Let X and Y denote the population of two different chemicals

Let X and Y denote the population of two different chemicals found in a sample mixture of chemicals used as an insecticide. Suppose X1 and X2 have joint pdf given by f(x,y) = { 2 for 0 LE x LE 1,0 LE y LE 1,0 LE x + Y LE 1, elsewhere Find Pr (X Le 2/3, Y LE 2/3); Find Pr (X LE 1/2, Y LE 1/4); Find Pr (X LE 1/2 Y LE 1/2); Find marginal densities for X and Y.

Solution

a)

P(X<=2/3, Y<=2/3)
= twice integrate over region x<=2/3, y<=2/3 of f(x, y) dA
= twice integrate over region x<=2/3, y<=2/3, 0<= x <= 1, 0<=y<=1, 0<= x+y <= 1 of 2 dA

= integral from 0 to 1/3 integral from 0 to 2/3 of 2 dydx
+ integral from 1/3 to 2/3 integral from 0 to 1-x of 2 dydx

= (1/3)(2/3)(2) + 2(1/2)(-1)[(1 - (2/3))^2 - (1 - (1/3))^2]
= 4/9 - (1/9 - 4/9)
= 7/9. Answer

2)

P(X<=1/2, Y<=1/4)
= twice integrate over region x<=1/2, y<=1/4 of f(x, y) dA
= twice integrate over region x<=1/2, y<=1/4, 0<= x <= 1, 0<=y<=1, 0<= x+y <= 1 of 2 dA
= integral from 0 to 1/2 integral from 0 to 1/4 of 2 dydx
= (1/2)(1/4)(2)
= 1/4. Answer

3)

P(X<= 1/2|Y <= 1/2)

= P(X<= 1/2, Y <= 1/2) / P(Y <= 1/2).

P(X<=1/2, Y<=1/2)
= twice integrate over region x<=1/2, y<=1/2 of f(x, y) dA
= twice integrate over region x<=1/2, y<=1/2, 0<= x <= 1, 0<=y<=1, 0<= x+y <= 1 of 2 dA
= integral from 0 to 1/2 integral from 0 to 1/2 of 2 dydx
= (1/2)(1/2)(2)
= 1/2.

P(Y<=1/2)
= twice integrate over region y<=1/2 of f(x, y) dA
= twice integrate over region y<=1/2, 0<= x <= 1, 0<=y<=1, 0<= x+y <= 1 of 2 dA
= integral from 0 to 1/2 integral from 0 to 1-y of 2 dxdy
= 2(1/2)(-1)[(1 - (1/2))^2 - (1 - 0)^2]
= -(1/4 - 1)
= 3/4.

Hence , , P(X<= 1/2|Y <= 1/2) = (1/2) / (3/4) = 2/3 Answer.