Question

1 3 4 9. Suppose the discrete random variables X and Y are jointly distributed according to the following table: Yl-1 0 1 0.1 0.1 0.1 0 0.2 0.1 0.2 0.1 0.1 a. Compute the expected values E(X) and E(Y), variances V(X) and V(Y), and covariance Cov(X,Y) of X and Y. (11) b. Let W = X - Y. Compute E(W) and V(W). [4] 10. Let X be a continuous random variable with probability density function h(x) ce* r > 0 *<0 ={c a. Verify that h() is a valid probability density function. [7] b. Compute the expected value E(X) and variance V(X) of X. (8)

Answer 1

X 9. gly) O 1 0.1 0.1 0.3 0-3 3 O 0.2 4 0-2 04 1 0.3 0.4 0.3 The marginals of x and y are calculated f (x) and gly) respectively (a) ELX) = Exf(x) - 1 x 0.3 + Ox 0.4 + 1xo.31 + - 0.3 + 0-3 M - 0-3 + O. 3 E(X) oo Et Ely) = Eygcy) 1x 0.3 3 x 0.3 + 4x o-4 M O-3 + 0.9 + 1-6 11 Ely) = 2-8 By the V(X) 2 definition of VLX) E (X2) - [E(x)]² calculated E (X) to calculate É (X²) he have need he

E (x2) = Ex» f (x) 2 (-1) ² x 0.3 + DX 0.4 + (1) 0.3 Ixo.3 + o t ( xo-3 0.3 t 0.3 E (x²) = 0.6 So, V(X) = ECx²) - [E(X)]² 0.6 So, o. 6 we we Vly) Ecy?) - [E (4) have calculated t(4) need calculate E(42) Ely 2) = & g 2 gly) = (1) ² x 0.3 + (3) 0.3 + (4)” x 0.4 1 x 0-3 t 9 x 0.3 16x 0 4 + 0.3 + 2.7 + 6.4 Ely2) = 9.4 2 So, vly) - ELY2) [ELY) (2.8) 2 9.4 - = g. 4 - 7.84 VLY) 1.56

By the definition of Lov(x, y) Cov(x, y) = Elxy) E(X). Ely) + 1x oxo.1 + 1x1xo.1 ELXY) 1x (1)xo.1 + 3x (-1) xo + 3x ox0.2 + 3x1ool + 4 x (-1) xo.2 + 4 xox o. l + 4 x 1 xool 0.3 - Elxy)= – 0.1 + 0 + 0 + 0 + 0 + - 0.8 tot 0.4 - Elxy) = o. 7 - 08 = Elx y) = -0.1 so, Lou (x, y) - Elx y) - E(X) Ely) - 0.1 0 x 2-8 = - CON (x, y) = (6) W = x - Y Taking expectation ElW) = E (X-Y) both sides = E(hl) = ELX) - E(Y) a) E (W) 0 - 2-8 from part (a) E (X) = 0 Ely) 2-8 Eln) 2-8

W = x-Y Taking variance both sides V(W)= V(x-4) =) VIW) = V(x) + vly) - 2 Lou (x, y) From part la) Vly) = 1.56 2 VLX) = 0.6 and Coucx, 4) = – 0.1 - =) Vlwla 0.6 t lost 2 x (-0.1) => vlw= 0.6 + 1.56 + 0.2 => VW) 2-36 10 TO be valid probability density function hlx) dx = (0) x so, Sxe using Inegration By parts method e du -k se e (-1) 0 x = 0 + Jet dx 0

у o 8 (-1) = (-1) Le eº] = (-1) Lo- 1] D + I So, it is a So, Ixe dx = 1 valid probability density function (6) By the definition of E(X) J x R (x) dx Jr. x ex dx s 2. 12 e c du e 3-1 W! e dx 13 O This is 3-1 8 - re 13 => E(X) = e B Gramma integral 1

= E(X) = B xl Elx) N1 3. [since (n-1)!] = E(X) 2! so, Elx) = 2 V(x) E (x²) - [E(X)] [E (x)] 2 2. e e dre so, E(X²) = x ng xe du 4-1 e dx O 4-) -K 14 e du 10 14 4-1 e dre - Gammo integral 1) of 4 TA 1

la xl = E(X²) = = E(X2) 14 = ) E (x²) 3! 3 x 2 - E (X2) so 6 2 And, V(x) É (X²) [F(x) 6 (2) ? 4 2. V (X) =