4
$\begingroup$

Suppose that my data $y \in \{0,0.1,\ldots,1\}$. What are the consequences of modeling that data as continuous, i.e., as if $y \in [0,1]$, by using the beta distribution? Is there a version of the beta distribution that can account for this?

$\endgroup$
6
  • 3
    $\begingroup$ When you say $\{0,0.1,…,1\}$ do you mean the data is all multiples of 0.1, or something else? If so, how does it arise - as ratios of counts, or some other way? Why do you want to use a beta, rather than something else? $\endgroup$
    – Glen_b
    Commented Feb 21, 2014 at 23:37
  • $\begingroup$ Yup, that's what I mean. $\endgroup$ Commented Feb 22, 2014 at 7:27
  • $\begingroup$ Is it a set of counts (out of 10) divided by 10, or rounded values, or something else? Why a beta? $\endgroup$
    – Glen_b
    Commented Feb 22, 2014 at 9:55
  • $\begingroup$ The reason why I am going for a beta is that the underlying interpretation of the variable is the chance that something will happen. $\endgroup$ Commented Feb 23, 2014 at 5:46
  • $\begingroup$ @BrashEquilibrium No, the beta distribution models the likelihood on a bias $p$ of a Bernoulli random variable induced by its realizations. It's not appropriate for this data. $\endgroup$
    – Neil G
    Commented Jun 10, 2017 at 11:04

2 Answers 2

9
$\begingroup$

You could model the variable $10Y \sim {\rm BetaBinomial}(n = 10, a, b)$. Specifically, if $X \sim {\rm BetaBinomial}(n, a, b)$, then $$\Pr[X = x] = \frac{\Gamma (b+1) \Gamma (a+n) \Gamma (n+x) \Gamma (a+b-x)}{\Gamma (a) \Gamma (n) \Gamma (x+1) \Gamma (b-x+1) \Gamma (a+b+n)}.$$ The only restriction on the parameters $a,b$ is that they be positive; $n$ must be a nonnegative integer; and $x \in \{0, 1, \ldots, n\}$. This PMF has many nice properties: see http://en.wikipedia.org/wiki/Beta-binomial_distribution

$\endgroup$
1
  • 1
    $\begingroup$ right after I asked this question, I did some digging and this is the solution I decided for myself. Came back and here it is. Check please! $\endgroup$ Commented Feb 22, 2014 at 7:28
4
$\begingroup$

It is rather easy to create a discrete version of the beta distribution, and in its standard interval from zero to unity. In your case the support is $Y \in \{0,0.1,\ldots,1\}$. So consider the probability mass function, with $j=\{0,1,...,10\}$

$$P(Y=j/10;\alpha;\beta) = \frac{(j/10)^{\alpha-1}[1-(j/10)]^{\beta -1}}{ \sum_{j=0}^k(j/10)^{\alpha-1}[1-(j/10)]^{\beta -1}} $$

but with $\alpha \ge 1,\;\; \beta \ge 1$, and using $0^0 \equiv 1$.

It is characterized by the same flexibility in shape than the continuous version (bar the "U-shape" which would require parameters smaller than unity).

$\endgroup$
2
  • $\begingroup$ Alecos, if you have an argument for why this is superior to @heropup's solution, I'd love to here it and would totally consider switching if your argument gets validated. $\endgroup$ Commented Feb 22, 2014 at 7:29
  • 3
    $\begingroup$ It is not. My answer is just a first step towards a direct discretization of the beta distribution, so it remains to be checked and studied, for results, moments, properties to be derived for it etc. On the other hand, the Beta-Binomial is an already well-studied distribution. On a first check, they seem to provide roughly analogous results regarding the shape of the pmf graph. $\endgroup$ Commented Feb 22, 2014 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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