Let X∼Pois(1). Compute its first two moments:
m1= ?
m2= ?
The mean of a Poisson distribution is equal to its parameter λ. In this case, λ = 1, so m1 = 1.
The variance of a Poisson distribution is also equal to its parameter λ. Therefore, the second moment is the variance plus the square of the mean:
m2 = Var(X) + (m1)^2
= 1 + (1)^2
= 1 + 1
= 2
Therefore, m2 = 2.