can you explain and give me a numerical example?

Theorem 3.6 The Poisson (a) random variable in Definition 3.9 has expected value E[X] = Q. Proof E[X] = xPx (x) = ) x- e (3.46) x! x=0 x=0 We observe that x/x! = 1/(x - 1)! and also that the x = 0 term in the sum is zero. In addition, we substitute a = a . at to factor a from the sum to obtain DO E[X] = a (3.47) x=1 x - 1)! Next we substitute l = x - 1, with the result E[X] = Q = Q. (3.48) =0 We can conclude that the sum in this formula equals 1 either by referring to the identity e" = Eno o' /!! or by applying Theorem 3.1(b) to the fact that the sum is the sum of the PMF of a Poisson random variable L over all values in SL and P[SL] = 1