Here are the numerical answers for Part I of Assignment 2:

**Question 1:**

a) Distribution for X: 1:0.3, 2:0.4, 3:0.3

Distribution for Y: 0:0.6, 1:0.4b) P(X <= 2Y) = 0.1+0.1 = 0.2

c) No. For example, P(X=2)P(Y=0) = 0.24 is not equal to P(X=2,Y=1)=0.3

d) E(X) = 1x0.3 + 2x0.4 + 3x0.3 = 2

E(Y) = 0x0.6+1x0.4 = 0.4e) E(X

^{2}+Y) = 1x0.2+4x0.3+9x0.1+2x0.2+5x0.1+10x0.2 = 5

or you can compute it as E(X^{2})+E(Y) = 1x0.3+4x0.4+9x0.3 + 0.4

E(XY-2) = E(XY)-2 = 1x0.1+2x0.1+3x0.2-2 = -1.1f) P(X=1|Y=0) = P(X=1,Y=0)/P(Y=0) = 0.2/0.6 = 1/3

P(Y<1|X>1) = P(Y<1 and X>1)/P(X>1) = (0.3+0.1)/(0.3+0.1+0.1+0.2) = 4/7g) VAR(X) computed directly is (1-2)

^{2}x0.3 + (2-2)^{2}x0.4 + (3-2)^{2}x0.3 = 0.6

Computing it indirectly:

W=E(X|Y) has distribution with P(W=11//6)=0.6 and P(W=9/4)=0.4

Z=VAR(X|Y) has distribution with P(Z=51/108)=0.6 and P(Z=11/6)=0.4

VAR(W) = 5/120 and E(Z) = 201/360, so VAR(X)=VAR(W)+E(Z)= 5/120+201/360 = 0.6h) COV(X,Y) = 0.1

**Question 2:**

a) Y_{i}has the geometric distribution with p=(7-i)/6

E(Y_{i}) = 6/(7-i)b) E(X) = 6/6+6/5+6/4+6/3+6/2+6/1=14.7

c) VAR(X) = 38.99