Respuesta :
Answer:
a) [tex] E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1+[/tex]
b) [tex]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[/tex]
c) [tex] E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2[/tex]
d) [tex]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[/tex]
e) [tex] E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1[/tex]
f) [tex] E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3[/tex]
g) [tex]E(u) = E(v) +E(X) = 0+1=1[/tex]
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]
Step-by-step explanation:
For this case we know this:
[tex] Y = 1+X +u[/tex]
[tex] u = v+X[/tex]
with both Y and u random variables, we also know that:
[tex] [tex] E(v) = 0, Var(v) =1, E(X) = 1, Var(X)=2[/tex]
And we want to calculate this:
Part a
[tex] E(u|X=1)= E(v+X|X=1)[/tex]
Using properties for the conditional expected value we have this:
[tex] E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1[/tex]
Because we assume that v and X are independent
Part b
[tex]E(Y| X=1) = E(1+X+u|X=1)[/tex]
If we distribute the expected value we got:
[tex]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[/tex]
Part c
[tex] E(u|X=2)= E(v+X|X=2)[/tex]
Using properties for the conditional expected value we have this:
[tex] E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2[/tex]
Because we assume that v and X are independent
Part d
[tex]E(Y| X=2) = E(1+X+u|X=2)[/tex]
If we distribute the expected value we got:
[tex]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[/tex]
Part e
[tex] E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1[/tex]
Part f
[tex] E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3[/tex]
Part g
[tex]E(u) = E(v) +E(X) = 0+1=1[/tex]
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]
