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I am reading the answer to this post (Prediction based on bayesian model) about new predictions, and I have a question. Could I find beta0 and beta1 distributions in regular way:

 beta0 ~ dnorm(0, 10)
 beta1 ~ dnorm(0, 10)
 sigma ~ dunif(0, 50)
 for (i in 1:N) {
    y[i] ~ dnorm(beta0 + beta1 * x[i], sigma)
 }

And then in R do something like this (beta0 and beta1 would be taken from the posterior distribution (regression) and x_new (1:40) would be an array where I need extrapolation):

beta0 beta1 x_new y_new
1.09  0.01  1     1.1
0.98  0.015 1     0.995
1.08  0.012 1     1.092
...
...
...
1.09  0.01  2     1.1
0.98  0.015 2     1.01
1.08  0.012 2     1.11
...
...
...
1.09  0.01  40     1.49
0.98  0.015 40     1.59
1.08  0.012 40     1.56
...
...
...

Can I use this approach to create Bayes inference? Is it wrong? Why do I need "sigma" in the answer from the link for y_new?

Thank you very much for explanation

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1 Answer 1

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Writing your code move verbosely:

for (i in 1:N) {
    mu[i] <- beta0 + beta1 * x[i]
    y[i] ~ dnorm(mu[i], sigma)
}

Your model is a Bayesian linear regression, so as standard linear regression, it predicts the mean $E[y_i|x_i] = \mu_i = \beta_0 + \beta_1 x_i$. If you are interested in the mean $\mu_i$, sure, take the predictions for the mean. However if you want to look at the samples from the conditional distribution of $y_i|x_i$ itself, then you sample from the posterior predictive distribution. In many cases, saving both the mean and samples from the posterior predictive distribution would be useful.

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  • $\begingroup$ Hi Tim, thank you very much for your answer. I see that my question was not clear enough; I edited it. So, instead of using the mean values for beta0 and beta 1, I was thinking to use beta0 and beta1 samples from the posterior distributions. Can you please see my edits (table) in the question? Thank you again. $\endgroup$
    – eod
    Commented Oct 28, 2021 at 13:19
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    $\begingroup$ @eod it was clear. mu would be samples from $E[y|x]$ while y would be samples from $y|x$ itself. $\endgroup$
    – Tim
    Commented Oct 28, 2021 at 13:23
  • $\begingroup$ Thank you again for clarification! $\endgroup$
    – eod
    Commented Oct 28, 2021 at 13:28

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