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I'm assuming that Y, although unit less, has a natural ordering to it. There's no specific rule on when you MUST use a specific type of model. Really, you could even use a standard OLS regression (I would even recommend starting off with this, and then progress to more complicated models). In general, If Y in continuous (i.e can take values 2.75 or 3.45), ...


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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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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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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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If you're only solving for a and c (rather than this being a simple toy example) then you'll only need two values for both x and y. Then each set of corresponding values in x and y represent an equation in a system of equations with two unknowns. This can be solved by substitution or elimination (see here ...


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I request you to look on Google before asking a question here hope the following helps For one dependent variable Multiple Regression http://en.wikipedia.org/wiki/Coefficient_of_determination#Adjusted_R2 $$\bar R^2=1-(1-R^2)\frac{n-1}{n-p-1}$$ is the easiest one to use for this you dont actually need to perform the second Regression. however if you ...


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I have found the answer by myself. It is an "ugly" formula but it works. This is the relationship between $R^2(i)$ and $R^2$: $R^{2}(i)=1-(1-R^{2})\left(1-\frac{r_{i}^{2}}{n-k-1}\right)\frac{(n-1)SS}{(n-1)SS-n(\bar{y}-y_{i})^{2}}$ Where: $R^2(i)$ is the $R^2$ of a regression where the i-th observeations has been removed. $R^2$ is the $R^2$ of a ...


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Hint: The formula for the value of $\hat{B}=(X'X)^{-1}X'Y$ has an inverse in it, but not all matrices are invertible...


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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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Well suppose you have 9 observations of Y: 0 0 0 0 1 1 1 1 1 then the 50th percentile will be 1 and you can similarly calculate percentiles as you would any other list of numbers. Also, remember that quantile regressions assign the check-fucntion to the residual, not Y directly.


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So let's assume the linear case: $M=\gamma_1N + \gamma_2BW$ $L=\beta_1N + C$ Now plug these into your regression equation: $Time = \alpha_0 + \alpha_1M + \alpha_2L + \eta$ which yields: $Time =(\alpha_0+\alpha_2C) + N(\alpha_1\gamma_1 + \alpha_2\beta_1) + \alpha_1\gamma_2BW + \eta$ In the linear case, it is easy to see that the minimum is a function of ...


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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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It could look like $$\underset{N,BW}{\texttt{min}} \ \ \sum_{i=1}^n \left[ T_i- \alpha_1(N\cdot n_{i}+ BW\cdot b_{i})-\alpha_2(N\cdot n_{i}+C )-\eta\right]^2 $$ s.t. $N\leq N_{max}$ $BW\leq BW_{max}$ $N,BW \in \mathbb N$ n is the number of observations. $T_i$, $b_i$ and $n_i$ are the observations. The parameters $N$ and $BW$have to be determined.


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The assumption of mean 0 is a normalization that must be made because you already have a constant term in the regression. It relates to the issue of identification - that you as the researcher cannot tell the difference between the constant term in the regression and the mean of the error term. Proof: Suppose that $\epsilon$ is not mean 0 Let ...


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Here's the general idea - someone who has a better background than I do in statistics could probably give a better explanation. So you have this linear regression model: $$Y = \alpha + \beta X + \epsilon $$ where $\epsilon$ follows a normal distribution with mean $0$. What exactly does random mean? My back ground in statistics is very low level, but I ...


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The question of adjusting a line or a plane to a set of points depends on how you are looking at it. Basically, you can fit by minimizing the "vertical" distance between the line/plane and the cloud of points or by minimizing the normal distance (perpendicular to the plane). In the first case you have a linear system to solve, in the second case it ...


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You can prove it just by setting the derivative to zero. You can find an example derivation in these slides And $X^T X + \gamma I$ is always positive definite, because $X^T X$ is positive semi definite for any matrix $X$. Adding a scaled identity just increases each eigenvalue, hence the matrix is positive definite.



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