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Tibor Pejić eNotes educator | Certified Educator

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I will answer first two i.e. (a) and (b) questions because eNotes rules forbid answering multiple questions.


Because of linearity of expectation `E[a X+bY]=aE[x]+bE[Y]`

we have



By definition `Var(X)=E[X^2]-(E[X])^2` hence we have



Here we've used expected value from part (a) and now we will use linearity of expectation.


` `  Now because `beta^2E[x_i^2]-beta^2(E[x_i])^2=beta^2Var(x_i)`

` `and `E[epsilon_i^2]=Var(epsilon_i)-(E[epsilon_i])^2=sigma^2-0^2=sigma^2` we have

` ` `beta^2Var(x_i)+2betaE[x_iepsilon_i]+sigma^2` which would be your solution.

If we assume that `E[x_iepsilon_i]=0` which will be the case if the estimator `beta` is unbiased and consistent, then you would have

`beta^2Var(x_i)+sigma^2`  <-- Your solution if `x_i` and `epsilon_i` are uncorrelated.  

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