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The questions "What do you want to predict?" and "What is the outcome or result here?" often have the same answer, but not always. The terminology of independent variables is widely considered overloaded in statistical sciences. Numerous writers and researchers -- over at least the last several decades -- have suggested using other terms, although there is ...

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One straightforward way to analyze this situation is to assume for testing purposes that the fix made no difference. Under this assumption, you may view the assignment of the (potential) 48 observations into pre-fix and post-fix groups as being random. If (hypothetically) all the post-fix outcomes work as expected, it means you have observed 44 expected ...

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@NickCox gave an excellent answer. A couple additions: You ask But in statistics, can the independent variables be regarded as the "things we are using to make the prediction," while the dependent variable is the "thing being predicted?" To give an explicit answer: Yes, that is often how the terms are used. I use them that way, myself. Second, the ...

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Great Question! The general approach is called a ARMAX model The reason for the generality of approach is that it is important to consider the following possible states of nature which not only provide complications BUT opportunities.. The big mac price might be predicted better using previous big mac prices in conjunction with activity in the two ...

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A better approach is to think hard about the model specification and to seldom assume linearity, as it is unusual for variables to be linearly related to each other. With very small sample sizes we must sometimes force a linearity assumption because we can't do much else without penalized maximum likelihood estimation. Making scatterplots isn't always a ...

4

Yes, yes it does: This is a trivial application of optimisation rules. So long as $n$ is constant (i.e., does not depend on $\theta$) then for any objective function $F$ and any other function $h$ you have: $$\underset{\theta \in \Theta}{\text{arg min }} F(\theta) = \underset{\theta \in \Theta}{\text{arg min }} h(n) F(\theta).$$ For your particular case ...

4

Predicted R^2 in Minitab is based on predicting an observation not fitted by the model. You're using a 5th degree polynomial (why you selected 5 is another question) to estimate approximately 10 data points. Removing one of the points will greatly change your estimated coefficients. In other words, you are incredibly overfitting your data and your model ...

4

If you are interested in prediction accuracy then in general I'd advise you to not be that concerned with the statistical significance of predictors. There may be certain predictors that you think should definitely be significant (for example, if you were modeling ice cream sales and had a variable for time of year), and if they are not, that may indicate ...

4

Quantile regression is not a special case of OLS. The use-case for quantile regression methods is to estimate a given quantile of the distribution $[Y|X]$, such as the median. $[Y|X]$ is the distribution of $Y$ conditional on $X$, which is maybe why you have heard quantile regression as a way of "predicting change in the dependent variable that is not the ...

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tl;dr Yes, you could build a voting classifier using the array of your trained estimators. I don't recommend that, especially as I don't know much about the use case. My typical workflow There could be multiple candidates to build a decent estimator in this use case (Let's say RandomForest, xgboost, SVM) Build an estimator using each of them with their ...

4

When you have time-varying covariates there are a couple of important things you need to consider. Whether the covariate is exogenous or endogenous. An exogenous time-varying covariate $x_i(t)$ ($i$ denotes the subject and $t$ the time) satisfies the following condition: $$p\{x_i(t) \mid \mathcal H_i^X(t), \mathcal H_i^Y(t)\} = p\{x_i(t) \mid \mathcal H_i^X(... 4 This totally depends on your use case. Here are some of the ways to look at this: What is acceptable to users/society/other stakeholders? One can actually just ask people. What is currently done (e.g. manually by humans) and how well does the current solution perform? What makes economic sense (e.g. cost of mistakes - which may differ depending on the type ... 4 You cannot. How can you check the truth of something when you have no information about whether it is true? 4 As you have correctly noticed, the term 'independent' has completely different meanings depending on context. Statistical independence is what you are describing between the weather and your dinner. These two events are independent in the sense that the value of one does not affect the other. There are more formal mathematical definitions of this ... 4 The difference between the two approaches is actually a difference between an Empirical Bayes versus a fully Bayesian approach to estimate the same thing. If you fit the mixed model using maximum likelihood, then you typically follow the first option, whereas, under the fully Bayesian approach from which you take posterior samples also for \theta, you ... 3 If you call predict.zeroinfl() without any parameters, it uses the default setting for the type parameter, which is type="response". You will then get a prediction for the mean, or the expected response. This expectation will typically vary much less than your actual observations, will not be integer, and will be larger than zero. To obtain a probabilistic ... 3 There are many measures of point prediction accuracy, like the mae, the rmse, or the mape. You may want to browse through our questions with these tags. Why use a certain measure of forecast error (e.g. MAD) as opposed to another (e.g. MSE)? is likely helpful, as may be What are the shortcomings of the Mean Absolute Percentage Error (MAPE)? and Mean absolute ... 3 You predict a class if it is given maximum probability by the estimated model. So when some classes are not predicted that is simply because the model never gave them maximum probability. That might just be correct, and not necessary a reason for concern. So I disagree with the answer by @Nitin, proposing oversampling. You might say: but they did occur in ... 3 Let$$ a:=TP,\quad b:= TN,\quad c:=TP+FN,\quad d:=TN+FP. $$Then accuracy and balanced accuracy are$$ Acc=\frac{a+b}{c+d},\quad BAcc=\frac{a}{2c}+\frac{b}{2d}, $$or$$ Acc=\frac{acd+bcd}{cd(c+d)},\quad BAcc=\frac{\frac{1}{2}ad(c+d)+\frac{1}{2}bc(c+d)}{cd(c+d)}. $$Therefore,$$ Acc<BAcc $$is equivalent to$$ acd+bcd < \frac{1}{2}ad(c+d)+\...

