Question

1. Consider the joint probability density function 0<x<y, 0<y<1, fx.x(x, y) = 0, otherwise. (a) Find the marginal probability density function of Y and identify its distribution. (5 marks (b) Find the conditional probability density function of X given Y=y and hence find the mean and variance of X conditional on Y=y. [7 marks] (c) Use iterated expectation to find the expected value of X [5 marks (d) Use E(XY) and var(XY) from (b) above to find the variance of X. [7 marks)

Answer 1

(a) density funcsion of y Marginal is given probability by byly) = - f(x,y) dx dr Oslo orci 2 x 1 Ilyo cole - The distribution of y is in fly) = w20,1) 1 oldal of conditional (B) probability By the density definition function f (x14=4) = f(x, y) Byly) So, - oly el Mean - E(X14=4) = f(x1y) dr.

- xdx | - | Ly - 0 ] 24 24 #1X/4= y) e AVO, E ( *2/4=y) - 9J x (x1y) dx р = I lys - 0] 34 Therefore, variance - VIXIY=y) = f(XlY=y) - [E (X/Y=y)]? = y 2 - 2 since ELXlY= y) =

4 4 ² - 342 12 42 12 (c) By the definition of iterated expectation Elx) = ET ELX14) ] ELxly) is calculated in bo pass E (X) = 1 - t Ely) = } 8.+ ly) dy I holdy [f(y) = 1 from (al part ] 2 Jydy = 1 [l- og

ELVcXl4] in (b) pars (d) Variance of x Vix) = VE ELXlY) I we have calculated ELX14) & 9 Vlxly) - 4² -> VIX) = V0 +E(42) = Ivly) & 1 Ely² 4 since, ya Locoil] vey) = 1 12 And & (92) = Jy² f (9) dy - 'S ². 1 dy [since fly) - 1 - Syr dy from a pary ] 4 12 VLX) - titties 1 + 1 4836 = 0.02 + 0.02 = V(x) = 0.04 0.04