Question

Please read the English parts as the Chinese parts are just a translation. I faced a difficulty from Question (c)

Question 1 Let X denote the total number of "heads” when 2 coins are tossed. Assume the probability of getting "head" is 0.4 on either coin. text& RETT **. Wit-R FREE*20.4. a) Derive the probability distribution of X? it u s b) Calculate the mean and variance of X, denoted as x, o3.it SCXW R . c) Given the result of (b), calculate E[(x – Hx) 1/. (This value is called skewness that measures whether the distribution of X is symmetric about expected value) (b) , it SIE[(x - Mx)}]/o; (iš 1 Day , * x 3 * F#2x17.)

Answer 1

Probability of getting a Head is 0.4 a] Probability distribution of a If we get both heads then x=2 P(x=2) = 0.4² If we get one head one tail then X=1 (HT, TH) P(X=1) = 0.4 x0.6 +0:6 *014 If we get both tails then x=0 X 0.6 PCX=) - 0.6 1P(x): 0.36 0.48 0.16 b ] Mean or expectation of x (lx) lx = {x.p(at) = 0 x 0.36 + 1x0.48 + 2x0.16 0.8

Variance Vix) = E(X2) - [EC%)]? E(X)= lx E(+2) = { ² pla) = 0? x 0.36 + 12 *0.48 + 2? *0.16 2 1.12 V(x) = 1.12 - 0.82 = 0.48 = 0.48 0x=0.69282 c] E [lx-ex)"] / E (x-lx)3 = E [x3_ 3x²4x + 3 4,8x - e}] = E(X3) - 3.4x E642) + 3 Ux? € (x) - Mi E (x3) - E243. pca) = 030.36 + 13 x 0.48 + 23 x 0.16 = 76 E (x-x)3 = 1.76 - 3x0.8 x 1-12 + 3x0.82x0.8 -0.89 = 0.096 : = 0.2885 E(x-ex)3 / 0x3 = 0.096 0.692823

Question 2 P(X) 0.6 Unemployed (Y=0) Employed (41) 0.36 0.24 0.4 0.32 Non-College grads (x2) college grads CX-1) PCy) 008 0.32 0.68 b ] The conditional probability function of y given x Ply1x) = P(y=y NX =) PUN P(X=3) Therefore we need to find Marginal distribution of Marginal distribution of x P(x) = = 0 0.6 p(y=01 *= 0) = P(y=on X=0) P(x=0) = 0.24 0.6 = 0.4 P (y=0 =1) = P(y =on X=1) P(x=1 = 0.08 = 0.2 conditional Probability distribution of y given & PLYIX) PLY=1/x=0)=P(y=1 n x=0) P(x=o) - 036 -0.6 0.6 0.4 0.6 P(Y=11x=1) = P(y=1n x=1) 0.2 P(x=1) 0.32 = 0.8 0.4 0-8 =

c] Conditional Expectation écy | X=0) E(91=0) = { y.P (4)X =0) = 0x0.4 + 1 X 0.6 = 0.6 conditional expectation of E(y)x=1) - & y.P(41x=1) 0x0.2 + 1 X 0 8 = 0.8

E CECY ()] = { € (41x) • P(X=) = E(y 1x=0). P(x=0) + E(Y)x=1).P(x=1) 0.6 X 06 + 018 30 = 0.68 Marginal distribution of y E(Y) = {y. Ply) = 0x0.32 + 180.68 = 0.68 | PCY) 0.32 0.68 ECECy1x)] = [(y) Hence proved.