Simplest way to realize this is that the difference can easily be negative but a poisson random variable can't be negative.
Why the difference of two independent Poisson random variables doesn't have Poisson distribution? Demonstrate it.
Simplest way to realize this is that the difference can easily be negative but a poisson random variable can't be negative.
I don't have emotions and sometimes that makes me very sad.
I only know how to do the sum of two independent Poisson random variable using moment generating function (by using theorem)but don't know how to do the difference. Can you please explain in some detail how to do the difference.
Are you asking how to show that the difference *isn't* poisson distributed or how to derive the distribution of the difference?
I don't have emotions and sometimes that makes me very sad.
For reference
http://en.wikipedia.org/wiki/Skellam_distribution
I have to prove that the difference of two independent Poisson random variables isn't Poisson distribution. I know that the difference of two independent Poisson random variables has a Skellam distribution but for homework I have to prove that the difference is not Poisson Distribution.
This is my homework question.
Determine if the difference of two independent Poisson random variables has a Poisson distribution. If it does, prove it. If it doesn’t, demonstrate why.
I already showed you an extremely quick easy way to show that it does NOT have a poisson distribution. The difference can be negative while the poisson distribution can't be negative. That's enough to show that the difference can't be poisson distributed. Is there something wrong with this?
I don't have emotions and sometimes that makes me very sad.
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