I suspect this is not an error of lm, but rather vif (from package car). If so, I believe you have ran into perfect multicollinearity. For instance x1 <- rnorm( 100 ) x2 <- 2 * x1 y <- rnorm(...

As it has been already pointed out, ''logistic'' comes from logistic curve/function/distribution (which is underlying logistic regression). So the question is: where is logistic coming in their names? ...

First of all, note that the term ''null event'' is not unambiguous: some sources use it in a sense ''an event that has zero probability'', while others understand it as ''empty set (as an event)''. As ...

The expectation of the square of any random variable is its variance plus its expectation squared, as $\mathbb{D}^2(X)=\mathbb{E}([X-\mathbb{E}(X)]^2)=\mathbb{E}(X^2)-[\mathbb{E}(X)]^2 \Rightarrow \... View answer 9 votes Here is an approach. Consider this model: $$Y = \beta_0+\beta_1 X_1 + \beta_2 X_2 + \varepsilon.$$ Now, the marginal effect of$X_1$on$Y$($Y$'s derivative w.r.t$X_1$) is simply$\beta_1$. I can ... View answer 7 votes It might worth adding another, perhaps more straightforward example to Stephen's excellent answer. Let's consider a medical test, the result of which is normally distributed, both in sick and in ... View answer Accepted answer 7 votes The confusion comes from the fact that these conditions (that you state under the label "can be easily proven") pertain to the$Y_t=\varepsilon_t-\theta_1\varepsilon_{t-1}-\theta_2\varepsilon_{t-2}$... View answer 6 votes First of all, days are not ordinal, but nominal. (The elegant way to include them in a regression is to convert them into a factor, you haven't done that, but in this particular case it doesn't matter,... View answer Accepted answer 5 votes I don't really see the problem. For me t.test(A,B,var.equal=TRUE) and summary(aov(c(A,B)~rep(c("A","B"),each=10))) and anova(lm(c(A,B)~rep(c("A","B"),each=10))) all result in$p=0.00268$. View answer Accepted answer 5 votes If you understand a graphical technique under ''comparison'' you should probably try a QQ-plot (qqplot under R). If you are thinking of an analytical way (i.e. statistical test), the two-sample ... View answer 4 votes We have$Y^n=(aX+b)^n=\sum_{k=0}^n \binom{n}{k}(aX)^k b^{n-k}$so$\mathbb{E}Y^n=\mathbb{E}(\sum_{k=0}^n \binom{n}{k}(aX)^k b^{n-k})=\sum_{k=0}^n \binom{n}{k} b^{n-k} a^k \mathbb{E}X^k. The rest ... View answer Accepted answer 4 votes While I am not aware of anything that can be called ''standard'' bimodal distribution, in this particular case, mixture normal distribution seems to be appropriate at first glance. The pdf of such ... View answer Accepted answer 4 votes You can use, for example, two-sample Kolmogorov-Smirnov test with kstest2. (If the other distribution is also available as a sample. If it is a prespecified distribution (e.g. exponential with a ... View answer Accepted answer 4 votes It depends on whether the generator outputs independent variates or not. (I assume from your question that it outputs identically distributed variates.) If it is independent and identically ... View answer Accepted answer 3 votes There are two differences for a usual linear regression model (lm) between AIC and extractAIC: AIC accounts for the estimation of the unknown variance of the error (i.e., scale) while extractAIC does ... View answer Accepted answer 3 votes R doesn't know that your variable is nominal (which I assume it should be, if it means sites). More precisely, it knows for Var1 because using letters will make it to have a type of character which ... View answer 3 votes If both unemployment and vacancy have a trend in the long-run, then regressing one against the other would be very misleading. (Every time series with strong positive trend could be well regressed ... View answer 2 votes I have no experience on this field, but what about method of moments? We could work it out: \begin{align*} \mathbb{E}\left(Z\right)&=\mathbb{E}\left[\mathbb{E}\left(Z\mid N\right)\right]=\mathbb{... View answer Accepted answer 2 votes You're right that the requirement is\mathrm{cov}\left(x_4,x_1\right)\neq0$. The important part is that$\mathrm{cov}$doesn't care if any of the variables (or both) is continuous or categorical. You ... View answer Accepted answer 2 votes No. The sum of independent$\chi$-squares is$\chi$-squared, so$Y_i\sim\chi_1$will work, and$\chi_1\neq\mathcal{N}\left(0,1\right)$. View answer 2 votes Yes, by the law of iterated expectations (with a slight generalization):$\mathbb{E}\left(X\mid Y\right)=\mathbb{E}\left[\mathbb{E}\left(X\mid Y,Z\right)\mid Z\right]=\mathbb{E}\left[0\mid Z\right]=0$... View answer Accepted answer 2 votes Following the suggestion of @BenBolker, I've asked this on the R-INLA mailing list, where I received an extremely rapid and detailed response from Håvard Rue. Briefly, log precision should be ... View answer 2 votes Just to add one remark to @Frans Rodenburg's answer: Overadjustment might also be an issue. I.e. you don't want to control for variables for which it is not meaningful to keep them fixed when the ... View answer Accepted answer 2 votes It is fairly straightforward to test such linear constraints on coefficients with a Wald-$F$test. For example, in R, you might use the glht function of the package multcomp: lmod <- lm(Fertility ... View answer Accepted answer 2 votes If you run 100,000 simulations, you'll get a very good estimate of the true (population) value of the parameter, but no information -- at least from the simulations -- on its sampling distribution (... View answer Accepted answer 2 votes First note that this is not a clustering problem in the usual sense of the word. The ''clusters'' what you described are actually called contingency table of Diabetes and CHF. To answer your first ... View answer Accepted answer 2 votes Your procedure is correct. The rank can be very well interpreted for a single vector as well: the (row) rank will be the number of linearly independent rows. Given that a row vector has a single row ... View answer 2 votes Try x1 = -5:.02:5; x2 = -5:.02:5; instead. The problem is that the distribution is quite - but not exactly - parallel to one of your axis. If you consider the$\left\{3.6,3.8\right\} \times \left\{-...

The answer in one word: multicollinearity. If two predictors are correlated it might happen that both is insignificant itself (i.e. with $t$-test), but they are jointly significant (with $F$-test). ...