Tag Info

Hot answers tagged

3

From an engineering and practical point of view, very good models have been proposed. For temperature dependency, Vogel proposed $$\Large\eta=a~ e^{\frac{b}{T+c}}$$ (ignore $c$ if you want to have a model which can be linearized to provide you good estimates for the nonlinera regression - but, in such a case, $T$ must be absolute) and for pressure ...


2

The $\frac{1}{m}$ is to "average" the squared error over the number of components so that the number of components doesn't affect the function (see John's answer). So now the question is why there is an extra $\frac{1}{2}$. In short, it doesn't matter. The solution that minimizes $J$ as you have written it will also minimize $2J=\frac{1}{m} \sum_i ...


1

Dividing by $2m$ ensures that the cost function doesn't depend on the number of elements in the training set. This allows a better comparison across models.


1

Generally I know 2 ways to deal with problem like yours. Replace your missing variable by some "nominal one". It might be zero if it makes sense (sometimes it does not if typical $X_1$ is far from zero.) It might be an average of $X_1$ on the past or on say last week. Introduce a categorical variable $Z(X_1)$ wich is one if $X_1$ is not missing and zero ...


1

You can use logistic regression, which is typically used to generate a multivariable model for the prediction of a dichotomous outcome. Independent variables in logistic regression can be continuous or categorical. Using a "stepwise" procedure, most statistical softwares provide a model including only significant predictors. Looking at the odds ratios given ...


1

There are several methods to approach the problem of adjusting estimated regression coefficients for rounding errors in the data. Most of these techniques perform an adjustment to the main diagonal of the sample covariance matrix of the variables. One of the most commonly used method is the so-called Sheppard's correction. This is based on a Taylor expansion ...


1

You want something simple, like $\left(10 - \dfrac{A}{x+B}\right)\%$ for suitable constants $A$, $B$. If you want to go through the points $(700,5\%)$ and $(10000,9.5\%)$, choose $A = \dfrac{15500}{3}, B = \dfrac{1000}{3}$.


1

Let's work with a unit circle, centered at the origin, and suppose the "eccentric" point $E$ is at $(0,a)$, where $a<1$. Let $c$ be the length of the line $L = PQ$, where $c > 1+a$. A little vector reasoning shows that $$ Q = P + \frac{c}{ \| E - P \| }(E - P) $$ If the point $P$ is on the circle, then it can be described by coordinates $P = ...


1

A very simple equation in polar is coordinates is obtained if the center of the system of axis is not located at the center of circle, but at the "exentric" point :


1

It's not 'wrong' to use that estimator, which is the mean of the ratios given by each point. But they are not the same; the first one is the unique value of $\beta$ which minimizes $$ E(\beta) = \sum_i \left( \beta x_i - y_i \right)^2 $$ and will give a different result in the general case. In the case where all the $x_i, y_i$ can be exactly fitted with a ...


1

The estimator you define also makes sense. But like others said this is the different estimator. Actually it solves the different problem: $$ y_i/x_i = \beta +\epsilon_i $$ and minimizes the different sum of squares $\sum_i(y_i/x_i-\beta)^2$. Of course $x_i$ cannot be zero in such case. So the main idea under choosing the estimator what error exactly you ...


1

The ordinary least squares estimator, $\widehat \beta_{OLS}=\sum \frac{x_i y_i}{x_i^2}$, as others have mentioned minimizes the sum of squares error $\widehat \beta_{OLS}=argmin_b \sum(y_i -bx_i)^2$. The major reason that it is so widely used is because it is BLUE, best linear unbiased estimator (http://en.wikipedia.org/wiki/Gauss%E2%80%93Markov_theorem) ...



Only top voted, non community-wiki answers of a minimum length are eligible