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One way to write the data generating mechanism for logistic regression is as follows
$$ \mbox{logit}(p) = X\beta $$
$$ y \sim \mbox{Binomial}(n , p) $$
From this formulation, we find that the linearity assumption is made on the log odds scale. So were we to plot the log odds of the outcome versus the predictor, we would see a straight line$^{1.}$
$^{1.}$ ...
6
A few things to start with. I agree that nature is rarely linear, but I disagree that the approach you've taken is a reasonable attempt to account for this. Polynomial terms do add non-linearity, but they are high bias models (because we can only estimate functions in the set of polynomials). A better approach to adjusting for non-linearity would be use a ...
6
One troubling situation is when the regressor variables are not independent from each other. This can make it look like there is seemingly (causal) relationship that is not truly present.
Example:
Say we have a model where $y$ is a function of $x$ described by an exponential term $E[y|x] = e^{0.6x}$ and some Gaussian noise with deviation $\sigma = 0.1$
$$y \...
5
It is not true in general that an insignificant variable has no effect on the response. A variable can be insignificant because the sample size is too low or the random variation too large to find a clear significant effect even if an effect in fact exists, or because it is correlated with other variables and the data cannot know how much of the effect of ...
4
As far as I understand, the assumption of Heteroscedasticity and Linearity are violated. As far as I understand, the first causes lower p values and imprecise coefficients, the second causes wrong coefficients?
If the "real" relationship isn't linear, then the linear coefficients aren't merely the "wrong coefficients", there aren't any ...
3
On the question: "Linearity assumption violated - can I still draw conclusions from my model?", my answer is a limited yes!
You may chose to verify my claim by constructing a simple non-linear model and assuming normal error generating model, project out. Further, fit a linear version of the model. I claim the Least-Square theoretical derived ...
3
There is no need to Z-transform a continuous predictor for standard types of regression models.* Some prefer to center such predictors about their mean values when there are interactions, as it puts the "main effect" coefficients into a range that is appropriate to the data. Otherwise, for example, a "main effect" coefficient for a ...
2
Isn't it divided by $\sqrt{(X^TX)^{-1}}$?
$\sqrt{(X^TX)^{-1}}$ is not a single number.
This is because $X$ is a $n$ by $p$ matrix and $X^T X$ will be a $p$ by $p$ matrix of which you need the $j$-th diagonal element.
But I am confused that why $$ \frac{N(0, (X^TX)^{-1})}{\sqrt{(X^TX)^{-1}_{jj}}}\sim N(0,I)?$$
This is not the case.
Yes, you do have that ...
1
When you select variables using your "crude" method, this amounts to placing a prior on your coefficients that says that one specific subset of coefficients is (a priori) certainly equal to 0, and the others are (a priori) entirely unknown (i.e. a uniform prior).
So the question is: is this a better prior than the Laplace prior that is implied by a ...
1
Their latent growth model is piecewise linear:
it has an intercept (level at week 1), a linear slope before the midterms,
and a linear slope after the midterms. Using latent growth structural equation model, rather than simple piecewise linear regression, makes it easy to explicitly model additional parameters of interest, such as the correlation between the ...
1
This is what in Econometrics goes under the name of "errors in variables model": see for instance the Wikipedia. The important point to bear in mind is that you may experience inconsistency if the errors in the variables are correlated iwth the regression disturbance. Unless this is the case, I think you can safely take a representative value of ...
1
The $ \beta_{i} $ elements are the correlation coefficients between $ {X}_{i} $ to all other elements.
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