Question

The Joint pdf of X and Y is f(x,y) = $\frac{5}{3}(1&plus;xy^{2})$ for 0<x<y<1, and 0 otherwise.

a) Find $P(X<\frac{1}{2})$

b) Find $E(X|Y=y)$

Answer 1

Given

$f(x,y)=\frac{5}{3}(1+xy^{2})\; \; \; for\; \; \; 0<x<y<1$

a)

P(X

Marginal probability mass function of X

fx(x) = { $(5, 1)dy = { [1 + xy") dy - [+]. - (+3) - (+-)] - [* *** 3 =4) 5 [(3 +1 -3.6 - 1+) ICT COIT y=r 3 = [3 – 22 – 24] for 0<x<1

P(x <3 - L 12 3 13 – 23 – z'] ds = [35-2 ) r=0 5 199 199 = 5 * 160 - 288

P(X<5) =

b)

fXlY=y(2) = f(x, Y = y) fy(Y = y)

sylu) = s(2, yde – L, $ [1 + xy"] de - [+- [+] - (20+4) for 0<yçi JT=0

f(x, Y = y) fy(Y = y) (1+xy?) (2y + y) 3(1 + ry?) 2y + y4 for 0<<1

nari E(X Y = y = | ) , 3(1+x fx|Y=yz] = - 2y+y r-t | | 32 | +, | T=0 - ਆ, + - ਜੈਤਾ । - ਚ + ਤੋਂ , - + | 31 | 2 + | 3+2j2 – 2(2y+)

E(X|Y = y) = 3+2y 2(2y + y)