# How are these priors generated? [closed]

I am trying going through an exercise, I don't understand how the information provided in the text below transitions into the parameters displayed in the beta priors. How are these informative priors generated?

This is the model:

• Can you please include the parts below that line in the screenshot? – Arya McCarthy May 12 at 3:00
• @AryaMcCarthy done, thank you. – frantic oreo May 12 at 6:24
• Please type your question as text, do not just post a photograph or screenshot (see here). When you retype the question, add the self-study tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. – kjetil b halvorsen May 12 at 7:46
• @kjetilbhalvorsen Thanks, I would have typed the question however am not well versed in writing equations in Latex. – frantic oreo May 12 at 8:03
• – kjetil b halvorsen May 12 at 22:20

The expert starts 90% shure that the expected survival time in absence of the contaminant is less then 40 and 90% shure that it is greater then 10.

This is the information on which we base the prior to $$\beta_0$$ because the absence of the contaminent means we have a $$conc_i = 0$$ and thus $$\beta_1$$ does not matter. We look for a normal distribution with 10% of the mass below log(10) and 10% of the mass above log(40). As the normal distribution is symmetric around the mean, the mean of that standard deviation will be in the middle between log(10) and log(40), so (log(10) + log(40))/2 = 2.9957. (Your screenshot says the mean was 3).

Now all we need is a standard deviation with qnorm(c(.1, .9), mean = 2.9957, sd = x) == c(log(10), log(40)) == c(2.302585, 3.688879).

Now there will be a number of ways to come to a standard deviation. For example a graphical approach:

curve(qnorm(.1, mean = 3, sd = x), ylim = c(1.5, 4.5),
ylab = expression(log(lambda)), xlab = expression(sigma),col = "blue")
curve(qnorm(.9, mean = 3, sd = x), add = TRUE, col = "blue")
abline(h= c(log(10), log(40)), lty = 3, col = "grey")
abline(v = .54, lty = 2, col = "grey")


Let's try the .54 from your book:

> qnorm(c(.1, .9), mean = 2.9957, sd = .54)
[1] 2.303662 3.687738


Close enough. I will leave the $$\beta_1$$ for you to try because after all this is a self-study question, even if it does not have a self-study tag.