Question about poisson distributed variables

In summary, the conversation is discussing the proof of X~Po(m) => 2X~Po(2m) and whether using the addition formula is the correct approach. The conversation also includes a discussion on the definitions of "~", X, and Po(m), as well as the implications of multiplying a random variable by 2. Different approaches, such as using the moment generating function technique or a goodness-of-fit test, are suggested for proving or disproving the implication. The conversation also touches on the difference between X + X ~ Po(2m) and 2X ~ Po(2m) and the use of dice as an example to illustrate this difference.
  • #1
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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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  • #2
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)?
 
  • #3
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.
 
  • #4
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
 
  • #5
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.
 
  • #6
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
 
  • #7
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.
 
  • #8
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:
  • #9
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.
 

1. What is a Poisson distributed variable?

A Poisson distributed variable is a type of discrete probability distribution that is used to model the number of events that occur within a certain time or space interval. It is often used when the probability of the event is low, but the number of opportunities for the event to occur is large.

2. How is a Poisson distributed variable calculated?

The probability of a Poisson distributed variable can be calculated using the formula P(x; λ) = (e^-λ)(λ^x) / x!, where x is the number of events, and λ is the average rate of events occurring in the given interval.

3. What are the key characteristics of a Poisson distributed variable?

The key characteristics of a Poisson distributed variable include a discrete and non-negative range of values, a single parameter (λ) that determines the shape and location of the distribution, and the assumption that the events occur independently of each other.

4. What are some practical applications of the Poisson distribution?

The Poisson distribution is commonly used in fields such as healthcare, finance, and telecommunications to model events such as customer arrivals, equipment failures, and disease outbreaks. It can also be used in quality control to monitor the number of defects in a production process.

5. How does the Poisson distribution differ from other probability distributions?

The Poisson distribution differs from other distributions, such as the normal distribution, in that it is discrete rather than continuous. It also differs in its assumption of events occurring independently, and its parameter, λ, represents the average rate of events rather than the mean of the distribution.

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