Stack Exchange Network

Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange

New answers tagged

0

Apart from endorsing Frank's comment that binning is seldom a good idea and certainly not to this extent the most likely explanation, as mentioned by Kjetil in a comment is that the model suffers from separation. For some combination of levels the outcome variable is all zero or all one. A typical symptom of this is the coefficients trying got go off to ...


1

If we focus on the primary purpose of a regression model as summarizing mean differences, it's easy to see there's no such thing as an omitted variable bias. In fact, some causal schools of thought (such as the one to which I belong) think that so long as there is any residual standard error, then variables have been omitted (no such thing as inherent ...


2

You're right that the requirement is $\mathrm{cov}\left(x_4,x_1\right)\neq0$. The important part is that $\mathrm{cov}$ doesn't care if any of the variables (or both) is continuous or categorical. You can calculate it, irrespectively of what is their nature! Concretely in your case: if $x_1$ is binary, and $x_4$ is continuous then $\mathrm{cov}\left(x_4,...


2

Whether a variable is binary or continuous has no bearing on its potential for omitted variable bias (unless it's a poorly measured version of a truly continuous variable). If the "true" model uses the binary variables as they are, then the same rules that apply for continuous predictors and continuous omitted variables apply here. Regression models don't ...


1

The following bullet points are correct, and equivalent: Keeping all other predictors constant then, the log odd ratio of survival for having an additional sibling decreases by 0.38 units (what does it mean?) Keeping all other predictors constant then, the odd ratio of survival for having an additional sibling is 0.68 times lower (less likely) To ...


0

The odds ratio is the multiplicative factor that converts one odds to another. For example, if one passenger's odds of survival is $1$ (i.e., a $50\%$ probability of survival), and another passenger has all the same covariate values as the first except that they have one more sibling, then the second passenger's odds of survival is $1\times 0.68 = 0.68$ (i....


0

The coefficient of determination is defined for the model as whole and not for individual variables. However, there is a technique called ANOVA which can roughly be thought of as breaking $R^2$ into contributions from each variable. Recall that the coefficient of determination is defined in terms of the sums of squares of residuals: $$ \begin{align} R^...


-1

The coefficient of determination is defined as $$R^2 = 1 - \frac{\sum(y_i - \hat{y_i})^2}{\sum(y_i - \bar{y_i})^2}$$ If I needed to figure out each weight's effect on the coefficient of determination, the most straightforward approach would be taking partial derivative w.r.t. the weights. $$\frac{\partial{(R^2)}}{\partial{w_i}}$$


1

We use the bootstrap when we want to know how some function will behave when applied to different datasets drawn from a particular population. Obtaining new datasets is expensive, so instead the bootstrap pretends that the dataset you have is the population, and simulates new sampled datasets by resampling from the dataset you have. Therefore, you want your ...


0

If the dataset is large, it shouldn't be a problem. In fact, if your dataset is large, repeated observations may even help you to get a better idea of the underlying true distributions. It is worth noting that even with the regular bootstrap, sooner or later there are bound to be repeated obsevations in the bootstraped dataset anyway.


1

Assuming the errors are i.i.d $N(0,\sigma^2)$ with $\sigma$ unknown. The mean response at $x=x_0$ is $$E(y\mid x_0)=\beta_0+\beta_1x_0$$ We estimate $E(y\mid x_0)$ from the fitted model by $$\hat\mu_{y\mid x_0}=\hat\beta_0+\hat\beta_1x_0$$ Show that \begin{align} \operatorname{Var}(\hat\mu_{y\mid x_0})&=\operatorname{Var}(\hat\beta_0)+x_0^2\...


0

The second difference is analogous to the second derivative of a function which, broadly, tells you how "spiky" a function is. The model you're specifying is basically regressing that function on covariates. I.e. how covariate influence the spikiness of the time series.


1

The property used here is that $\mathbb{Var}(\text{const}+X) = \mathbb{Var}(X)$. If you expand the terms out formally, from the second step to the third, you get: $$\begin{equation} \begin{aligned} \mathbb{Var} (\hat{\beta}_1) &= \mathbb{Var} \left(\frac{\sum_i (x_i - \bar{x})(\beta_0 + \beta_1 x_i + u_i )}{\sum_i (x_i - \bar{x})^2} \right) \\[6pt] &...


1

If $X$ and $Y$ are independent random variables, then $$ \textrm{Var}(X+Y)=\textrm{Var} X+\textrm{Var} Y.\qquad (1) $$ This is because $$ \mathbb E\left[(X+Y)^2\right]=\mathbb E[X^2]+2\mathbb E[XY]+\mathbb E[Y^2]\overset{indep!}=\mathbb E[X^2]+2\mathbb EX\mathbb EY+\mathbb E[Y^2]=\left(\mathbb EX+\mathbb EY\right)^2. $$ Similarly if $a$ is deterministic and $...


0

In a regression model with a binary independent variable, the coefficient is just the difference between the means of the two levels. In fact: If you have a linear regression model with only a binary categorical variable, this is equivalent to a two-sample t-test assuming equal variances. Switching the reference level therefore just switches the sign of the ...


0

You cannot compare the unstandardized coefficients. If you standardize the raw scores (i.e. convert to z-score) before calculating the correlation coefficients, they will be comparable. This assumes that the raw scores are continuous or pseudo-continuous.


1

Interpretation of $\alpha$ : one percent change in $X_{it}$ is associated with $\alpha * \ln\left(\frac{101}{100}\right)$ change in $Y_{it}$. Interpretation of $\beta$ : an increase of one-unit in the Height would result in $\left(e^\beta - 1\right)*100\%$ change in $Y_{it}$. For more information, you can see this document


5

Yes. The estimated standard error gives the analyst (or reader) an idea of how precise the parameter estimate (estimated coefficient/slope) is: the larger the standard error, the less precise the estimate. To help you see this, recall that the standard errors are directly tied to confidence intervals of parameter estimates in simple multiple regression by ...


1

Given that your outcome/depended variable is a proportion between zero and one, it would be more appropriate to fit a Beta mixed effects model. The linear mixed model may give results/predictions outside the permissible (0, 1) interval. A Beta mixed model can be, for instance, estimated with the GLMMadaptive package; for example, check here.


0

I've found the answer to my own question, in case it'll be useful to someone I post it here. Once the HC regression has been fitted, I estimated a regression line separately for each individual of the other group. I then used a F statistic to test, all together, whether the linear combination of the parameters of the new regressions are statistically ...


-1

The the house price of having lets say: Neighborhood Age 44-64% is 0.008 more than having Neighborhood Age >64%. Take note, that when adding dummy variables in a linear model: $y = \beta_0 + \beta_1X_1 + \beta_2X_2 + \beta_3D $ where $D$ is a dummy variable, can be re-written into : $y = \beta_0 + \beta_1X_1 + \beta_2X_2 + \beta_3 $ when D = 1 $y = \...


Top 50 recent answers are included