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" />

answer: the 4 kilogram bag

step-by-step explanation: this is because if you divide the number of kilograms into the total it will tell you how much each individual kilogram would be worth. 2 kilos is 7 and change 3 would be around 6 ish and the 4 kilo would be 4.6 and change and the 5 kilos would be like 4.88

hope this .

r3t40