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Okay so I have gotten down to

$$b=\beta + \frac{\sum_{i=1}^{n}x_i \varepsilon_i}{\sum_{i=1}^{n} x_i^2}$$

but I cannot figure out how to show that second term is $0$.

Michael Hardy
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Gamecocks99
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  • Can you provide all details? I guess you are looking at something like $Y=X\beta +\epsilon$ and for sure you have some assumptions on $\epsilon$. – math Nov 08 '13 at 11:21
  • Exactly where are you stuck? If your regressors are deterministic take the expected value of $b$ and since they are deterministic, they do not have an expected value, only the error term has. If the regressors are stochastic, then take the expected value of $b$ conditional on the regressors. – Alecos Papadopoulos Nov 08 '13 at 17:34

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Use that the expectation of a constant times a random variable is the constant times the expectation of the random variable. From which it follows that you need only show that the numerator is zero. And then use that the expectation of a sum is the sum of the expectations, from which it follows that you need only show that each term in the numerator is zero. And then use (again) that the expectation of a constant times a random variable is equal to the constant times the expectation of the random variable to obtain that you need only show that the $\epsilon_i$ all have expectation zero. And this is the assumption underlying the use of least squares!

All this assumed that the $x_i$ are fixed. In settings where the $x_i$ are random, one first conditions on all of the $x_i$, so that the what is needed is that the conditional expectation of the $\epsilon_i$, given the $x_i$ are all zero. Which is the assumption underlying least squares when the $x_i$ are random.

dan
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