Question about poisson distributed variables

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Discussion Overview

The discussion revolves around the relationship between a Poisson-distributed random variable and its transformation when multiplied by 2. Participants are exploring whether if X follows a Poisson distribution with parameter m, then 2X also follows a Poisson distribution with parameter 2m. The conversation includes attempts to prove or disprove this implication, along with clarifications on terminology and properties of the Poisson distribution.

Discussion Character

  • Debate/contested
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant proposes to prove that if X~Po(m), then 2X~Po(2m) using the addition formula for Poisson distributions.
  • Another participant questions the validity of the approach, noting that adding random variables is not the same as multiplying a random variable by 2.
  • Clarifications are requested regarding the notation used, specifically the meaning of "~", X, and Po(m).
  • Some participants point out that 2X can only take even non-negative integer values, which raises concerns about it being Poisson distributed.
  • It is noted that the Poisson distribution has non-zero probabilities for all non-negative integers, while 2X would have zero probabilities for odd integers.
  • A suggestion is made to use moment generating functions as a potential method for analysis.
  • One participant expresses confusion about the distinction between X + X and 2X, indicating a need for further clarification on their equivalence.
  • A later reply suggests that a goodness-of-fit test might be useful in this context.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether 2X can be Poisson distributed. There are multiple competing views regarding the implications of transforming a Poisson-distributed variable and the properties of the resulting distribution.

Contextual Notes

Participants highlight limitations in understanding the implications of transformations on distributions, particularly regarding the nature of the Poisson distribution and the specific characteristics of the random variables involved.

farbror
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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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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)?
 
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
 
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, let's 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
 
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.
 
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 ([tex]2\lambda[/tex]) when X is Poisson distributed ([tex]\lambda[/tex])?
ie [tex]X\in Po(\lambda)\Rightarrow 2X\in Po(2\lambda)[/tex]

Thank you [tex]\LaTeX[/tex]

/farbror - feels silly that he can't grasp this
 
Last edited:
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.
 
  • #10
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
 
  • #11
mathman said:
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.
 
  • #12
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.
 
  • #13
Maybe a goodness-of-fit test could help.
 
  • #14
I'm sorry, I misread the question; please ignore my previous post.
 

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