Question

The joint probability density function of two continuous random variables X and Y is

fxy(x,y) = { C, 0 <y < 2.y < x < 4-y 10, otherwise

Find the value of c and the correlation of X and Y.

Consider the same two random variables X and Y in problem [1] with the same joint probability density function. Find the mean value of Y when X<1.

Answer 1

Yondom lovciable X andy Sol"! Here we have given point of g turo Continuous ;O{y <2, J&X {9-4 lo, ow Now we chow cria y owe poffcm,9) Exy (x,y) = ELS Y (2012) X+1=4 Y=X Ye2 ((2, 2) X B of(0,0) (4,0) the Shaded area is pdf crea help lenow that fore joint pdf g Continuous vandem voce. st xx(x, y) = 1 clxdy => cs 1 dx dy zl => c f S dx dy = 1 = (xf xxx2 = 1 jarea of thange 2}x Basex Height triangle BC -1 [C=4 area of ora Now to find Covelation we have to formelo Cor Conx, 4) Ju(x) Vich

cou (x,y) = E(X4) - E(X) E(0)| So JE (XY) = s ry fry 3,y) dxdy - o 2 9-1 ५ (4-29ीव५ way of drdy = $*** [ca-y)-yol day [(4-y+Y) (9-2y)dy Szydy - 1 y d y = E(Y) = f (x)dc 8 2 2 - 4 4 - 8 3 4 3 - So wie find Marginal of x by which wie find lodf of x with full range of y.

fcm-For dy = Da con = 22 O<<<2 11 0 Similarly fy (%) = -4-8 & Fru (nychcz I (4-24) కార = 1- ghi 0422 jow Now E(X)= 5 * feconda - xdx 2 [E(y) = y fugly dy E(4)2 S ² (2 - 4 dy = 2-8 = 9 = 2/3 Elaphe E649)-D. Cos gay day 5**8-TXIC کے 1. 02 21 x 8 = 1 9 E(X²) p2c²dx < B= Sol V(x) = E(X?) - E(X) / / - 1 = 1 / 3 3 3 4 3 9 z 9 VCy)= 3 & 9 So Cov (8,42 = 9/23 - 1x ² = 2 3

こ こ Corr. (X14) = 2/3 2 x 43 x 3 J13 J8/g 2 x 252 = 11.5 onl corr(x,y) = 1 I ole; there is a perfect Corvelation between x andy Next we have to find Mean valul of Y when XL 1

3 Т- 2 ## - [t] = RPG R - 4), - fpts-) Л 3 2 2 Rр xptx = (1>Х'5) 3 E