Let Y follows a Poisson distribution with mean lambda that i

Let Y follows a Poisson distribution with mean lambda: that is, Y approx equal Poisson(lambda). Show that E(Y) = lambda. Hint: you may use one of the following two methods Method 1: use exponential series e^x = 1 + x + x^2/2! + x^3/3! + .,.. Method 2: Use the 2nd property of a valid pmf that summation y p(y) = 1

Solution

P(x; ?) = (e^-?) (?^x) / x!

xP(x) = x (e^-?) (?^x) / x! =  (e^-?) (?^x) / (x-1)!= (e^-?) (?) (?^x-1) / (x-1)!

Mean = sum of all xP(X)

= (e^-?) (?) e^??

= ?

Thus mean of Poisson distribution is its parameter.

Or E(Y) = Lemda