New answers tagged

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| ...


0

If the least-squares line is $y = ax + b$ then for any value of $x$, the value $ax+b$ is an estimate of the average of all $y$-values for members of the populations that have the specified $x$-value.


0

IMHO, the "actual meaning" is not a mathematical question. I.e., if you understand the technical aspects of the changes in the coefficients, then anything else is just kind of philosophy. Namely, in a classical regression analysis you assume that the "real" underlying model that explains a poverty rate ($Y$) is the GDP ($X$) that is given by $Y = \beta_0 ...


2

$$\left(\sum x_i\right)^2 = \sum x_i^2 + 2\sum_{i<j}x_i x_j$$


2

$$\sum_{i=1}^n x_i^2=x_1^2+x_2^2+\cdots+x_n^2$$ $$\Big(\sum_{i=1}^n x_i\Big)^2=(x_1+x_2+\cdots+x_n)^2$$


1

For example, let $x_i$ represent $i^{th}$ natural number, and n = 4. Then $\left(\sum_{i=1}^n x_i\right)^2$ = (1 + 2 + 3 + 4)^2 = (10)^2 = 100 and $\left(\sum_{i=1}^n x_i^2\right)$ = (1^2 + 2^2 + 3^2 + 4^2) = (1 + 4 + 9 + 16) = 30.


1

Let us suppose that you have $n$ data points $(A_i,B_i,I_i)$ ($i=1,\cdots,n)$ and you want to find the "best" $x$ and $y$ for matching the model $$I=x A+y B$$ the genral method, as already answered by G-Cam, is ordinary least square which consist in the minimization of $$F=\sum_{i=1}^n \big(x A_i+y B_i-I_i\big)^2$$ Computing the derivatives and setting them ...


1

You are looking for a technique called "Least-squares regression". This method allows for overdetermined systems (more equations than variables) to be fit to the polynomial that minimizes the sum of the distances (actually the sum of the squared distances) from each point to that polynomial. You specifically want the matrix version that looks like this ...


0

Matrix multiplications is indeed of complexity $\mathcal{O}(n^3)$ (well, for school book implementations). It is cubic when we multiply two matrices together. However, in my personal opinion, it is quite rare to perform this operation in machine learning algorithms, hence perhaps that is why we rarely hear people blaming the complexity of algorihms on matrix ...


1

Your interpretation of a prediction interval is incorrect. A 90% prediction interval will contain 90% of the probability of the true underlying distribution on average (not always nor "at a minimum"). What you are thinking about is a less-taught concept: a Tolerance Interval. A tolerance interval is specified by a confidence and a coverage: The coverage is ...


0

During typical linear regression, our goal is to chose $\alpha,\beta$ to minimize the objective function $$J(\alpha,\beta) = \sum_{i=1}^n (y_i - (\alpha + \beta x_i))^2$$ (the ordinary least square). This corresponds to the equation $y = \beta x + \alpha$ which will best fit the data set. An affine transformation of $Z$, means we should replace $y_i$ by ...


1

Let $\eta_j(i)$ denote the $i$th element of vector $\eta_j$ with $j =1,2$. By definition, we have $\text{cov}(\eta_1,\eta_2) = \left( \begin{array}{cc} cov(\eta_1(1),\eta_2(1)) & cov(\eta_1(1),\eta_2(2)) \\ cov(\eta_1(2),\eta_2(1)) & cov(\eta_1(2),\eta_2(2)) \end{array} \right),$ where direct substitution gives $\text{cov}(\eta_1,\eta_2) = ...


1

Yes, it is meaningful to imagine the regressors are sampled from a theoretical distribution. The quantity $P(X_1=0)$ would then refer to the theoretical distribution for $X_1$, not to the observed values $x_{1,1},\ldots,x_{1,p}$ of $X_1$. Note that in elementary treatments of regression we assume the regressors are non-random, i.e., they are known constants ...


-1

Pedro, => partial fractions $$\log(1 - \frac{a}{b})$$ $$1 - \frac{a}{b} = \frac{b}{b} - \frac{a}{b} = \frac{b-a}{b},$$ $$\log(1 - \frac{a}{b}) = \log(\frac{b-a}{b}) = \log(b-a) - \log(b)$$


2

The model being $$y=a+bx+\frac c {\sqrt x +d}$$ it is nonlinear with respect to the parameters (because of $d$) and adjusting coefficients $a,b,c,d$ will require nonlinear regression and this will also require some reasonables estimates of the parameters for starting it. What you could notice is that, for a fixed value of $d$, the model is linear. For a ...


