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).

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" alt=" E(v) = 0, Var(v) =1, E(X) = 1, Var(X)=2" />

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" />

Here is the answer to the given problem above. Given that the business owes $8000 on a loan and every month, it pays 1212. So for the first month, the remaining is 6,788. On the second month is 5,576, and on the third month, it is 4,364. In the fourth month, the remaining would be 3,152, and in the fifth month is 1,940. Therefore, on the sixth month, if the business pays 1,212 still, the remaining would be $728 only. Hope this answer helps.