Beta distribution questions 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.
 A: 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 $b$. 
A: "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.
