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1

No there is nothing from that output that could have told you about the possibility of model misspecification. The formal way to see if there is misspecification (test if the model is indeed nonlinear) would have been to run a "Ramsay reset test". This mean to estimate the following: $$ 1: y = a_0 + a_1 x + e $$ calculate: $$ \hat{y} = a_0 + a_1 x ...


1

If I understood correctly you have observations $(x_i, y_i)_i$ from the joint distribution $p_{X,Y}$. It seems you are asking about the quantity $$\mathrm{Supp}(p_{X\mid y}) = \{ x\mid p_{X\mid Y}(x\!\mid\!y) > 0 \}$$ You should be able to estimate the joint pdf $p_{X,Y}$ of your data. This can be done via a good ol' KDE estimation or via the use of ...


2

I'll use $\mu_i = \eta_1 - 2\theta\eta_2x_i + \eta_2 x_i^2$ for convenience. If we're thinking of $\mu_i$ as a function of $\theta$, so only $\eta_1$ and $\eta_2$ are parameters, then we can write this as $$ \mu_i = \eta_1 + \eta_2(-2\theta x_i + x_i^2) = \eta_1 + \eta_2 z_i $$ for $z_i = -2\theta x_i + x_i^2$. This is just a simple linear regression now so $...


2

Is this correct? Yes, except that it's not the kernel trick. There are two equivalent approaches to a regression problem. One is to write the response variable $y$ as a linear combination of basis functions $\phi(x)$ as $y(x) = \sum_i \omega_i \phi_i(x)$, $i=1,2$ (2 in this case, you could have infinite). This is what you are doing, and what you are calling ...


2

In terms of the relationship alone $ y = a \exp(-b x) $ implies $\ln y = \ln a - b x$ so that you could regress $\ln y$ on $x$ and accordingly expect that the coefficient of $x$ will be negative. Exponentiating the estimated intercept will give you an estimate of $a$ in the original equation. But whether this is a good thing to do depends on the variability ...


0

My understanding from what you wrote is that you want to adjust the estimates for v=1 or 2 because you observed 0 of them in the simulation and you know the probability has to be greater than 0. When v=1 or 2 and you observe 0 of those out of 1 million, you can be pretty confident that the probability mass function at those two points is less than $3 \times ...


3

A partial answer follows, contributions are welcome, and the question remains open. Some of the questions that posed substantial roadblocks toward an understanding of what $R_{adj}^2$ is, are dealt with here in order to address the question "Would the real adjusted R-squared formula please step forward." What it is not is a proof of anything, it is ...


8

Standard implementation in R One potential confusion comes from the role of the intercept in the correction factor used by lm. If present, it shows up in both the numerator and the denominator of the correction factor. Quoting from an answer in the linked question: Extracting the most relevant line you get: ans$adj.r.squared <- 1 - (1 - ans$r.squared) * (...


0

In the first case, $r$ (or really $r^{2}$) is a measure of the strength of a linear relationship between the variables, so it would have no bearing on your ability to use nonlinear regression (and perhaps is indicative that you should). In the second case, there's no problem using categorical predictors in regression of any type. In the third case, you ...


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