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1

It suffices to show that $X^{T}X (X^{T}X)^{g^{T}}$ is the identity matrix, which is true because you can interchange the generalized inverse and matrix transpose. Then the result follows by noticing that $X^{T}X$ is a symmetric matrix.


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The data $(x_i,y_i)$ are given, so although it looks like you have a quadratic because of the $x_i^2$, in fact this is just a constant. You're solving for $\alpha_0, \alpha_1$ and $\alpha_2$, and the equation is linear in these terms.


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It is not true that $J(I-H)=0$ unless the rows of $J$, each of which is a row of $1$s, are in the row space of $X'$. That will certainly happen if one of the columns of $X$ is a column of $1$s, as happens in typical multiple regression problems. If you're not using the fact that one of the columns of $X$ is a column of $1$s, or failing that, that a column ...


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As Paul said, the model is totally nonlinear with respect to its parameters and then nonlinear regression will be required. However, admit for the time being that you know $a_4$. So, you can write the model $$y -a_4 = \exp(a_1 \log (x) + a_2 \log (z) + a_3)$$ $$\log(y -a_4) = {a_1 \log (x) + a_2 \log (z) + a_3}$$ So, for given value of $a_4$ you can compute ...


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$$ e^{(a1 log (x) + a2 log (z) + a3)} + a_4$$ $$ =e^{(log (x^{a_1}) + log (z^{a_2}) + a_3)} + a_4$$ $$ =e^{(log (x^{a_1})}e^{+ log (z^{a_2})}e^{a_3)} + a_4$$ $$=x^{a_1}z^{a_2}e^{a_3}+a_4$$ Without $a_4$, you could do a linear regression in the log-transformed variables, but with it there is no way. So it would have to be a non-linear regression.


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Here are some matrix algebra equations you can try. I'm suggesting this approach because you mentioned that there may be future cases where you have more than three points to work with and in such cases the linear regression solution is what you want. Starting with your equation for $s$, define the following quantities: $$\underline s = \begin{bmatrix} ...


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The equation for the elasticity is: $e=\frac{dy}{dx}\frac{x}{y}$ In your case, we have $ln(y)=5-0.1x$, Now take the derivative of both sides and multiply by x: $\frac{dy}{dx}\frac{x}{y} = -0.1x$ Therefore, at $x=10$, the elasticity is $-1$.


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Do know about log-log graph and semi-log graph? In log-log graph, both x and y axes are labeled as x and y but log(x) vs log(y) are plotted. They label x and y on the axes but scales are changed from 0,1,2,3... to 1,10,100,1000,.... So in your case, if you label $x_1$ on the x-axis, then use normal scaling 0,1,2,3,... and write on side of the graph about ...


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I think we had some interaction on Stack Overflow a few days ago regarding F-stats. The concept of the F-test is fairly widespread in statistics, so it is not an innovation by Michael Newville (lmfit developer). Basically there are two types of F-tests to perform. Both of which deal with comparing two models, because an F is simply a ratio of chi-squares ...


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You are looking for an equation $y=mx+c$. Ideally it would pass through all of the given points: that is, you would have $-6m+c = -1$, and similarly for other points. This is a set of four equations with two unknowns $m,c$. Its matrix representation is $$ \begin{pmatrix} -6 & 1\\ -2 & 1 \\ 1 & 1 \\ 7 & 1 \end{pmatrix} \begin{pmatrix} m \\ c ...


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Rename $\beta \cdot \gamma = \delta$ and you have a regression in $\alpha$ and $\delta$ with $\epsilon_i$ denoting the model error.


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The model you want to fit on the basis of $p$ data points $(x_i,y_i)$ is $$y=ax+b\sin(x)$$ To make life easier, define $z_i=\sin(x_i)$ so the model becomes $$y=ax+bz$$ which corresponds to a multilinear regression with no intercept. What you want to minimize is $$F=\sum_{i=1}^p n_i(a x_i+b z_i-y_i)^2$$ So, the derivatives are $$\frac{dF}{da}=2\sum_{i=1}^p ...



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