PDA

View Full Version : Question about poisson distributed variables


farbror
Nov14-03, 09:22 AM
Hi,

I'm trying to prove if X~Po(m) => 2X~Po(2m)
But I'm not sure how to prove or disprove it.
I'm thinking about using the addition formula, but is this the right approach?

X_1~Po(m)
X_2~Po(n)
X_1+X_2~Po(m+n)
n=m => X_1=X_2 => 2X_1~Po(2m)

Any help is appreciate.

Thanks
/farbror

mathman
Nov14-03, 05:42 PM
I'm trying to prove if X~Po(m) => 2X~Po(2m)
I might be able to help you if you could explain your terminology. Specifically, what is the definition of "~", what is X, what is Po(m)?

Hurkyl
Nov14-03, 06:19 PM
Adding two random variables is definitely not the same as multiplying a random variable by 2.

Not convinced? Suppose your random variable is rolling a 6 sided die. The sum of two of these random variables could be anything from 2 through 12, with an uneven distribution. Multiplying the result of a roll by 2 can only be even numbers from 2 through 12 with an even distribution.

farbror
Nov15-03, 10:42 AM
Okay, Hurkyl, I can understand your reasoning there.

So my idea of the proof is no good. Any other ideas how I should be able to prove/disprove this implication?

Thanks!

/farbror

Hurkyl
Nov15-03, 12:05 PM
It would be nice to know what your notation means. [6)]


Anyways, I imagine you want to use the fact:

P(2X < &alpha;) = P(X < &alpha; / 2)

to prove that 2X has the right cumulative distribution.

farbror
Nov16-03, 07:45 AM
Okay, lets see if I can explain my notation...

X is my random variable with a poisson distribution

I'm trying to verify if the statement
X in Po([lamb]) implies that 2X in Po(2[lamb])

I hope this is clear enough

/farbror

mathman
Nov16-03, 06:45 PM
For starters, X can only assume non-negative integer values. 2X is then restricted to EVEN non-negative integer values and cannot have a Poisson distribution.

farbror
Nov20-03, 08:44 AM
Ok, I'm not really sure if I follow your reasoning there.
The beginning is ok, due to the fact that X is a discrete random variable.

2X will only give us even non-negative values; I'm still with you. But in the next step, I'm lost.
Why can't 2X be Poisson distributed (2\lambda) when X is Poisson distributed (\lambda)?
ie X\in Po(\lambda)\Rightarrow 2X\in Po(2\lambda)

Thank you \LaTeX

/farbror - feels silly that he can't grasp this

mathman
Nov20-03, 05:55 PM
Poisson distribution is very specific. One feature is that random variables have non-zero probabilities for ALL non-negative integer values. 2X will have probability 0 for all odd integers.