Question

Question 4: Let X and Y be two discrete random variables with the following joint probability distribution (mass) function Pxy(x, y): a) Complete the following probability table: Y 2 f(x)=P(X=x) 1 3 4 0 0 0.08 0.06 0.05 0.02 0.07 0.08 0.06 0.12 0.05 0.03 0.06 0.05 0.04 0.03 0.01 0.02 0.03 0.04 2 3 foy)=P(Y=y) 0.03 b) What is P(X s 2 and YS 3)? c) Find the marginal probability distribution (mass) function of X; [f(x)]. d) Find the marginal probability distribution (mass) function of Y; [fo)]. e) Find conditional distribution of X given Y=y; (fxy). f) Find expected value of X [E(X)] and variance of X [Var(X)]. g) Find expected value of X given Y=2 [E(X[v=2)]. h) If Z=3X-2Y and Cov(X,Y)=w, find expected value of Z and Variance of Z (as a function of o).

Answer 1

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(a) Jl) plx=x) Х 2 Y3 b (3 4 0 0.01 0.02 0.08. 0.0.6 0.05 0.02 0.07. 0.08 0.07 0.03 0.25 2 3 0.06 0.12 0.05 0.03 0.26 0.06 10.05 0.04 0.03 0.18 12 0.28 0.33 0.24 0.15 0.03 (0:04 0.10 - fly)= P(Y=y) 0.21 (b) Of 25.07 PCX <2 and y<3) 07.33 = 0.08 +0.07 +0.06 +0.06 t 0.06+ 0.08 t 0.12 + 0.05 +0.05 +0.07 +0.05 +0.04 (si nove brio 0.79 (S.XIV us-t-ivo to = (c) marginal inal prodani üty distribution (mox ) function gx vai SI - (in f(x) { flavy) Wom y 1601) 0.28 IX=0 0.33 2.=) 0.24 N = 2. 0.15 x=3 0 othes wise 2 Scanned by TapScanner

d) y. margina) probab? city distribution (mass function of f(y) = { fray) X fly): 0 ya → = 5 0.21 0.25 0.26 0.18 0.10 y =) 2 y = 2 • v-3 7 2 i otherwise of Ce conditional distribution x given y = y; fxly using the famula flaly)= f(x,y) Jy cy) with x=0, 1, 2, 3 and y = 0, 1, 2, 3, 4 Y 0 July) 2 3 y 0 21 7/25 6/26 6/18 110 8/25 (12/26 s/18 2110 7/25 5/26 4/18 |3110 3/25 3/26 | 37,8 420 Х 6/21 5/21 2/21 2 3 1 1 1 T Scanned by TapScanner

(8) f(x) = Ex. fx(x) X 0X0.28 + 1X 0.33 + 2 x 0.24 + 3x0.15 ot 0.33 to,u8 +0.45 - 1.26 こ Vas (x) = f(x2) - (Ex))² So, for this f(x²) = 4x² fxlx) x = 0X0.28 + 1² x 0.33+ 2² x 0.24 + 3 ² x 0.15 of 0.33 + 0.96 + 135 : 2.64. So, Vas (X) = f(x2) - Lecx))? 2.64 (1.26) ² 2.64 - 1.5876 = 11.0524 . 3 (9) E(X/4-2)= 2.3(2,2) 2-0 - ㅗ 10.160,2) + 1. 401,2) +2.412,2+3.463,2) 1 24 (2) - OX 0.06 + 1 X 0.12 + 2X0.05+3 X 0.03) 0.26 ) 0 +0.12 + 0.10+ 0.09. Scanned by TapScanner ܘ ܝܘ ܐ ܝ 0.26

Given. 2= 3x - 24 and Cov (x, y) = w E (2) = E [3x-24 8 = 3 E (X) - 2 E (4) 381.26 - 2x 1.71 van 3,78 -3.42 = 0.36 E (4) = Ey Ey. fyly) walking note. = 0 X 0.2) + 1 X 0.25 + 2X0.26 + 3 X0-18 + u X 0.10 ot 0.25 +0.52 +0.54 +0.40 = 1.71 1 Vas 127 and vas (2) = vas (3x-24) = Cov (3X-24 , 34-24) = Cow (3x, 3x) + cow (3x, -24) + Cou(-24,3x). + Cov(-24, -24) qv(x) - 6 cov (x, y) - 6 covlx, 4)+ 4 vey) = a (x) -12 lov lx, y) + 4 V14) - g (1.0524) - 12 W + 4 (1.5859 15.8152-120 (142) = 0x0.21 + 12x 0.25 +. 2² x 0.26 + 32x0.18 + y²x0,10 0 +0.25 +1.04 + 1.62 + 1.6 - 4.5) So, vas(4): EC42) (E14))? = 4.51-(1.72 = 4.5) - 2.924 Scanned by TapScanner walking Notte