Question

2. Probability model of X and Y discrete random variables is given below in the table. Find the marginal probability mass function, mean and variance of X and Y. correlation (Txy), covariance (Oxy) and correlation coefficient (Pxy). y = 0 y = 1 0 Pxy(x,y) x = 1 x = 2 x = 3 y= -1 0.3 0.1 0.1 0.2 0 0.1 0.2

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Answer 1

10 kay My Y = -11 9.0 9.1 | +(x-x) **-10- 300 x=20-10-20 10-14 x=3 | Ollol 0- 2003 ply- yos 10:210-31 Marginal of x PLx1)=0:34 0+0=0-3 p (2=2) = 0.1to-2t 0.1=0.4 Þ(x-3) = 0.1 to 70-23 0 3 7 Marginal of Y bly. -1) = 0 3+01oob ply:o) 0:40:2+ 0 = 0.2 þ (y =)) - O+ oil+02= 0 3 •) E(X)= En plx= x) - (10:3) + ( 2*0.4) (3x0.3 ) = 0.3+0.8 +0.g. 2.0 Mean of x = 2.0

Fly) - E ply=9) - 1-100.5+ ( 0x0.2)+(1803) i - 0.2 - Mean of Y. -) (x) E (X2) - E 2 (X.) F(x) = ? plX= n) - 1130.3) + (22x0.4) + (320.3) - 0.3 + 9067 257 4-6 vex - 4.6 -12.02 o. 6 - variance of x. - VIY) -, Ely) - F?CYJ & (V2) - 2 y 2 p (44) -((-1)2x0.5)+(0?*0:2) i (1? X0:3) = 0.8 17- 0.8 - (-0.282 = 0.76. vaname d y = 2.76 Now, condation (Sxy) and correlation coeffsicient are all the same. The formalo is - Pxy = COV(x, y) ſux) July

Coulx, y) = f(xy) - E(X) E(X.) $(XY) = 3 ay P1X- *, y-y) - (1 - 180.3) + (1xoxo) + (1x1xo) +(?-180-1) + (2x0002) + (2x180.1) (3x - 1x0.17 (3xoxo) & (3x1x02) in – 0.3 -0.2 t o. - 0-3 toob : 0 - cov(x, y) = o- (2x=0.2)=0.4 ..) correlation coefficient = 0.5923 Pxy = 0.4 No.6 Voozt Since the correlation coefficient is o Pxy=05923. so we conclude there is moderately Strong correlation between x and you