What properties of random variable leads to modeling it with Beta Distribution?

Context: If you ask the same question about Bernouli distribution, the answer would be the distribution of a random variable that has only two outcomes and the probability of one outcome is fixed at $p$. Now how can we answer a similar question about Beta distribution?


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  • $\begingroup$ First I might ask you What properties of random variable leads to modeling it with Normal Distribution?. This question lacks context until you write that in your post! Note that Bernoulli Distribution is Discrete. By definition it takes $0$ and $1$....(I used Normal df since it is well known, nothing special) $\endgroup$ – MAN-MADE Aug 19 '17 at 2:45
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    $\begingroup$ @MANMAID. It is unclear why you want to drag the normal distribution into this to 'provide context' for a perfectly reasonable question about using beta distributions. $\endgroup$ – BruceET Aug 19 '17 at 6:17

One of the most important uses of beta distributions is in Bayesian statistical analysis of binomial data. Beta distributions are used for prior and posterior distributions. I will give an elementary example below.

In Bayesian statistics a beta distribution is used as a model for the binomial success probability, which I will denote as $\theta$ (instead of $p$). [Bayesian analysis treats $\theta$ as a random variable, rather than as an unknown fixed parameter.]

Suppose you are wondering whether Proposition A will get a majority of the vote at the next election. The circumstances are that Prop A had to have some support in order to get on the ballot. Also a similar proposition passed with about 65% of the vote at the last election. That may be good or bad: perhaps good because it may show voters favor such propositions; perhaps bad because such propositions require a slight increase in property taxes and it may be too soon to ask again. Everything considered, you think Prop A may be favored by slightly more than half of the voters, but that the election is likely to be close.

Because $0 \le \theta \le 1$ a beta distribution seems appropriate because beta distributions have support $[0,1].$

There is no precisely correct prior distribution for $\theta,$ because it should reflect your personal opinion. Suppose you choose the prior distribution $\mathsf{Beta}(\alpha_0 = 330, \beta_0=270).$ The 'kernel' of its PDF (omitting the constant of integration) is $f(\theta) \propto \theta^{\alpha - 1}(1-\theta)^{\beta-1},$ where the symbol $\propto$ indicates omission of the constant. It has mean 0.55, mode 0.5502, and median 0.55006. Also, $P(0.51 < \theta < 0.59) \approx 0.95,$ as computed in R statistical software.

qbeta(.5, 330, 270)                         #  'qbeta' is inverse CDF
## 0.5500556
pbeta(.59, 330, 270) - pbeta(.51, 330, 270) #  'pbeta' is CDF
## 0.9513758

Now suppose that a well-run poll of $n = 1000$ randomly chosen likely voters shows $x = 620$ in favor of Prop A and $n - x = 380$ opposed. The likelihood function corresponding to these results is $f(x|\theta) \propto \theta^x(1-\theta)^{n-x}.$

Bayes Theorem states that the posterior distribution $f(\theta|x)$ is found by multiplying the prior and likelihood functions: $$f(\theta|x) \propto \theta^{\alpha_0 - 1}(1-\theta)^{\beta_0-1} \times \theta^x(1-\theta)^{n-x}.$$

In our example, $$f(\theta|x) = \theta^{\alpha_0 + x -1}(1-\theta)^{\beta_0 + n - x -1} = \theta^{\alpha_n - 1}(1-\theta)^{\beta_n - 1},$$ where $\alpha_n = 950$ and $\beta_n = 650.$ We recognize $f(\theta|x)$ as the kernel of $\mathsf{Beta}(950, 650).$

We say that the beta prior and the binomial likelihood are 'conjugate' (mathematically compatible). [Without conjugacy we would not be able to recognize the posterior distribution so easily, and we would have to use a different form a Bayes' Theorem in which a denominator might need to be integrated by numerical methods.]

Finally, a 95% posterior probability interval for $\theta$ is $(0.570, 0.618).$

qbeta(c(.025,.975), 950, 650)
## 0.5695848 0.6176932

A melding of the information in the prior and the likelihood has given a somewhat more optimistic estimate of the chances Prop A will pass, than did our prior distribution.

Notes: (1) If we had used the noninformative prior distribution $\mathsf{Unif}(0,1) \equiv \mathsf{Beta}(1,1),$ then the posterior distribution would have been $\mathsf{Beta}(621,381)$ with a 95% posterior probability interval $(0.589, 0.650).$ This is numerically the same (to three places) as a frequentist Agresti-style 95% confidence interval for $\theta$. However, Bayesian and frequentist interval estimates have somewhat different interpretations.

(2) This example is similar to Example 8.1 in Suess (2010).

  • $\begingroup$ Thanks so much for the good and detailed answer @BruceET. $\endgroup$ – user25004 Aug 19 '17 at 17:05
  • $\begingroup$ Also thanks for introducing the reference Suess (2010). That is very helpful. $\endgroup$ – user25004 Aug 19 '17 at 17:06

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