Beta distribution questions

Question

Just a simple beta distribution question just to be sure that I understand it.

Say we do experiments, and we expect a proportion $\theta$ of people having a specific property (which means $\theta \in [0,1]$)

Assume we have a prior beta "belief" that $\theta = 0.3$ and we are very certain. The question now is, what should $a$ and $b$ be, if we define

$$f(\theta, a, b) = \frac{\theta ^{a-1}(1-\theta)^{b-1}}{B(a,b)}$$

I would say, using the fact that the mean is $m = \frac{a}{a+b}$, solving for $a$ and $b$ : $$ a=mn; b=(1-m)n$$

If I am very sure that $\theta$ is $0.3$, then $a= mn = 0.3 \cdot25 =7.5$ and $b$ would be $0.7\cdot 25 = 17.5$

Is this good reasoning? I took the number $n$ quite randomly, but the idea is I guess that if you are very sure, $n$ should be big whereas if you are not that sure, $n$ should be quite low.

Answer 1

Elaborating on @Michael Hardy answer, "very sure" that $\theta = \bar \theta$ can be translated as $E(\theta) = \bar \theta$ and also that Var$(\theta)$ is really "small". Translate this as a small standard deviation, expressed as distance from the mean, perhaps more conveniently as a percentage deviation : $\sigma = \bar \varepsilon\bar \theta$.

The mean and variance of the beta distribution are

$$ E(\theta) = \frac {a}{a+b}$$ $$ \text {Var}(\theta)= \sigma ^2= \frac {ab}{(a+b)^2(a+b+1)} $$

Then you have a $2 \times 2$ system of equations to solve

$$\frac {a}{a+b} = \bar \theta\;, \qquad \frac {ab}{(a+b)^2(a+b+1)} = \bar \varepsilon^2\bar \theta^2$$

From the first we obtain

$$ b= \frac {a(1-\bar \theta)}{\bar \theta} \qquad \frac {b}{a+b} = 1-\bar \theta$$

Then we have $$\frac {ab}{(a+b)^2(a+b+1)} = \frac {\bar \theta(1-\bar \theta)}{(a+\frac {a(1-\bar \theta)}{\bar \theta}+1)} = \frac {\bar \theta^2(1-\bar \theta)}{a\bar \theta + a(1-\bar \theta)+\bar \theta } = \frac {\bar \theta^2(1-\bar \theta)}{ a+\bar \theta }$$

Then the second equation becomes

$$\frac {\bar \theta^2(1-\bar \theta)}{ a+\bar \theta } = \bar \varepsilon^2\bar \theta ^2\Rightarrow a = \frac {(1-\bar \theta)}{\bar \varepsilon^2} - \bar \theta$$

Once you quantify your prior beliefs by specifying numerical values for $\bar \theta$ and $\bar \varepsilon$, you obtain the corresponding numerical value for $a$ and then you can solve for $b$.

Answer 2

"Very sure" is a vague term but it means the probability distribution should be concentrated very close to $0.3$, which means the variance should be small, and that means the parameter you've called $n$ should be big. Probably $n=25$ is in accord with most common-sense interpretations of this, but that could depend on further context that's not mentioned here.