Sum of two independent Poisson random variables

In summary: In particular, when p0 is one of the $s, you get a triangle like: # $ * * * *If p0 is one of the *'s, you get a trapezoid like # * * * * * * * * * * * * * * * * * * * *The values of q that you sum over will be a function of p0 that changes in a stepwise fashion. That is, for p0 = #, q ranges from 1 to 1, for p0 = $, q ranges from 1 to 2, for p0 = *, q ranges from 1 to
  • #1
andresc889
5
0
Hello!

I am trying to understand an example from my book that deals with two independent Poisson random variables X1 and X2 with parameters λ1 and λ2. The problem is to find the probability distribution of Y = X1 + X2. I am aware this can be done with the moment-generating function technique, but the author is using this problem to illustrate the transformation technique.

He starts by obtaining the joint probability distribution of the two variables:

f(x1, x2) = p1(x1)p2(x2)

for x1 = 0, 1, 2,... and x1 = 0, 1, 2,...

Then he proceeds onto saying: "Since y = x1 + x2 and hence x1 = y - x2, we can substitute y - x2 for x1, getting:

g(y, x2) = f(y - x2, x2)

for y = 0, 1, 2,... and x2 = 0, 1,..., y for the joint distribution of Y and X2."

Then he goes ahead and obtains the marginal distribution of Y by summing over all x2.

My question is this. How did he obtain the region of support (y = 0, 1, 2,... and x2 = 0, 1,..., y) for g(y, x2). I can't for the life of me understand this.

Thank you for your help!
 
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  • #2
andresc889 said:
How did he obtain the region of support (y = 0, 1, 2,... and x2 = 0, 1,..., y) for g(y, x2). I can't for the life of me understand this.

I don't understand which aspect of the support you are asking about. Is your question about the restriction of [itex] x_2 [/itex] to 0,1,..y ? If you had a non-zero probability for a point like [itex] y = 3, x_2 = 4 [/itex] that would imply that [itex] x_1 = -1 [/itex] since [itex] y [/itex] is defined as [itex] x_1+ x_2 [/itex].
 
  • #3
Stephen Tashi said:
I don't understand which aspect of the support you are asking about. Is your question about the restriction of [itex] x_2 [/itex] to 0,1,..y ? If you had a non-zero probability for a point like [itex] y = 3, x_2 = 4 [/itex] that would imply that [itex] x_1 = -1 [/itex] since [itex] y [/itex] is defined as [itex] x_1+ x_2 [/itex].

Thank you for your reply!

Yes. My specific question was about the restriction of [itex] x_2 [/itex] to 0, 1 ,... y. It makes more sense when you give an example. What I'm going to ask next might be dumb and might show that I don't fully understand the transformation technique. Could he have described the support differently? For example, letting [itex]x_2[/itex] be 0, 1, 2,... and restricting [itex]y[/itex] instead somehow?
 
  • #4
andresc889 said:
that I don't fully understand the transformation technique. Could he have described the support differently?

Actually, you can help me by explaining what (in general terms) the "tranformation technique" is and what it's used for. When I took probability, the books didn't identify a particular method called "the transformation technique". Of course, it was taken for granted that you could do a change of variables.

As long as y and x2 are defined as they are, then the non-zero values of the joint density for a fixed y value occur at x2 = 0,1,2...y. If you define y differently, then the set of x2's non-zero values for a given y could change.

I don't like the terminology that the "support" of x2 is 0,1,2...y. I prefer that the "support" of a random variable be defined as the set of values for which it's density is non-zero. The set {0,1,2...y} should be described as the "support of x2 given y" or something like that.

If you've had calculus, what is going on amounts to the usual change in the limits of integration when you change variables. Here, the "integrals" are sums. (In advanced mathematics there are very general definitions for integration and sums actually are examples of these generalized types of integrals.)

Suppose you have a discrete joint density f(i,j) defined on a rectangle where the (i,j) entries are in a 3 by 4 pattern like

# $ * *
$ * * *
* * * *

Then if you want to compute the marginal density for a given i0, you sum f(i0,j) for j = 1 to 4.

Suppose you change variables so the indices become (p,q) and pattern is changed to
a parallelogram like:

#
$ $
* * *
* * *
* *
*

If you want to compute the marginal density of a given p0, you must adjust the indices of q that you sum over depending on the value of p0.
 
  • #5


Hello there!

The region of support for g(y, x2) is obtained by considering the possible values that y and x2 can take on based on their definitions. Since y = x1 + x2, it makes sense that y can only take on values that are equal to or greater than x2, as x1 cannot be negative. Similarly, x2 can only take on values that are less than or equal to y, as x1 cannot be greater than y. This leads to the region of support for g(y, x2) being y = 0, 1, 2,... and x2 = 0, 1,..., y.

I hope this helps clarify things for you. The transformation technique can sometimes be a little tricky to understand, but once you understand the reasoning behind it, it becomes much easier to apply in various situations. Let me know if you have any other questions and I'll be happy to help. Happy studying!
 

Related to Sum of two independent Poisson random variables

1. What is the formula for calculating the sum of two independent Poisson random variables?

The formula for calculating the sum of two independent Poisson random variables is:
P(X+Y=k) = ∑ P(X=i)P(Y=k-i) k/i=0
Where X and Y are two independent Poisson random variables with parameters λ1 and λ2, and k is the desired sum value.

2. How is the sum of two independent Poisson random variables different from the sum of two dependent Poisson random variables?

The sum of two independent Poisson random variables follows a Poisson distribution with parameter λ12. This means that the probability of a certain sum value is equal to the product of the probabilities of each individual variable.
On the other hand, the sum of two dependent Poisson random variables does not follow a Poisson distribution and the calculation of probabilities becomes more complex.

3. Can the sum of two independent Poisson random variables be greater than the individual variables' maximum possible values?

Yes, it is possible for the sum of two independent Poisson random variables to be greater than the maximum possible values of the individual variables. This is because the Poisson distribution is a continuous distribution and has no upper limit.

4. Is it possible for the sum of two independent Poisson random variables to have a mean of 0?

No, the sum of two independent Poisson random variables will always have a mean greater than 0. This is because the mean of a Poisson distribution is equal to its parameter λ, and since the sum of two independent Poisson random variables has a parameter of λ12, the mean will always be greater than 0.

5. How can the sum of two independent Poisson random variables be applied in real-life scenarios?

The sum of two independent Poisson random variables can be applied in various real-life scenarios, such as:
- Estimating the number of customers in a certain time period, where one variable represents the arrival rate of customers and the other represents the service rate of the business.
- Modeling the number of defects in a manufacturing process, where one variable represents the number of defects in one component and the other represents the number of defects in another component.
- Predicting the number of accidents on a road, where one variable represents the number of accidents caused by one factor (e.g. weather) and the other represents the number of accidents caused by another factor (e.g. distracted driving).

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