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(since I can't add a comment I will repost a answer I posted here) There is a simple proof that requires only linear algebra. First notice that if you take $\beta=(\beta_0,\beta_1,\ldots,\beta_{n−1})^T$ and the matrix $X=[\mathbb 1,x_1,x_2,\ldots,x_{n−1}]$, where $\mathbb 1=(1,1,\ldots,1)^T$ and $x_i=(x_{i1},\ldots,x_{i(n−1)})^T$, you can write the model ...


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It is easier to expand the brackets first \begin{align} \mbox{RSS}(b) =& (y-Xb)'(y-Xb) = y'y - y'Xb-b'X'y+b'X'Xb\\ =& y'y-2b'X'y+b'X'Xb \end{align} then \begin{align} \frac{\partial}{\partial b}\mbox{RSS}(\hat{b}) = -2X'y+2X'X\hat{b} = 0 \end{align} by rearranging the equation you get $$ X'X\hat{b}=X'y. $$ Assuming no (complete) ...


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You probably meant that a $\textit{pair}$ of independent variables are highly correlated ($0.9$). Basically, although technically it's fine, this may indicate that both variables measure (almost) the same factor. Hence, generally speaking, bring the same information to the model, as such may cause redundancy. The fact that both variables are significant may ...


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Probabilities are restricted to being in the interval $[0,1]$, which means they can lead to unacceptable results if used in extrapolating linear regression, which can suggest answers in the range $(-\infty, +\infty)$ Meanwhile, suppose you want to say that doubling the probability of some outcome from $1\%$ to $2\%$ is roughly as substantial an effect as ...


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You know all the values $F_i = F(x_i)$, $f_i = f(x_i)$ and $g_i=g(x_i)$ in the model $F_i = Af_i + Bg_i$ so we can find an estimate for $A$ and $B$ by least-squares fitting. That is we try to find the values of $A,B$ that minimize the sum of squared errors $$S =\sum_i (F_i - A f_i - Bg_i)^2$$ The relevant equations / the way to derive them can be found on ...


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Actually, $\hat{\beta}_0$ isn't constant - it depends on the $y_i$ (which follow a normal distribution). Note that in most cases, this variance would be likely computed for a prediction interval since you're working with a new $x$-value. Using properties of variances and covariances, $$\begin{align} \newcommand{\Var}[1]{\text{Var}\left(#1\right)}\newcommand{...



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