1. assume y=1+x+u, where x, y, and u=v+x are random variables, v is independent of x; e(v)=0, var(v)=1, e(x)=1, and var(x)=2.

Var(u|X=1) = 1

Var(Y|X=1) = 1

Var(u|X) = 3

Var(Y|X) = 9

Step-by-step explanation:

First, we need to identify the properties of the variance, so:

1. Var(a) = 0 , if a is constant

2. ,  Where a is constant and X is a random variable

3. , Where a and b are constants and X is a random variable.

4. , Where X and Y are random variables and are independents.

Then, if Y=1+X+u, where X, Y, and u=v+X are random variables, v is independent of X; E(v)=0, Var(v)=1, E(X)=1, and Var(X)=2, the variance of the following cases are calculated as:

Var(u|X=1) = Var( v + X | X = 1) = Var( v + 1 )    

                 = Var(v) + Var(1)                            

                  = 1 + 0 = 1

Var(Y|X=1) = Var( 1 + X + u | X = 1 )

                 = Var (1 + X + v + X|X=1)

                 = Var(1 + 1 + v + 1)                      

                 = Var(3) + Var(v)                      

                 = 0 + 1 = 1

Var(u|X) = Var ( v + X | X)

             = Var ( v + X)

             = Var(v) + Var(X)

             =  1 + 2  = 3

Var(Y|X) = Var ( 1 + X + u | X)

             = Var ( 1 + X + v + X | X)

             = Var ( 1 + 2X + v)

             = Var(1) + Var(2X) + Var(v)

             = 0 + 4*2 + 1 = 8 + 1 = 9

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.