Question

I am having problem with this stats question, help appreciated. (I can do A but need help with B).

The random variable X has a Poisson distribution with mean 4. The random variable Y is defined by Y = aX + b, where a, b are positive constants.
(a) Given that the mean and variance of Y are both equal to 16, find the value of a and the value of b. [6] [answer is that a = 2, b = 8]
(b) Bill states that, because the mean and variance of Y are equal, Y has a Poisson distribution. Give a reason why Bill’s statement cannot be true.

Answer 1

@Zarkon statistics fun

Answer 2

So \(X\sim\)poisson(4) and \(Y=2X+8\)

the support for a poisson distribution is \(\{0,1,2,\ldots\}\)

the support for \(Y\) is \(\{8,10,12,\ldots\}\)

thus \(Y\) is not poisson.