I'm pretty sure what the author is saying is:

E(X) =:= \sum_t(t * P(X = t)) is the definition

another important note is P(X^2 = t^2) = P(X = t) - because it's the same distribution.

E_a(X) is a bit sloppy, but consider X_a aka Alice's latency "experience" distribution. The argument is:

P(X_a = t) = t * P(X = t) / \sum_u(u * P(X = u)) - i.e. scale the probability up by t but make it sum to 1.

Then

E(X_a) = \sum_t(t * P(X_a = t)) = \sum_t(t * t * P(X = t) / \sum_u(u * P(X = u))

aka

E(X^2) / E(X)

Then (from wikipedia)

Var(X) = E(X^2) - (E(X))^2

And we get

E(X_a) = (Var(X) + (E(X))^2) / E(X) = E(X) + Var(X) / E(X)