Let X be a geometric random variable with parameter p. Find E[min(X, 3)]
My work:
min(X, 3) = X if X < 3 and 3 if X ≥ 3 so
E[min(X, 3)] = 0* P(X = 0) + 1* P(X = 1) + 2*P(X = 2) + 3* P(X ≥ 3)
= 1*p + 2*p*(1-p) + 3* (1-p)^2
as P(X ≥ 3) = P(X > 2) = (1-p)^2
Is my answer of = 1*p + 2*p*(1-p) + 3* (1-p)^2 correct?