# Question about poisson distributed variables

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

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mathman
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
Staff Emeritus
Gold Member
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.

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
Staff Emeritus
Gold Member
It would be nice to know what your notation means.

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.

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
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.

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

Last edited:
mathman
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.

My suggestion is that you should try the moment generating function technique,ofcos u will need the theorem on limiting mgfs,bt it is the easiet way,or do a transformation

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.
The sum of 2 poissons is obviously poisson just like the sum of 2 binomials is binomial,ur reasoning is flawed,if X ε Poi(λ) and Y=X+X,THEN Y ε Poi(2λ),it follows quite simply evn by general reasoning.

Are you trying to show that X + X ~ Po(2m) or that 2X ~ Po(2m)?

And do you understand why these are not the same? (The earlier dice nicely shows you that they aren't so understand that)

So if Z = X + Y, then P(Z=1) is the probability of the events {X=1} and {Y=0} or {X=0} and {Y=1}.

Where as if Z = 2X then P(Z=1) is the probability of the event that {X=1/2} which can't occur.

Maybe a goodness-of-fit test could help.