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Let the matrices satisfy $$ \|A-A'\|<\epsilon,\quad AB=BA=I,\quad A'B'=B'A'=I, $$ then $$0=BA-B'A' = B(A-A')+(B-B')A',$$so $$B-B' = B(A'-A)B',$$ or $$d = \|B-B'\|\le \epsilon \|B\|\|B'\|\le \epsilon \|B\|(\|B\|+d),$$ hence $$d\le \frac{\epsilon \|B\|^2}{1-\epsilon \|B\|}.$$ The only condition is that your matrix norm satisifies $\|CD\|\le \|C\|\|D\|$. ...


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Take matrix $$ A = \begin{pmatrix} 10^{-n} &0 &0\\ 0 &10^{-n} &0\\ 0& 0 &10^{-n}\\ \end{pmatrix} ,\ B = \begin{pmatrix} 10^{n} &0 &0\\ 0 &10^{n} &0\\ 0& 0 &10^{n}\\ \end{pmatrix} , \\A' = \begin{pmatrix} 2\cdot10^{-n} &0 &0\\ 0 &2\cdot10^{-n} &0\\ 0& 0 &2\cdot10^{-n}\\ \end{pmatrix} $$ ...


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We know that the values of $a^*$ and $b^*$ are simply the regression coefficients (by definition) $ a^* = \frac{cov(X,Y)}{Var(X)}$ $ b^* = \bar{Y} - a*\bar{X}$ It should be straight forward to plug these values into $E[(a^{*}X + b^{*} - Y)]$ and solve for the given identity


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From the paper referenced, the cost function to minimise is $$ \ C=\sum_{u,i}c_{ui}(p_{ui}-x_u^Ty_i)^2+\lambda(\sum_u\|x_u\|^2+\sum_i\|y_i\|^2)$$ Differentiating with respect to vector $x_u$ results in (note: as the vectors $x_u$ and $y_i$ are real vectors, their scalar product is commutative, see red) $$\begin{align}\frac{\partial C}{\partial x_u} &= ...


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With the least squares method you try to 'solve' an system of linear equations $Ax = b$ for $x$, but if $A$ is not square (thats why you cannot solve exactly) the least square approach is $A^tAx = A^tb$. If you solve this straight forward this can generate huge numericl errors (everything is finde though if you solve it exactly, but Matlab can't do that.). ...


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Hint I wonder if the problem is not to minimize $$SS=(\hat{\alpha}-\alpha)^2+ (\hat{\beta}-\beta)^2+(\hat{\gamma}-\gamma)^2$$ subject to the condition $\hat{\alpha}+\hat{\beta}+\hat{\gamma}=180$. If this is the case, extract $\hat{\gamma}=180-\hat{\alpha}-\hat{\beta}$ from the constraint to get $$SS=(\hat{\alpha}-\alpha)^2+ ...


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$R^2$ is the square of the correlation between TESCO and FTSE100, so you can rely on the symmetry in the definition of correlation. Your formula for $R^2$ ought to say $R^2 = 1-\dfrac{\sum(y_i-\hat y_i)^2}{\sum(y_i-\overline y)^2}$. So we have $$ R^2 = \frac{\sum(y_i-\overline y)^2- \sum(y_i-\hat y_i)^2}{\sum(y_i-\overline y)^2}.\tag 1 $$ Now ...



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