Intereting Posts

$1 + 1 + 1 +\cdots = -\frac{1}{2}$
Showing that the sequence $x_n=\frac{1}{3}x_{n-1}(4+x_{n-1}^3)$ where $x_0=-0.5$ quadratically converges
Another simple rule satisfied by the Fibonacci $n$-step constants?
$\mu$ measurable functions and separable metric spaces
Linear Transformations on Function Spaces
The group of roots of unity in the cyclotomic number field of an odd prime order
Show $p(X)$ (over a field) is irreducible iff $p(X+a)$ is irreducible
variant on Sierpinski carpet: rescue the tablecloth!
If $m \tan(\theta – \pi/6) = n \tan(\theta + 2\pi/3)$ then find $\cos 2\theta$
Two problems on number theory
$p(x)$ irreducible polynomial $\iff J=\langle p(x)\rangle$ is a maximal ideal in $K$ $\iff K/J$ is a field
Arranging books on the shelf.
Showing the inequality $|\alpha + \beta|^p \leq 2^{p-1}(|\alpha|^p + |\beta|^p)$
Universal properties (again)
The equation $X^{n} + Y^{n} = Z^{n}$ , where $ n \geq 3$ is a natural number, has no solutions at all where $X,Y,Z$ are intergers.

What I know to begin with is that the sum will be 0 if there is a y-intercept b0 , why is that? my book doesnt say and can’t figure it out. I also know that an importantant assumption for the OLS estimators to be BlUE is that x and the erros can’t be corralated otherwise the estimators would be biased. This correlation assumption may entail problems so we take one step further and we assert that by combining E(u)=0 ( an assumption that the book doesnt explain where it comes from) + E(u|x) – the average value of u doesnt depend on the value of x) , we get E(u)=0=E(u|x). so this assumption is somehow related to that the sum of residuals should be 0. But how do I mathematically prove that?

Sorry if I have been a bit unclear but I am a quite confused, it seems all this topic quite redundant to me , like of the assumptions that jusify the method are infered from the method itself..

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I have answered the other question of yours about why the OLS residuals are zero, when an intercept is included in the regression.

The stochastic assumptions on the *error term*, (not on the residuals) $E(u) = 0$ or $E(u\mid X) = 0$ assumption (depending on whether you treat the regressors as deterministic or stochastic) are in fact justified by the same action that guarantees that the OLS *residuals* will be zero: by including in the regression a constant term (“intercept”).

Consider the specification

$$y_i = b_0 + b_1x_i + u_i\,,\; E(u) = 0$$

Consider the alternative specification

$$y_i = b_1x_i + \varepsilon_i\,,\; E(\varepsilon_i) = b_0 $$

The expected value of $y$ in these two different specifications *is the same* -and it is the expected value of $y$ that we are essentially estimating in an OLS regression(see this post).

In other words, the constant term plays the role of, (or absorbs together with other constants coming from the theory behind the model) the non-zero mean of the error term, and permits us to assume safely that the “remaining” error-term has a zero mean (and this is why it is strongly advised from every corner to always run a regression including a constant term).

And then, it also gives us that the sum of OLS residuals will be zero. What more to ask from a constant?

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