# Tag Info

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The point is that among those who score as high as the 97th percentile or as low as the 16th percentile, there are some who are having an unusually high-ESP-score day (in the former case) or an unusually low-ESP-score day (in the latter). Those will regress to the mean. Tomorrow will be someone else's day to have an unusually low-score day or an unusually ...

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Well, I guess, you are saying - there is a big data set, and you divided the whole data set into multiple small parts. Then you ran linear regressions on each of these small subsets of data and calculated the correlation. But I am trying to guess the purpose. At least, I can think of one use of it you may thought of. I guess if the correlation values are too ...

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Such correlations are guaranteed if you have not standardized your predictors to take the value 0 at their means. However, this correlation is not really a mathematical/statistical problem, per se, but it may be easier to interpret the coefficients if you first standardize the variables. Therefore, the short answer is no, such a correlation is not a ...

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Standard errors affect your t-tests and F-tests and your confidence intervals. To treat the regression as violating one of the stochastic assumptions while it doesn't (i.e. use robust SE while heteroskedasticity is absent), is no less of a mistake than treating the regression as not violating the assumption while it does (i.e. use non-robust SE while ...

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The data set provided by William has a very low scatter and the points are numerous (500) and well distributed. As a consequence, the method of "regression with integral equation" gives a very accurate result (attachment : points in black, fitted curve in red)

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The regression procedure aims to find the most suitable parameters for the following problem: $$(\alpha_1,\dots,\alpha_n)\ \mbox{s.t.}\ F((x;\alpha_1,\dots,\alpha_n))\approx f(x),$$ where $F(\cdot)$ is the regression function and $f(x)$ is the function that you want to approximate. Therefore, since you shown a "perfect" linear relation between $x$ and ...

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HINT: Let's assume your model is $y=\theta x$, then the least-squares solution will be $\hat{\theta}=\min_{\theta} \|Y-\theta X\|_2^2$, where $X$ and $Y$ are column vectors made up of the data points (that you don't have, you only have some sufficient statistics that are extracted from them). We have: $\|Y-\theta X\|_2^2=(Y-\theta X)^T(Y-\theta ... 1 Yes, curve fitting and "machine learning" regression both involving approximating data with functions. Various algorithms of "machine learning" could be applied to curve fitting, but in most cases these do not have the efficiency and accuracy of more general curve fitting algorithms, finding a choice of parameters for a mathematical model which gives "best ... 1 From the least squares estimation method, we know that $$\hat{\beta}=(X'X)^{-1}X'Y$$ and that$\hat{\beta}$is an unbiased estimator of$\beta$, i.e$E[\hat{\beta}]=\beta$. Moreover, the linear model $$$$Y=X\beta +u$$$$ has the assumption that $$Y\sim N(\mu=\beta_0+\beta_1x,\sigma)$$ or equivalently that$u \sim N(\mu=0,\sigma)\$. ...

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As you can see in this link (page 5, col.2), Hoerl (presumably the inventor of ridge regression) "gave the name "ridge regression" to his procedure because of the similarity of its mathematics to methods he used earlier i.e., "ridge analysis," for graphically depicting the characteristics of second order response surface equations in many predictor ...

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