# Linear Regression: Expectation Proof

I found the following proof in my notes:

$E(Y_i) = E[\beta_0 + \beta X_i + \varepsilon_i] =\cdots= \beta_0 + \beta X_i$. This does not seem right to me, however. Why would $E(\beta_1 X_i) = \beta_1 X_i$? I wonder if i might have written it down incorrectly, with the actual proof meaning to be for the estimated value Yi hat (I don't know how to code this unfortunately). Does anyone recall this property of linear regression?

• Perhaps you meant $\beta_i E X_i$? Your question needs some context... – copper.hat May 19 '13 at 15:56
• It seems like what we are trying to show is that the expectation of a linear regression is a linear function of expectations but we may also be trying to show the proof shown on page 4 here: web.njit.edu/~wguo/Math644_2012/Math644_Chapter%201_part2.pdf – user78504 May 19 '13 at 16:06
• That proof shows the expectation of $Y_{i}$ is equivalent to the expectation of its estimate Yi hat – user78504 May 19 '13 at 16:07
• I actually think my answer is a bit more complete than the "accepted" one. – Michael Hardy May 19 '13 at 19:00

In this sort of regression problem, $X_i$ may be random in the sense that if you take another sample, all the $X_i$ values change, but one behaves as if one seeks the conditional expected value of $Y_i$ given $X_i$, so that in effect $X_i$ is treated as if it were constant rather than random. And the $\beta$s are also being treated as constant.
If $Y_i=\beta_0+\beta X_i+\epsilon_i$, where $\beta_0$ and $\beta$ are constants and $\epsilon_i$ is an "error" random variable with mean $0$, then $E(Y_i)=\beta_0+\beta E(X_i)$.
• If there is a systematic error, we would probably incorporate it in the $\beta_0$ term. Linearity of expectation would work in any case, we just use $E(\epsilon_i)$. – André Nicolas May 19 '13 at 16:15