Question

Let X and Y have the joint pmf defined by (х, у) (1,2) (0,0) (0,1) (0,2) (1,1) (2,2) 2/12 1/12 3/12 1/12 1/12 4/12 Pxy (x, y) Find py (x) and p, (y) а. b. Are X and Y independent? Support your answer. Find x,y,, and o, С. d. Find Px.Y

Answer 1

x f y have joint pm f (x,y) 110,0) (0) (0,2) 1 (1.1)|(1, 2) (2, 2) Exy (2,7) 1/12 | 2%2 /23/214/21/12 1 2 Total | 3/12 1/12 5/12 6/12 112 2 Total 2/12 V12 4/12 4/12 7/12 / L /12 @ no Find marginal P(x) and distribution Py(4) of x Py(x) 14/ 4/12 7/12 /12 marginal Y lo Py (9) 12 distribution of y 12 I 5/12 16/12

E(X) = 2 I x, P(X=2 ED - 506) + 7) + ?(id) 7 + 2 12 12 E(X) = 9 2 E(T) = { Yip(4=Y) i=0 (2) + '() +27 ELY) = 17 12

F(X.V = Į < xil PlXiyi) 120 ico - -otot otot 3/12 to tot 8/12 + 4/12 E(XY)= 15 = 1.25 12 If E( XY) then we variables - E(X). Ely) say that X fy are independent

6)+Gi+6) 0=1 (.*=X)D 2?x = (x) (1/2 + ( 2 ) + ( 4 )o = (!x=)!* 3 = (X) =xm © 21golon tuopuadopu! tou jeb tox : (27 (X)7 € (ax) 1.0625 I E(X). Ely) = 기린 러 Esi= tl x 6 = Ch . ( X 7

12 TE(x²) = 1 12 6x² = variance (x)=E(x2) - [E(x)]2 T 81 = 11 - 144 = 132 - 81 144 = 132-81 144 16x2 = 51 = 0.3542 L 6x =0. 5951 - standard deviation

lly = E(Y) - EyiRcy=yi) 120 =0442) +1 (9) + 2C) My = ECY) 11 12 E(Y2) = {y: 2 Ply=4;) = 0 ( 12 ) + ( 5 ) + 466 + 24 12 - 5 12 E(Y2) = 29 12 & Gy2 = var (4) = E(Y2) /E(Y)J² - 29 -1172 12 (12) = 29 - 289 12 44

= 348 - 289 144 144 = 348 - 289 - 144 = 59 @ Find 8y= & 0.6403 Pxy = correlation (X,Y) corr(x,y) = bxy 6x64 Oxy = E(XY) – El X). ElX). - i (92) (10) 15 - 153 12 144 1.25 - 1.0625 oxy = 0.1875

corr(x, y) = Px.y = 66.64 = 0.1875 0.5951 x 0.6403 0.1875 0.3810 Corrlx,y)= P x,y =0.4921