If the poisson arrival process and the work process are independent,
then have a look at Wald's law in (almost) any probability book. For
example, the mean amount of work is then simply the product of the means
of each RV, in your case:
E(amount of work in a fixed time interval)=v*E(U) where U is your
lognormal RV.
Jacek Gomoluch wrote:
>
> In a stochastic process the number of customers which are arriving at a
> server (during a time intervall) is desribed by a Poisson distribution:
>
> P(n)=exp(-v) * (v^n)/(n!)
>
> Each arriving customer has a task to be carried out of which the size (in
> units) is described by a lognormal distribution:
>
> f(u)= exp(-(ln u)^2 / (2*a^2)) / (u*a*SQRT(2*PI))
>
> Question: What is the total number of units (i.e. size of all tasks)
> requested during the time intervall ?
>
> I wonder how these distributions can be concatenated, and if there is a
> formula for this.
>
> Thanks for any help!
>
> Jacek Gomoluch
--
Peter Rabinovitch
=================================================================
Instructions for joining and leaving this list and remarks about
the problem of INAPPROPRIATE MESSAGES are available at
http://jse.stat.ncsu.edu/
=================================================================
