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?
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asked Feb 21 '14 at 22:39
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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 Feb 21 '14 at 23:37
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$\begingroup$ Yup, that's what I mean. $\endgroup$ – Brash Equilibrium Feb 22 '14 at 7:27
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$\begingroup$ Is it a set of counts (out of 10) divided by 10, or rounded values, or something else? Why a beta? $\endgroup$ – Glen_b Feb 22 '14 at 9:55
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$\begingroup$ The reason why I am going for a beta is that the underlying interpretation of the variable is the chance that something will happen. $\endgroup$ – Brash Equilibrium Feb 23 '14 at 5:46
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$\begingroup$ @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 Jun 10 '17 at 11:04
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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
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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! $\endgroup$ – Brash Equilibrium Feb 22 '14 at 7:28
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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).
answered Feb 22 '14 at 0:43
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$\begingroup$ 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. $\endgroup$ – Brash Equilibrium Feb 22 '14 at 7:29
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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. $\endgroup$ – Alecos Papadopoulos Feb 22 '14 at 7:57
