\wedge Y$ to mark the times $t$ when $Y(t)\ge x$, then $x\wedge Y$ is found to be Poisson? And you are then speculating that the process $x_1\wedge Y$ for $x_1\ge x$ will also be Poisson? $\endgroup$
    – whuber
    Nov 16, 2011 at 17:48
  • $\begingroup$ Thanks for your reply. However here I could not understand, how a Stochastic process concept comes here? Just consider a fixed 10 years interval. In this fixed interval, there could '0' number of such events or could be '10' number of such event and so on. I am not talking anything on "time evaluation/process" $\endgroup$
    – Bogaso
    Nov 16, 2011 at 17:54
  • $\begingroup$ I am struggling to find a probability model that accords with your description. It practically requires a stochastic process of some sort. But let's try this. You draw a random natural number $N$ from some distribution. Then you draw a sequence $Y_1,Y_2,\ldots,Y_N$ where the $Y_i$ have some common distribution. For any number $z$ you can form the random variable $X(z)$ equal to the number of $Y_i$ that exceed $z$. Taking a particular value $x$, you claim $X(x)$ has a Poisson distribution and you would like to know the distribution of $X(x_1)$ for any $x_1\ge x$. Is this correct? $\endgroup$
    – whuber
    Nov 16, 2011 at 18:01
  • $\begingroup$ YES! Convinced. This mathematical description is mimicking my problem :) Thanks for that. $\endgroup$
    – Bogaso
    Nov 16, 2011 at 18:12
  • $\begingroup$ Hi, is there any update? Atleast can you please show some hints on how I should proceed? Thanks, $\endgroup$
    – Bogaso
    Nov 20, 2011 at 16:43

1 Answer 1

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Unfortunately, in order to calculate the distribution of the number of events at different levels of loss, you need to know the distribution of the losses. To see this, imagine that the underlying event that can cause monetary loss is such that you either lose 0 or 10. If we define an "event" to be "I lose > 1", perhaps that has a Poisson distribution with mean $\lambda$, let us assume so. But the event "I lose > 9" has the same Poisson distribution, not one with mean $(1/9)\lambda$, and the event "I lose > 11" has 0 probability of occurring altogether. On the other hand, if the underlying event is such that you lose either 0 or 100, then even if the distribution of the event "I lose > 1" remains Poisson with the same mean $\lambda$ as the previous example, the distribution of "I lose > 11" is different than in the previous example - it is now also a Poisson with mean $\lambda$, instead of an event with probability zero.

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