Question

1. Suppose that N = {1,2,3} and let X be a random variable such that P(X = 1), P(X = 2) and P(X = 3) are all 1/3. So the probability mass function for X is p(1) = P(2) = P(3) = 1/3. Then, for each n e N= {1,2,...}, we have 3 E[X"] - Ý k"p(k) 1" + 2 + 3" 3 (1) k=1 Calculate E[X], E[X2] and var(X).

Answer 1

3 2) 3 Kol kal 0 Given that Pli) = P(2) = P(3) 2 3 and E[an] =Ž Kn M+29 30 plk) ... Putting sal in ③ we get E [A] - & Xpek) = 8 K PCK) I.PLU) + 2.P(2)+3.P(3) slodz +243 + 3 3 1/3 + 2/3 3 + 1 [using o Putting na 2 in ③ we get E[x2] = { K² p(x) = 12p(1) + 2² P(2) + 3²p(3) atentat 4.43 + 9 + 3 22 K 1+4+9 3 2 3 14] ule know that var ok) E[M] - {e [x 14 3 (294 14 - 4 Hence E[x] - Ķ, E(X) = 3) E(X)= and var(x)=3