Final answer:
The question is about calculating the expectation E[(X - 1)^2] for a Poisson distributed random variable X with mean μ. The computation shows that E[(X - 1)^2] equals the mean μ of the Poisson distribution.
Step-by-step explanation:
The student is asking about the Poisson distribution, which is a discrete probability distribution that gives the likelihood of a given number of events occurring in a fixed interval of time or space, assuming these events occur with a known constant mean rate and independently from each other. Specifically, if X is a Poisson random variable with mean μ, then the probability of X taking on a certain value k is given by the formula:
P(X = k) = μke-μ/k!
To find E[(X - 1)^2] for a Poisson random variable, we would calculate:
E[(X - 1)^2] = E[X^2] - 2E[X] + 1
Since for a Poisson distribution E[X] = Var(X) = μ, we can find E[(X - 1)^2] by substituting μ for E[X] and Var(X) resulting in:
E[(X - 1)^2] = μ + μ^2 - 2μ + 1 = μ + μ - 1 = μ
This shows that E[(X - 1)^2] is equal to the mean μ of the Poisson distribution. Note that the question is malformed, and the variables and the meaning of 'show that () = .' might be a typo, so we proceed assuming the task is to link the expectation of a transformed Poisson variable to its mean μ.