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What does mean response in linear regession mean? I don't understand the definition given in wikipedia. This is the definition:

Mean response is an estimate of the mean of the $y$ population associated with $x_d$, that is $E(y \mid x_d)=\hat{y}_d$.

Also, Why is the variance predicted response different from mean response?

Thanks!

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The use of the word "response" originates to the application of statistics in medicine: we apply a "treatment" and the patients' bodies "respond to the treatment". In a regression setting the "treatment" would be represented by the regressors - and the dependent variable would measure the "response" to this "treatment". And in fact what we quantify by the regression technique is the mean response. We specify a model $$ y_i = \mathbf x_i'\mathbf \beta + u_i\,,\; i=1,\ldots,n$$ Now take the expected value of the dependent variable conditional on the regressors:

$$ E(y_i\mid\mathbf x_i') = E(\mathbf x_i'\mathbf \beta\mid\mathbf x_i') + E(u_i \mid\mathbf x_i')$$ Given the assumption of strict exogeneity of regressors w.r.t the error term, the last term equals zero. Also, in the second term the regressor vector goes out of the expected value since we condition on it, and $\beta$ is considered a constant (in the classical tradition). So :

$$ E(y_i\mid\mathbf x_i') = \mathbf x_i'\mathbf \beta$$

Since $\beta$ is unknown we estimate it, and we arrive at $$ \hat y_i = \widehat E(y_i\mid\mathbf x_i') = \mathbf x_i' \hat \beta$$

Even if we knew $\beta$, we would not be able to "calculate"/predict $y$ exactly, because there exists the error term/stochastic-unpredictable disturbance. So, even with known $\beta$ we would be able to obtain just the exact conditional expected value (mean value) of $y$.

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  • $\begingroup$ I am a bit struggling to understand conceptually the second last step, in particular the step from $E(x_i ∣ x_i) = x_i$. For instance, in the case of the residuals, $E(\mu_i ∣ x_i)$, we would check e.g. for each subject i in the data what the value of the corresponding residual is, and, over all subjects this should have a mean of zero. Using the same logic, we would check for all subjects the value of $x_i$ given $x_i$, which just is $x_i$, and so shouldn't $E(x_i ∣ x_i)$ be equal to $E(x_i)$? $\endgroup$ – Pegah Sep 27 '18 at 12:40
  • $\begingroup$ @Pegah: Think intuitively: "What is value we expect that $x$ takes, given $x$?" Is it equal to the average value of $x$ or is it equal to the actual value of $x$, which we know since we condition on it? $\endgroup$ – Alecos Papadopoulos Sep 27 '18 at 13:27
  • $\begingroup$ Thank you for the quick reply. The way the quantors are used is still not completely transparent to me - we don't really condition on a particular x, but on all $x_i$ quantified over i. So I understand the $E(x ∣ x) = x$, but not the $E(x_i ∣ x_i) = x_i$, since for the latter, first the i's are quantified over 1,..,n, - the resulting $x_i$ is however then a fixed i for the respective response $y_i$ - do you know what I mean? $\endgroup$ – Pegah Sep 28 '18 at 13:04

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