intuitive interpretation of canonical parameterization of beta distribution For exponential family, e.g. Beta distirbution, someone argues that the canonical parameterization is better than the traditional $Beta(\alpha,\beta)$ way. The canonical parameters are defined as 
$n^{(0)}=\alpha+\beta$ 
 and 
$y^{(0)}=\frac{\alpha}{\alpha+\beta}$.
 Note the upper index ^(0) means prior.
So the canonical parameters can be interpreted as:


*

*$y^{(0)}$ is the prior expected value.

*$n^{(0)}$ is the prior confidence (strength) in this $y^{(0)}$.


But my problem is, the range of $n^{(0)}$ is $[0,+\infty]$, how can I intutively express my strenght of belifs in $y^{(0)}$ on a scalar $[0,+\infty]$ (I am thinking say I am 90% confident the prior mean is $y^{(0)}$, then $n^{(0)}$ should be in the range $[0,1]$). So I cannot see any advantages of such canonical parameteration or am I missing something? Or is there anyway to rescale $n^{(0)}$ into $[0,1]$ to have a good intutive interpretation of it?
Hope I am making sense..
 A: I think this reparameterization makes the most sense when we're using the Beta distribution as the conjugate prior of the Binomial distribution.
Let's say we are trying to estimate the unknown parameter $p$ of a Binomial distribution. If we start from a uniform prior, then our initial belief is that $p \sim Beta(1,1)$. Now, if we observe $a$ failures and $b$ successes, we can update that belief to $p \sim Beta(1+a, 1+b)$.
In general, given a prior $Beta(a_0, b_0)$ and new information in the form of $a_1$ observed successes and $b_1$ observed failures, we can update that to $p \sim Beta(a_0 + a_1, b_0 + b_1)$.
So far all we've done is recap the definition of "conjugate prior" and give intuition of what the two shape parameters in the usual $Beta(\alpha, \beta)$ parameterization mean. So now consider: if I observed $a$ successes and $b$ failures, how many total observations did I make? That's right: $n = a + b$. And what was the sample proportion in my latest batch of observations? $p = \frac{a}{a+b}$. So your "cannonical parameters" can be interpreted as "sample size" and "best estimate sample proportion" respectively.
Now, its fairly intuitive that the "strength" of your beliefs should be captured by the total number of observations. Regardless of whether we're talking about priors, updates, or posteriors, the "strength" of our belief should be a function of $n = \alpha + \beta$. We can also observe a little more formally that 
$$Var[Beta(\alpha, \beta)] = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)} = \frac{\alpha\beta}{n^2(n+1)} \approxeq \frac{p(1-p)}{n^2} $$
This expression decreases with increasing $n$: in other words, the probability distribution of our estimate of the Binomial parameter $p$ is getting narrower and narrower as $n$ increases, which is similar to saying the "strength" of our belief is increasing.

A: The parameterisation you are looking at is a way of expressing the beta prior distribution based on the prior mean $y_0$ and the prior strength $n_0$.  The latter is also commonly regarded as the number of "pseudo-data points" attributable to the prior belief, for reasons that will become obvious below.
To see the use of this parameterisation more clearly, consider a Bayesian conjugate analysis where we observe a binomial random variable with a beta prior for its probability parameter.  Using the parameterisation in your question, this yields the conjugate model:
$$\begin{equation} \begin{aligned}
\text{Prior} \quad \quad \quad \theta &\sim \text{Beta}(n_0, y_0), \\[10pt]
\text{Sample} \quad \quad X|\theta &\sim \text{Bin}(n,\theta), \\[6pt]
\text{Posterior} \quad \quad \theta|x &\sim \text{Beta} \Big( n_0+n, \frac{n_0}{n_0+n} \cdot y_0 + \frac{n}{n_0+n} \cdot \frac{x}{n} \Big).
\end{aligned} \end{equation}$$
As can be seen from this model, the prior begins with $n_0$ pseudo-data points and then the data adds $n$ actual data points to this, giving posterior strength $n_0+n$.  The posterior mean is updated as a weighted average of the prior mean $y_0$ and the observed sample proportion $x/n$, where the weights depend on $n_0$ and $n$.
