Question

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.

Calculate ​E(u ​| ​X=​1), ​E(Y ​| ​X=​1), ​E(u ​| ​X=​2), ​E(Y ​| ​X=​2), ​E(u ​| X​), ​E(Y ​| X​), ​E(u​) and ​E(Y​).

Answer 1

a)
`E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1+`

b)
`E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2`

c)
`E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2`

d)
`E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6`

e)
`E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1`

f)
`E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3`

g)
`E(u) = E(v) +E(X) = 0+1=1`

h) E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

Explanation:

For this case we know this:

`Y = 1+X +u`

`u = v+X`

with both Y and u random variables, we also know that:

`[tex] E(v) = 0, Var(v) =1, E(X) = 1, Var(X)=2`

And we want to calculate this:

Part a

`E(u|X=1)= E(v+X|X=1)`

Using properties for the conditional expected value we have this:

`E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1`

Because we assume that v and X are independent

Part b

`E(Y| X=1) = E(1+X+u|X=1)`

If we distribute the expected value we got:

`E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2`

Part c

`E(u|X=2)= E(v+X|X=2)`

Using properties for the conditional expected value we have this:

`E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2`

Because we assume that v and X are independent

Part d

`E(Y| X=2) = E(1+X+u|X=2)`

If we distribute the expected value we got:

`E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6`

Part e

`E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1`

Part f

`E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3`

Part g

`E(u) = E(v) +E(X) = 0+1=1`

Part h

E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]