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...
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
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
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