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

idk

step-by-step explanation: