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Let's Think About Restrictions If you want to the predictions of each quadrant to sum to the total, you have to incorporate that restriction into the model. Presently, there is nothing relating the 4 regressions and hence there is no requirement that the restriction be satisfied. Consider an alternative model. The total at timestep $i$ is $$ N_{total, i} = ...


5

SARIMA is a good first benchmark. I would suggest three more ones, all of which use seasonality, which I would assume to be relevant for rainfall: Seasonal exponential smoothing. If you use forecast::ets(), it will attempt to automatically fit a "good" model with additive or multiplicative trend and/or seasonality. A "seasonal naive" ...


4

You don't explicitly mention but it seems like you chose the first model yourself. You could try and see what the auto.arima() function in the forecasting package gives you. That function will test multiple models for you and come up with the best one it can find based in AIC, AICc, or BIC values. Setting the argument stationary to FALSE, it will include ...


3

Other answers say that traditional regression models, like linear or logistic regression, already does this, as regression is to model conditional expectations (or conditional probability, conditional hazard, conditional ...). As soon as you are calculating a prediction interval with some regression model, you are entertaining some kind of probabilistic ...


2

The predicted probabilities that Y=1 are given by p1 = nu/(1+nu) So just predict nu and then transformation nu to p1.


1

I am fairly sure it's valid to treat regression model as a joint probability distribution. Let $s,h,w$ be salary, height, weight, then a linear regression $$ s = \beta_0 + \beta_1h + \beta_2w + \epsilon \\ \epsilon \sim N(0, \sigma^2) $$ where $\beta_0, \beta_1, \beta_2$ are regression coefficients posits the distribution $$ s \sim N(\beta_0 + \beta_1h ...


1

I see you have any questions. Q&A sites like this are usually not best suited for such cases, I'd encourage you to ask one question at a thread next time. Also given the number of questions, it'd be probably a good idea to start with one of many great handbooks on machine learning. Nonetheless, I feel like the questions are fairly easy to answer, so let ...


1

You are right, you can't simply take the exponent. It seems that it is a log normal distribution. If $Log(Y)$ is normally distributed with $\mu, \sigma^2$, then $Y$ is Log normally distributed with these parameters. The expectation of a log normally distributed variable is: $exp({\mu +\frac{\sigma^2}{2}})$ So you will have to learn $\sigma$ as well. You can ...


1

I find it hard to think of scenarios, where predictions with random effects set to zero are of interest (there are of course inference situations for contrasts and comparisons where the random effects are irrelevant/cancel out). If there is variation and unexplained differences between random effects levels, why would we ignore that when making predictions? ...


1

I think you are confused about what cross validation is and why you use it. Quite simply, cross validation in this context is just repeated use of the learning and validation sets you describe above. Instead of splitting your data once into learning and validation you do it multiple times and take the average of your results to decrease the variance ...


1

A big one is generalisability. Good prediction requires that the conditional distribution of $Y$ given the features $X$ is the same in future use as it was in the training set. If the training set were a simple random sample from the population of future use this would be guaranteed by sampling, without any causal assumptions, but that's relatively unusual. ...


1

It appears you are missing the test statistic when calculating the margin of error in your first example. Recall the general formula : Since n=32 records, you will have 30 df. I'll assume a significance level = 0.05. Try: geom_point() + geom_smooth(method = 'lm', color = 'blue', se = TRUE) + geom_abline(intercept = 37.2851 + qt(0.975, 30)*...


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