# Tag Info

0

Suppose that we have $n$ observations $(x_1,y_1),\ldots,(x_n,y_n)$ from a simple linear regression $$Y_i=\alpha+\beta x_i+\varepsilon_i,$$ where $i=1,\ldots,n$. Let us denote $\hat y_i=\hat\alpha+\hat\beta x_i$ for $i=1,\ldots,n$, where $\hat\alpha$ and $\hat\beta$ are the ordinary least squares estimators of the parameters $\alpha$ and $\beta$. The ...

1

Approximation: As you have observed the problem is just fulfilling the sign constraints. The partial derivatives are: $\frac{\partial f}{\partial x}=cbx^{c-1}$ and $\frac{\partial f}{\partial y}=dey^{d-1}$. Since $x$ and $y$ are positive the derivatives are positive if $cb > 0$ and $de>0$. To make sure that $f>0$ as well, you actually need all ...

1

It turns out there is some literature available on this subject that I didn't see before. One trick is to first fit a cylinder to the data, then assume the helix lies on that cylinder. For all future helix-fitters out there, check out these links. http://www.geometrictools.com/Documentation/HelixFitting.pdf ...

0

1) By definition, a system of the form $x_{k+1} = A x_k + B u_k + c$ is called "affine linear". As said by A.G., one can find a steady state solution $\bar{x} = A \bar{x} + B \bar{u} + c$ and create a new system with dynamics $\tilde{x}_{k+1} = A \tilde{x}_k + B \tilde{u}_k$ where $\tilde{x}_k = x_k - \bar{x}$ and $\tilde{u}_k = u_k - \bar{u}$. ...

0

So you want to fit your given points with the curve $f(x) = a + b\,x + c\,x^{\,2}$. There are many "fitting" methods to choose from depending on considering the nature of the data and of the underlying error. I suppose you refer to the most common "least square" method (in the most common acception which assumes that the error is just on the $y$ ...

0

The sample correlation $r$ can take values in $(-1,1)$, where negative values indicate negative linear association between the two data vectors and positive values indicate positive linear association. The sample coefficient of determination $r^2$ takes values in $(0, 1),$ where larger values indicate increasing linear association. (The Wikipedia article on ...

3

We can solve this using the method of least squares: $$\overline X = \frac{\sum_{n=1}^{8}x_i}{8} = 2011.5\\\overline Y = \frac{\sum_{n=1}^{8}y_i}{8} = 974.375$$ The equation of the line of best fit is $y= mx + b$ where $$m = \frac{\sum_{n=1}^{8}(x_i-\overline X)(y_i-\overline Y)}{\sum_{n=1}^{8}(x_i-\overline X)^2}\approx 28.1622024$$ and $$b = \overline Y ... 0 A good way to do this is through Excel. We have your data:$$\begin{array}{c|c} \text{Year} & \text{Hats}\\ \hline 2008 & 875\\ 2009 & 923\\ 2010 & 913\\ 2011 & 958\\ 2012 & 1023\\ 2013 & 978\\ 2014 & 1015\\ 2015 & 1110\\ \end{array}$$We can plot this data and retreive a linear function:$$y = 29.655x - 58678$$We ... 0 Put \mathbf{X}_i's next to each other to form a matrix. Oh, in fact you did. Yes, it is X. Notice that$$\mathbf{X}_i^{\top}\mathbf{\Omega}^{-1}\mathbf{X}_i$$is the i'th element on the diagonal of$$\mathbf{X}^{\top}\mathbf{\Omega}^{-1}\mathbf{X}$$So, since you're summing these up,$$ \sum_{i=1}^N ...

0

It is $\sum (x_i-\overline x)^2$ First you have multiply out the brackets: $\sum (x_i^2- 2\left( x_i\right)\overline x+\overline x ^2)$ Each summand in the brackets gets a sigma sign. The constants $2, \overline x$ and $\overline x^2$ can factored out. $= \sum x_i^2-2\overline x\sum x_i+\overline x^2\sum 1$ $\sum x_i=n\cdot \overline x$ and $\sum 1=n$ ...

0

If $y_i=\alpha + \epsilon_i$ and $var(\epsilon_i) = \sigma^2x_i$, then in order to estimate $\alpha$ properly (i.e., in order to satisfy the iid assumption of the noise terms), you should estimate GLS which is WLS in this case. Namely, $y_i^* = y_i/x_i, \quad x_i \neq 0$. Then your model becomes $$y_i^* = \frac{y_i}{x_i} = \frac{\alpha}{x_i} + ... 2 You may observe that$$ \hat{\beta}_1 =\frac{\sum_{i=1}^n x_i y_i - n \bar{x}\bar{y}}{\sum_{i=1}^n x_i^2 -n\bar{x}^2} =\frac{\frac1n\sum_{i=1}^n x_i y_i - \bar{x}\bar{y}}{\frac1n\sum_{i=1}^n x_i^2 -\bar{x}^2}.\tag1 $$Then you may prove that$$ \frac1n\sum_{i=1}^n \left(x_i -\bar{x}\right)^2=\frac1n\sum_{i=1}^n x_i^2 -\bar{x}^2 \tag2 $$and that$$ ...

0

The covariance is $s_{xy} = \frac{\sum (x_i - \bar x)(y_i - \bar y)}{n-1},$ where the sum is taken over $i = 1, \dots, n$ and $n$ is the sample size. Then the correlation is $r_{xy} = \frac{s_{x,y}}{s_x s_y},$ where $s_x$ and $x_y$ are the two standard deviations. If you have the regression line y = 13.555 -0.1688842 x. then you might say (over the ...

0

In order to find the intersection point of a set of lines, we calculate the point with minimum distance to them. Each line is defined by an origin ${a}_{i}$ and a unit direction vector, ${n}_{i}$. The square of the distance from a point $p$ to one of the lines is given from Pythagoras:  d_{i}^{2}={{\left[ \left| \left| p-{{a}_{i}} \right| \right| ...

Top 50 recent answers are included