@ wyantjs: Did you write that to Joey? Re: Risk?? Are we wrong?? new Posted by: wyantjs (IP Logged) [hide posts from this user] Date: January 20, 2008 12:53AM Why are you such an @#$%&? You constantly attack people on this forum. Your Phd may be “better”, but that does not mean that you shouldn’t respect mine. Get a life dude. Please just be a nice guy.
@ Joey tse tse tse
I’m glad I stayed away from this heated discussion …
I love when the discussion starts to get hot!!! Wyantjs, Below you can find the solution (of course cfaok is right). If you think the mathematical proof is wrong…please prove it…if you can 
NOTE: wyantjs poisson in French means fish…so you can try to prove empirically the below theorem summing up cod fishes. X, Y independent Poisson r.v., find distribution of X+Y. Let X be Poisson distributed with parameter lambda, and Y likewise with parameter mu. Then P(X+Y=n) = Sum_{k=0}^n P(X=k, Y = n-k) (these events are disjoint (mutually exclusive, hence the sum) = Sum_{k=0}^n P(X = k)*P(Y = n-k) [using independence] = Sum_{k=0}^n e^{-lambda}*(lambda^k / k!) * e^{-mu}* (mu^{n-k} / (n-k)!) [using the def. of Poisson] = e^{-(lambda+mu)} Sum_{k=0}^n (lambda^k * mu^(n-k)) / k!(n-k)! = e^{-(lambda+mu)}/n! Sum_{k=0}^n (n choose k)* lamda^k mu^(n-k) = e^(-(lambda+mu))/n! (lamda+mu)^n [using Newton’s binomium theorem] So X+Y has a Poisson distribution with parameter lambda+mu.
@maratikus We have high respect for your thoughts, so if you want to share you are really welcome. @strangedays Thanks for your support… But i do not understand why you show: x ~ P Y ~ P both independent (of course) --> x + y ~ P That’s what wyantjs said as well (in a more a general way), but that was not the topic of our discussion.
Cfaok, I just wanted to write down both matematical concepts just to be clear.
Then i don’t understand why you wrote the prove for the same statement as wyantjs (and my statement is just accepted, without saying much about it)? May be you or someone else can prove or reject the following: The mean of independent continous time poisson processes is again a poisson process. That would be interesting…
The distributionof the mean of any distribution (Poisson or otherwise) is approximately normal. Are you arguing with that, cfaisok? I’m lost in this mudslinging parade.
Is this “easy quant question” debate ever going to end? Chill out guys Milos
“The distributionof the mean of any distribution (Poisson or otherwise) is approximately normal.” This is no valid general statement (n<30) and that’s not what i am talking about. Nevermind, dreary, don’t take this thread to important for your exam
You are right Milos, there is a diminishing marginal use (d²u/dx²<0, but: dU/dX>0) in this discussion.
If the sample size < 30, anything can happen, and I’m not sure you can prove anything. Is this what you are trying to prove?
Nope…
Cfaisok, have a look at the below link…it will help to clarify. http://nrich.maths.org/askedNRICH/edited/1219.html The reported result is the follows: “Although the mean of X - Y is just E(X)-E(Y), X - Y itself is NOT a Poisson distribution in general and cannot be modelled as such.”
@dreary: The “discussion” has two topics at the moment. a) The mean of discrete poisson rv’s doesn’t cant be a poisson again. - That’s what i am pretty sure of and nobody could show the opposite) b) The mean of independent continous time poisson processes is again a poisson process. @ all Since i think it’s not too relevant for cfa exam, i will prove/reject (b) on my own. Thread can be closed, if nobody has further questions or comments.
I agree…should we start a new quanto question about derivatives?
Re: Eays Quant Question Posted by: strangedays (IP Logged) [hide posts from this user] Date: May 15, 2008 01:24PM Cfaisok, have a look at the below link…it will help to clarify. http://nrich.maths.org/askedNRICH/edited/1219.html The reported result is the follows: “Although the mean of X - Y is just E(X)-E(Y), X - Y itself is NOT a Poisson distribution in general and cannot be modelled as such.” ------------------------------------------------------- Dear strangedays, everybody here in the discussion (including wyantjs and Joey) knows about poisson reproductivity. And of course X-Y is not poisson (you can see that easily because X-Y could get <0, you don’t need a web link to prove that ) But this is not the topic. We are talking about means and not about differences. To be honest: I am really tired of these comments that are not linked to the discussion. To reply to your post, strangedays, in the same way: My car is silver and the sun is shining. May be this helps clarify
Thread closed.
> b) The mean of independent continous time poisson processes is again a poisson process. Good luck disproving the CLT
Are you starting like the others? ( wyantjs or strangedays)