0

\begin{align} cov(\eta_2, \eta_1) &= \mathbb{E} \eta_2^T \eta_1 - \mathbb{E}\eta_2\mathbb{E}\eta_1\\ &=\mathbb{E} \begin{bmatrix}\xi_1^2 + \xi_1\xi_2,&\xi_1\xi_2\\ \xi_1^2 - \xi_2^2,&\xi_1\xi_2-\xi_2^2 \end{bmatrix}\\ &=\begin{bmatrix}1,&0\\-1,&-2 \end{bmatrix} \end{align}


1

For a coxph object (the Cox regression) you can use the extractAIC command from the stats package. For more details see here.


1

A few more ideas, in addition to the ones you mentioned: Direction. Fit a straight line to the curve, and then look at the direction of the line. Direction can be expressed by two angles. Moments of inertia. These measure the "distribution" of matter in the curves, in some vague sense. Some things with bounding boxes. For example, the ratio of the ...


1

The studentized residuals are $$t_i=\frac{\epsilon_i}{\hat\sigma\sqrt{1-h_{ii}}}$$ Where $\epsilon_i$ is the residual, $h_{ii}$ the leverage and $\hat\sigma$ is the estimate of the standard deviation of residuals, that is $$\hat\sigma^2=\frac1{n-m}\sum_{i=1}^n\epsilon_i^2$$ Where $n$ is the number of observations (here $4$) and $m$ the number of ...


0

The model being $$Y=\frac{Ke^{-\theta s}}{s(\tau^2s^2+2\zeta\tau s + 1)}$$ it is obvious that nonlinear regression is required. But this implies to have some réasonable estimates fo parameters $K,\theta,\tau,\zeta$. To me, the first step would be to consider that $\theta$ is given a value. Now, rewrite the model as $$\frac {e^{-\theta s}}{s Y}=\frac ...


1

$Cov(X,Y) = \sigma_X\sigma_Y\rho_{X,Y}.$ Where $\rho$ is the correlation. Usually, you see this written the other way round to define correlation. $\beta_{X/Y} = \frac{\sigma_X}{\sigma_X}\rho_{X,Y}$ $\beta_{A/B}\beta_{B/C} = \frac{\sigma_A}{\sigma_C}\rho_{A,B}\rho_{B,C}$ $\beta_{A/B}\beta_{B/C} = \beta_{A/C}$ would imply that $\rho_{A,B}\rho_{B,C} = ...


1

It does not seem to be correct. Let $A,B,C$ be random variables such that $P(A=1)=P(A=0)=P(A=-1)=1/3$, $C=A$, and $B=A^2$. Observe that $E[A]=E[C]=0$ and $E[B]=2/3$. Then, letting $\rho$ denote the covariance, $$ \rho_{A,C} = Var(A)= \frac{1}{3}(1-0)^2+\frac{1}{3}(0-0)^2+\frac{1}{3}(1-0)^2=\frac{2}{3} $$ On the other hand, $$ \rho_{A,B} = \rho_{B,C} = ...


0

I used to have the same question. You have to divide by the unit you want. So, your slope is based in USD/sq ft. And the same for the input (sq ft or $$ft^2$$). One simple workaround is to divide both slope and input by 0.092903 ( 1 $$ft^2 = 0.092903 m^2 $$) There you are. You have converted sq ft to sq m for the input, and USD/sq ft to USD/sq m for the ...


0

To check the definition, you want to bring $\widehat β_1$ to the form $$\widehat β_1=c_1y_1+\dots+c_ny_n$$ So, to simplify notation, let $d_i:=\dfrac{1}{n}\dfrac{1}{x_i-\bar x}$ for $i=1,2,\dots,n$ and write \begin{align}\hat β_1&=\frac{1}{n}\sum_{i=1}^n\frac{y_i-\bar y}{x_i-\bar x}=\sum_{i=1}^nd_i(y_i-\bar ...


1

If $z=y'+y''$, $$\frac{1}{n}\sum_{i=0}^n \frac{z_i-\bar{z}}{x_i-\bar{x}}= \frac{1}{n}\sum_{i=0}^n \frac{(y'_i+y''_i)-\overline{y'+y''}}{x_i-\bar{x}}= \hat\beta'_1+\hat\beta''_1 $$ If $z=\lambda y$, $$\frac{1}{n}\sum_{i=0}^n \frac{z_i-\bar{z}}{x_i-\bar{x}}= \frac{1}{n}\sum_{i=0}^n \frac{\lambda y_i-\lambda \bar{y}}{x_i-\bar{x}}= \lambda\hat{\beta_1} $$ ...



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