Thought about it. If x1 ~p(a) and x2~p(b) then x-bar not ~ p(…) (hint: x-bar not from N| - and I am sure that you can prove with a chi’square test or a kolmogoroff that x-bar is not poisson distributed)
Joey, did I convince you?
I seriously doubt you will convince a man with a PhD in statistics that you are correct about multiples of poisson rv’s. He is correct, and I’ll join in the argument if necessary.
wyantjs: If it’s that easy- 1. prove that the following is wrong: If x1 ~p(a) and x2~p(b) then x-bar not ~ p(…) 2. Remember, i am not talking about simple reproductivity of sum of rvs’, but about the mean of rv’s. (for a normal distribution it clearly works) will be happy to learn something from you or joey…
@wyantjs: Are you really impressed by Ph.D.? How do you know that i have none?
wyantjs Wrote: ------------------------------------------------------- > I seriously doubt you will convince a man with a > PhD in statistics that you are correct about > multiples of poisson rv’s. He is correct, and > I’ll join in the argument if necessary. Wait a minute, he might be right. I haven’t started much of an argument here. I didn’t think much about that before I wrote it to begin with. Let’s see… a possion plus a poisson is certainly poisson which means a mean of poissons is a constant times a poisson. But that’s as far as it goes. cfaisok is right.
Hi Joey, thanks for a statement. It’s nice to see that you have such a reputation here. and it’s interesting what the other guys (not beeing too deep in quant) write in that forum…
cfaisok Wrote: ------------------------------------------------------- > Hi Joey, thanks for a statement. It’s nice to see > that you have such a reputation here. > > and it’s interesting what the other guys (not > beeing too deep in quant) write in that forum… Correction - i had a misspelling in the last post - it must be: Hi Joey, thanks for a statement. It’s nice to see that you have such a reputation here. and it’s interesting what the other guy (not beeing too deep in quant) write in that forum…
That guy (not too deep in quant) has a masters in math finance, and 6 years experience as a quant. If they are independent poisson rv’s, the sum is also poisson. The same holds for exponential as well.
and?
That was not the question, if you carefully read our discussion.
I am reading directly from Larson’s Probability Theory text…"if X_1, X_2,…X_k are independent Poisson rv’s, where X_i has parameter lamda_i, then the sum of the X_i’s has the Poisson distribution with parameter sum of lamda_i’s. Poisson is completely discribed by its mean and variance, both of which are the same…lamda x time.
wyantjs Do yourself a favor and read the thread. Again and again. You don’t have a clue about what we are talking…
Sorry I didn’t want to be rude, but I am tired of these nowhere leading discussions from people who don’t read careful the thread and who blame me of beeing clueless.
cfaisok Wrote: ------------------------------------------------------- > @JoeyD > I love your explanations, > couldn’t say it in a better way, exept one thing. > > "If the X(i)'s are binomial their mean can’t be > binomial. Only normals, poissons, and mixtures of > those have that property. " > > The mean of poisson distributed rvs don’t have to > be poisson distributed. If the sum of them is Poisson, then the sum of them divided by some number is also Poisson. What exactly is it that I am missing here?
Poisson is element N| The mean of poissons is not element N| --> The mean can’t be poisson. If you want you can to some monte carlo and figure out with some Kolmogoroff or chi²-test
Whatever you say dude.
if you use a lot of poissons it could lead (in some case) to good normal approximation of the mean
To explain in an other way: Sum up 50 poissons and then calculate the prob that the mean is between 0.3 and 0.7. A poisson distribution would always respond with a prob of 0, although it is clearly bigger than 0. ----------------------------------------------- and do me a favor - don’t call me dude.
would you rather me call you ass?