Answer:
a)
b)
c)
d)
e)
f)
g)
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:
with both Y and u random variables, we also know that:
And we want to calculate this:
Part a
Using properties for the conditional expected value we have this:
Because we assume that v and X are independent
Part b
If we distribute the expected value we got:
Part c
Using properties for the conditional expected value we have this:
Because we assume that v and X are independent
Part d
If we distribute the expected value we got:
Part e
Part f
Part g
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]