2
$\begingroup$

I've heard that the $\alpha$ and $\beta$ parameters of the Beta distribution intuitively represent the number of successes and failures, respectively.

1) If so, what's the purpose of subtracting $1$ from them in the exponent?

2) If so, why would then $\alpha=1$ and $\beta=1$ represent a uniform distribution? Do we assume that there is one success and failure initially? Does this have anything to do with Laplace's Rule of Succession?

3) How can we intuitively understand positive real-valued shape parameters for the distribution?

$\endgroup$
2
$\begingroup$

For your initial question, see the link in the comments.

For 1), it is not a valid density integrating to 1 otherwise. Of course, one could let $\tilde\alpha=\alpha-1$ and $\tilde\beta=\beta-1$ and write $$ \pi \left( \theta \right) =\frac{\Gamma \left( \tilde\alpha+\tilde\beta+2\right)}{\Gamma \left( \tilde\alpha+1\right) \Gamma \left( \tilde\beta+1\right) }\theta^{\tilde\alpha}\left( 1-\theta \right) ^{\tilde\beta} $$

For 2), just plug in: the beta density $$ \pi \left( \theta \right) =\frac{\Gamma \left( \alpha+\beta\right)}{\Gamma \left( \alpha\right) \Gamma \left( \beta\right) }\theta^{\alpha-1}\left( 1-\theta \right) ^{\beta-1} $$ becomes $$ \pi \left( \theta \right) =\frac{\Gamma \left( 2\right)}{\Gamma \left( 1\right) \Gamma \left(1\right) }\theta^{1-1}\left( 1-\theta \right) ^{1-1} =1$$

For 3) While, again see the comment, one may interpret the coefficients in relation to prior failures and successes, that is not a necessity and real parameter values are also possible.

$\endgroup$
  • $\begingroup$ 1) Yes, but it's already restricted to an interval and normalized. Could we not just remove the $-1$ and renormalize it? Is it simply a matter of mathematical convenience? i.e. use of the $\Gamma$ function 2) I know mathematically it works out, but is there any way we can interpret using $(1,1)$ to represent a uniform distribution in the context of successes and failures? 3) What would a non-integer $n$ mean in the "fictitious sample interpretation"? Edit: 2) After looking at your answer again, you mention that it represents $\alpha_{0}-1$ successes, so $0$ successes. $\endgroup$ – John Dec 9 '16 at 13:26
  • $\begingroup$ ad 2) Yes, the uniform prior (see many posts on that on this site) is considered an "uninformative" prior, so it is intuitive that it corresponds to not having observed to prior successes or failures. $\endgroup$ – Christoph Hanck Dec 9 '16 at 13:34
  • $\begingroup$ ad 3) little, I guess. As I said, you do not have to interpret the beta in terms of the fictitious sample interpretation. $\endgroup$ – Christoph Hanck Dec 9 '16 at 13:34
  • $\begingroup$ ad 1) see my edit, if that is what you refer to? $\endgroup$ – Christoph Hanck Dec 9 '16 at 13:36

Not the answer you're looking for? Browse other questions tagged or ask your own question.