Mathematical Methods for Physics and Engineering - Matematica.NET

PROBABILITY◮A card is drawn at random from a normal 52-card pack and its identity noted. The cardis replaced, the pack shuffled and the process repeated. Random variables W, X, Y, Z aredefined as follows:W =2 if the drawn card is a heart; W =0otherwise.X =4 if the drawn card is an ace, king, or queen; X =2if the card isajackorten;X =0otherwise.Y =1 if the drawn card is red; Y =0otherwise.Z =2 if the drawn card is black and an ace, king or queen; Z =0otherwise.Establish the correlation matrix for W, X, Y, Z.The means of the variables are given byµ W =2× 1 = 1 , µ 4 2 X = ( ) ( )4 × 313 + 2 ×213 =16, 13µ Y =1× 1 = 1 , µ 2 2 Z =2× 6 = 3 . 52 13The variances, calculated from σU 2 = V [U] =E [ U 2] − (E[U]) 2 ,whereU = W , X, Y orZ, areσW 2 = ( ) (4 × 1 4 − 1) 22 =3, 4 σ2 X = ( 16 × 313σ 2 Y = ( 1 × 1 2)−( 12) 2=14 , σ2 Z = ( 4 × 652) ( ) (+ 4 ×2 − 16)−( 31313) 2=69169 .13) 2=472169 ,The covariances are found by first calculating E[WX] etc. and then forming E[WX]−µ W µ Xetc.E[WX]=2(4) ( ) (352 +2(2) 2)52 =8, Cov[W,X]= 8 − ( 1 16)13 13 2 13 =0,E[WY] = 2(1) ( )14 =1, Cov[W,Y]= 1 − ( 1 1)2 2 2 2 =1, 4(E[WZ]=0, Cov[W,Z]=0− 1 3)2 13 = −3, 26E[XY ] = 4(1) ( ) (652 +2(1) 4)52 =88, Cov[X,Y ]= − ( 16 1)13 13 13 2 =0,E[XZ] = 4(2) ( 652)=1212, Cov[X,Z]= − 1613 13 13( 3)13 =108, 169(E[YZ]=0, Cov[Y,Z]=0− 1 3)2 13 = −3. 26The correlations Corr[W,X] and Corr[X,Y ] are clearly zero; the remainder are given by(Corr[W,Y]= 1 3 × ) 1 −1/24 4 4 = 0.577,(Corr[W,Z]=− 3 3 × ) 69 −1/226 4 169 = −0.209,Corr[X,Z]= 108169Corr[Y,Z]=− 326( 472169 × 69) −1/2169 = 0.598,( 1 × ) 69 −1/24 169 = −0.361.Finally, then, we can write down the correlation matrix:⎛⎞1 0 0.58 −0.21⎜ 0 1 0 0.60 ⎟ρ = ⎝0.58 0 1 −0.36⎠ .−0.21 0.60 −0.36 11204