5-1. First, f(x,y) ≥ 0. Let R denote the range of (X,Y).

Then,

a) P(X < 2.5, Y < 3) = f(1.5,2) + f(1,1) = 1/60;

b) P(X < 2.5) = f (1.5, 2) + f (1.5, 3) + f(1,1) = 1/8 + 1/4 + 1/4 = 5/8

c) P(Y < 3) = f (1.5, 2) + f(1,1) = 1/8 + 1/4 = 3/8

d) P(X > 1.8, Y > 4.7) = f ( 3, 5) = 1/8

e) E(X) = 1(1/4) + 1.5(3/8) + 2.5(1/4) + 3(1/8) = 1.8125E(Y) = 1(1/4) + 2(1/8) + 3(1/4) + 4(1/4) + 5(1/8) = 2.875V(X) = E(X^{2}) − [E(X)]^{2} = [1^{2}(1/4) + 1.5^{2}(3/8) + 2.5^{2}(1/4) + 3^{2}(1/8)] − 1.8125^{2} = 0.4961V(Y) = E(Y^{2}) − [E(Y)]^{2} = [1^{2}(1/4) + 2^{2}(1/8) + 3^{2}(1/4) + 4^{2}(1/4) + 5^{2}(1/8)] − 2.875^{2} = 1.8594

f) marginal distribution of X

g) and *f _{X}* (1.5) = 3/8. Then,

h) and *f _{Y}*(2) = 1/8. Then,

i) E(Y|X = 1.5) = 2(1/3) + 3(2/3) = 2 1/3

j) Since f_{Y|1.5}(y) ≠ f_{Y}(y), X and Y are not independent

5-3. *f*(*x*, *y*) ≥ 0 *and*

a) *P*(*X* < 0.5,*Y* < 1.5) = *f _{XY}*(−1,−2) +

b) *P*(*X* < 0.5) = ...

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