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Validation sets and overfitting Let us start with your question about why we use a validation set. We use this to get an estimate of the true out-of-sample error of our model. We use a separate set here because, as you know, looking at in-sample residuals will give us a too optimistic idea of our performance. Now, suppose we apply our model $M_1$ to the ...

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I think, there is a misunderstanding regarding the meaning of the R^2 values. From the help of the package, you get the following information: Marginal R_GLMM² represents the variance explained by the fixed effects, and is defined as: R_GLMM(m)² = (σ_f²) / (σ_f² + σ_α² + σ_ε²) Conditional R_GLMM² is interpreted as a variance explained by the entire model, ...

3

I redrew the corrected figure, following the suggested link. The corrected R code is as follows. RndEff <- unlist(ranef(fitm)) FixEff <- fixef(fitm) for(i in 1:length(RndEff)) { prm <- exp(FixEff[1] + FixEff[2]*nd$cHEIGHT + RndEff[i]) par(new=T) lines(nd$cHEIGHT, prm, lwd=.1, col="red") } Then I got this (the right graph is the ...

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In the context statistical model building, terms predictor, explanatory variable, independent variable, and input (among some others) have been used interchangeably. I would not see the use of the term predictor as an issue in your case, regardless of whether the aim of your study was to build a descriptive or a predictive model. And because the terms have ...

3

That is totally case-dependent. Say I have a problem where have a 90% incidence (90% of my data would take value 1). Then, if my model is right 85% of the time, then I'd be better calling 1 every time, because that would give me a 90% accuracy. Now, if I have a phenomenon that occurs only 1% of the time, then if my model is right 10% of the time, that could ...

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To give a general answer on the background and the concept behind series, time series can be used to predict both long term and short term, the problem is what you are trying to predict and how: sometimes time series theory itself will tell you that some series are indeed not predictable, especially in the long term (because their long term moments are not ...

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If you only look at the last five years, then year alone can be used to predict TKP using a straight line:

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You have reported that you have transaction data . Since you don't have observations every day, you will need to aggregate/bucket your data to a time series i.e. some level of accumulation ...where non-zero observations are present. An example might be weekly or monthly buckets or even quarterly buckets. Buying patterns are seldomly weekly-based with ...

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If you know that this next observation comes from a specific group $i$, then the best prediction would be $$\exp(\hat \beta_0 + \hat b_i).$$ However, if you do not know from which group it comes from, then it would make more sense to use as prediction the average over the groups, which in your specific random intercepts model would be $$\exp \Bigl ( \hat \... 3 Great question - you've hit a great realization about PCA. Even though the important feature in the model is X2 because of the large coefficient, because X2 has such low variance, the PCA is essentially ignoring it by giving it a low weight. PCA is great for capturing the greatest variance among variables, but that is a different task than regression, ... 3 You have the right idea in spirit, but there is no magic in replacing the functional minimization with a scalar minimization. In particular, the middle part of the equation you wrote,$$ f(x) = \operatorname*{argmin}_{f \in \mathcal{H}} \mathrm{E}_{Y|X}[L(Y,f(X))| X=x],  does not type-check, so to speak. The left-hand side is a single point in $\mathcal{Y}$...

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