Question

Let X1 and X2 be two discrete random variables, where X1 can attain values 1, 2, and 3, and X2 can attain values 2, 3 and 4. The joint probability mass function of these two random variables are given in the table below: X2 X1 2 3 4 1 0.05 0.04 0.06 2 0.1 0.15 0.2 3 0.2 0.1 0.1 a. Find the marginal probability mass functions fX1 (s) and fX2 (t). b. What is the expected values of X1 and X2? c. Are X1 and X2 independent random variables? Justify your answer. d. Find the following conditional density function: fX1|X2 (s|X2 = 2). e. Find the conditional expected value of X1 given X2 = 2: EX1|X2 [X1|X2 = 2].

Question 2. Let X1 and X2 be two discrete random variables, where X1 can attain values 1, 2, and 3, and X2 can attain values 2, 3 and 4. The joint probability mass function of these two random variables are given in the table below: X1 2 | 1 || 0.05 2 | 0.1 3 | 0.2 X2 3 0.04 0.15 0.1 4 0.06 0.2 0.1 a. Find the marginal probability mass functions fx,(s) and fxz(t). b. What is the expected values of X, and X2? c. Are X1 and X2 independent random variables? Justify your answer. d. Find the following conditional density function: fx,\x2 (s|X2 = 2). e. Find the conditional expected value of X1 given X2 = 2: Ex,\x2[X1/X2 = 2].

Answer 1

Solution. et x, & X₂ have joint pot al fox(s, t) x2 fx (5) P(X₂, X2) = P(x, = 5) - soos dos 0.06 0.15 x 2 0.10 0.15 0.20 0.45 3 10.20 0.10 volo 0.40 1. P(x2=t) /0.35 0.29 0.36 Efart). I b) E(X, ) = {x, fx (s) - 1 Coris) +260.45)+300.40) = 0.15to.g+1.2. (E(X) = 2.25 E(X2) = EX2fxrt) =260135)+3 (0.29) + 4(0.36) = 0.77 0. 87+1.45 E(+2) = 3.01 e) fx : (I) = 045 & fx (2) = 0.35 fx, (2) fa (2) = 0.15 (0.35) = 0.0525 fa a lfx (2) & fxx, xz (112) ie , & Xz ax not independent. d) conditional pdf of X / X=2 Ex x(512=2) = fsir (2, 2) to (2) Fx (51x2 =2) o. 1429/0.2857/0.5714 e) Exix [x, 1x =2] = {x, fax (sl x=2] = 1 (0.429 +200.2857) + 3 (0.5714) Exix [x, 1x2 =2] - 2,4285