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So, the idea is that I use linear regression and I get an equation y = a * x + b. So when I give a value x I get a predicted value y. However, what I want to achieve is instead of a single value y to get a range of values, such as [val_1 - val_2].Can this be done? Also I have found this question: Regression analysis with range (instead of single value) as independent variable which did not help...

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    $\begingroup$ Can you explain what your val_1 and val_2 would be? $\endgroup$
    – Stephan Kolassa
    Sep 10, 2022 at 10:45
  • $\begingroup$ The values of the range: [min to max]... $\endgroup$
    – just_learning
    Sep 10, 2022 at 10:46
  • $\begingroup$ Hm. Linear regression usually assumes normally distributed errors, which go from minus to plus infinity, and your regression will not tell you that there is a max value your $y$ cannot exceed, so the best you can do would be to always output $(-\infty,\infty)$. This is probably not helpful. Would a prediction-interval work for you, i.e., an interval that contains the future realization with (say) 95% probability? $\endgroup$
    – Stephan Kolassa
    Sep 10, 2022 at 10:54
  • $\begingroup$ @Stephan Kolassa: If I calculate the MSE of my model and get the SD = sqrt(MSE) so a range: [y-SD, y+SD] ? Do we say the same thing? $\endgroup$
    – just_learning
    Sep 10, 2022 at 10:57
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    $\begingroup$ Yes, that is a prediction interval. It will be too narrow, though, since it only includes one source of uncertainty, namely residual uncertainty - but not parameter uncertainty. Let me write up an answer. $\endgroup$
    – Stephan Kolassa
    Sep 10, 2022 at 10:58

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What you are looking for is a prediction interval, i.e., an interval $[a,b]$ such that a new observation falls within it with a certain prespecified probability, e.g., 95%. (Of course, there are multiple such intervals. The typical convention is to have "prediction interval" refer to a symmetric PI, such that $P(y<a)=P(y>b)=0.025$.)

You can find the formulas and explanation in Faraday (2002), section 3.5. Note that Faraday calls PIs "confidence intervals for predictions", which I find unfortunate, because it risks confusion with "normal" confidence intervals, which pertain to unobservable parameters. There is a difference between CIs and PIs.

Your favorite statistics package can probably give you prediction intervals. For instance, here is some toy data in R, and we calculate a central 95% PI at a IV of $x=0.7$:

set.seed(1)
xx <- runif(20)
yy <- 1+xx+rnorm(20)
model <- lm(yy~xx)
xx_pred <- 0.7
predict(model,newdata=data.frame(xx=xx_pred),interval="prediction",level=0.95)
#        fit        lwr      upr
# 1 1.527059 -0.4707293 3.524847

You will get a prediction of the expectation of $y$ of $1.53$, and a prediction interval of $(-0.47, 3.52)$.

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