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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?

kjetil b halvorsen
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Brash Equilibrium
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    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? – Glen_b Feb 21 '14 at 23:37
  • Yup, that's what I mean. – Brash Equilibrium Feb 22 '14 at 07:27
  • Is it a set of counts (out of 10) divided by 10, or rounded values, or something else? Why a beta? – Glen_b Feb 22 '14 at 09:55
  • The reason why I am going for a beta is that the underlying interpretation of the variable is the chance that something will happen. – Brash Equilibrium Feb 23 '14 at 05:46
  • @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. – Neil G Jun 10 '17 at 11:04
  • Beta is not appropriate, but the beta-binomial is (more) appropriate. See checked answer below. – Brash Equilibrium Jun 12 '17 at 16:27

2 Answers2

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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

heropup
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    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! – Brash Equilibrium Feb 22 '14 at 07:28
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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).

Alecos Papadopoulos
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  • 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. – Brash Equilibrium Feb 22 '14 at 07:29
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    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. – Alecos Papadopoulos Feb 22 '14 at 07:57