# Tag Info

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A nice treatment of linear models with vector-valued outcomes is given in John Fox's notes. The competing approaches are equation-by-equation OLS and Zellner's SUR (FGLS) estimator. In special cases (in particular, for one that applies to you, the case of same explanatory variables in each equation), there is no gain in joint estimation over ...

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"This scenario" possibly refers to one of two things: a test of whether Y is associated with X (with no assumptions of their functional relationship) or a test of whether that relationship is linear. The Bayesian framework is much better for arguing in favor of one model versus another, so this seems adequately suited to the latter problem. It's worth ...

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Since GAM are nonparametric models, you will need to look at the literature on Bayesian Nonparametrics in order to find analogous models. This theory might be slightly more difficult to digest, though, given that the priors have to set on infinite dimensional spaces. If you are brave enough to dig into this area, I would recommend the following book as a ...

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Regarding mixed-effects models, in my opinion, the best applied book is: Fitzmaurice, G.M., Laird, N.M., & Ware J.H. (2011). Applied Longitudinal Analysis. Wiley. For more on fitting them with different software, West, B.T., Welch, K.B., & Galecki, A.T. (2006). Linear Mixed Models: A Practical Guide Using Statistical Software. Chapman ...

2

You say that you'd calculate the slope as follows: So normally you would calculate $$S_{XX} = \sum_i (x_i-\bar x)^2\\ S_{XY} = \sum_i ((x_i-\bar x)(y_i-\bar y))\\ S_{YY} = \sum(y_i-\bar y)^2$$ Then $b_2 = S_{XY}/S_{XX}$. So imagine you have a set of x-values and y-values: y x 1 2.3 0.36772 2 5.3 1.64873 3 6.5 7.38910 Step 1: ...

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It's fine to have both in the model and quite common for non-linear relationships. Essentially, when you take logs you're looking at the proportional change in the variable rather than the level. In your model 2 the $\beta_2$ coefficient tells you the levels change in $Y$ for a proportional change in $X$. For example, $\beta_2 = 2$ means that a 1% change in ...

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The red decay is an exponential function with negative exponent (as you might discern from sashkello's link, which I highly recommend). The green is the complement of that (they add to 1). That is, the green curve will be of the form $f(t)=1−\exp(−\alpha t)$ as a function of $t$, for some constant $α>0$, which is related to the half-life ...

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From the question, it's a little difficult for me to tell whether you're interested in comparing group means, or instead in estimate the effect of some other covariate for each group. I'll assume the latter. The right choice is going to depend a bit on the features of your dataset, but my first thought would be to try estimating a model with an indicator ...

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In connection to what jmbejara is saying, you can convert many models to a markov model by simply increasing the dimension of the state space and then precomputing the nonlinearities to make it a linear model. As far as I know this technique cannot be applied under conditions when number of parameters is not deterministic but stochastic. Or the process' ...

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You've got to distinguish the condition "linear in parameters" from "linear in variables." Often it is the case that a nonlinear relationship between variables can be transformed into a linear relationship in transformed variables. For example, $$y = a + bx + cx^2 + \epsilon$$ can be transformed into $$y = a + bx + cz + \epsilon,$$ with $z=x^2$ as a new ...

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To put the argument of @alecos-papadopoulos in graphical terms: Say his $a$ is -1 and his $b$ is 0.5 and $x$ ranges between 0 and 2. If we were to graph that relationship we could type in Stata: twoway function Pr = invlogit(-1+0.5*x), range(0 2) lwidth(medthick) That would result in the following graph: which looks pretty linear. However, if we were ...

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To get a sense of what might be going on here, consider a simple logistic regression model with one continuous regressor $$P(Y=1\mid X) = \Lambda (g(x)) =\Big(1 + \exp\{g(x)\}\Big)^{-1} =\Big(1 + \exp\{a+bx\}\Big)^{-1}$$ The marginal effect is $$\frac {\partial P}{\partial X} = \Lambda (1-\Lambda)b$$ The 2nd-order Maclaurin series of $\Lambda$ (its ...

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