Say that person $A$ thinks that a certain proportion is $0.3$, person $B$ thinks that a certrain proportion in a population is $0.7$. We have a $Beta(4,4)$ prior. How can one mathemtically prove that the prior does not favor the pre-knowledge of either of the persons?
I would say that a beta PDF with $a=b$ is symmetric.
Also If we fill in $p(x; a,b)= 1/\beta({a,b}) \cdot x^{a-1}(1-x)^{\beta -1}$ for $a=b=4$, and $x=0.3, x=0.7$ respectively we have that $p(0.3)=p(0.7)$. But I'm not totally sure what this demonstrates.
Can anybody help?
It is a matter of interpretation, meaning, how one links mathematical properties to aspects of the phenomenon under study.
For a Beta distribution with both coefficients equal, the expected value is equal to the mode, and it is always equal to $\frac 12$. More over both parameters equal lead to a symmetric distribution around the mean, as one can verify by checking the skewness coefficient of the distribution. So expected value = mode = median. One could then reasonably argue that since these three characteristics of the distribution coincide, the predominant belief embodied in such a prior is expressed by this point.
The values $0.7$ and $0.3$ are equally distant from the common point. This means that the value of the density function, evaluated at $0.3$ and at $0.7$ is equal. And although in continuous distributions the values of the density do not represent probabilities, in combination with symmetry around the median, it is immediate that $P(0.3-\varepsilon\le y \le 0.3+\varepsilon)$ = $P(0.7-\varepsilon\le y \le 0.7+\varepsilon)$ $\forall \varepsilon >0$. So they are "equally likely" given this prior, and in that sense the prior does not favor either one.