# How do I choose parameters for my beta prior?

Suppose today I'm going to flip a coin. I believe that 9 of 10 flips will come up heads. I flip the coin and 8 of 10 are heads. Is my distribution of belief

1. beta(9+8, 1+2)
2. beta(1+9+8, 1+1+2)
3. beta(m+8, n+2) where I choose m and n based on graphing beta(m,n) and deciding if it feels right. (This is not a facetious answer, it's a real suggestion.)
4. Hidden option 4.

If I use #1, I believe that is an improper prior, but Wikipedia claims some statisticians use them.

This choice is not obvious to me.

• If you want your prior to include the information of a hypothetical sample of 10 flips out of which 9 came out as heads, then your prior is $\mathrm{Beta}(9,1)$. The mean of the prior is $0.9$. After flipping the coin 10 times out of which 8 are heads, your posterior is $\mathrm{Beta}(17,3)$. The posterior mean is $0.85$. You can use an improper prior, that is no problem. The Beta distribution is the conjugate distribution of the binomial. Check out this post for a fully worked example using the Beta distribution. Jul 19, 2013 at 15:54
• Choice #1 is not an improper prior. Jul 19, 2013 at 15:58
• @QuantIbex Your right, of course (+1). Sorry, my wording was inaccurate: I didn't want to imply that #1 was improper, I just wanted to say that an improper prior is not a problem automatically, at least not mathematically. Jul 19, 2013 at 16:10
• For the Beta distribution, the improper prior is $\lim_{a \to 0}\rm{Beta(a,a)}$. Improper means it isn't a probability density, e.g. doesn't integrate to 1. You might want to consider a non-informative prior, which is flat for all possible values of your parameter: the uniform distribution is suitable here, which is a special case of the Beta distribution, i.e. $\rm{Beta(1,1)}$. See online beta graphing. Jul 19, 2013 at 16:10
• Improper priors are acceptable so long as they produce proper posteriors. For the beta prior with a binomial likelihood, any improper prior will produce a proper posterior provided you observe one (or more) heads and one (or more) tails.
– Sycorax
Jul 19, 2013 at 18:27

If your prior belief is that 9 of the 10 coin flips will come up heads, then you want the expectation of your prior to be 0.9. Given $X \sim \mathrm{Beta}(\alpha,\beta)$ (for conjugacy in the beta-binomial model), then $E[X] = \alpha/(\alpha+\beta) = 0.9$, so you can use this as your first constraint. Obviously this leaves you with an infinite number of possibilities, so we'll need a second constraint which will represent your subjective confidence in this prior belief: the variance of your prior, $\operatorname{var}[X] = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}\$
Decide how you want to set your variance and solve the system of equations for $\alpha$ and $\beta$ to define the parameters for your prior. Justifying your choice of variance here may be difficult: you can always err on the side of a wider (i.e. less informative) variance. The wider you set the variance, the closer your prior will approximate a uniform distribution.