# Mean and variance of the Beta distribution using identities of exponential families

I was studying the part of exponential families from Statistical Inference (George Casella, Roger L. Berger) and they give the following definition of an exponential family:

$$f(x|\pmb{\theta}) = h(x)c(\pmb{\theta})\exp\left( \sum_{i=1}^k w_i(\pmb{\theta})t_i(x)\right)$$

and the following identities to calculate some expectations and variances from the exponential families:

1. $$\text{E}\left( \sum_{i=1}^k \frac{\partial w_i(\pmb{\theta})}{\partial \theta_j}t_i(X)\right) = - \frac{\partial}{\partial\theta_j}\log{c(\pmb{\theta}})$$

2. $$\text{Var}\left( \sum_{i=1}^k \frac{\partial w_i(\pmb{\theta})}{\partial \theta_j}t_i(X)\right) = -\frac{\partial^2}{\partial \theta_j^2} \log{c(\pmb{\theta}}) - \text{E}\left(\sum_{i=1}^k \frac{\partial^2 w_i(\pmb{\theta})}{\partial \theta_j^2}t_i(X) \right)$$

Now, in exercise 3.30 they ask to calculate the mean and variance of the beta distribution, but, the only thing i was able to get was:

1. $$\text{E}(\log(X)) = \psi(\alpha) - \psi(\alpha + \beta)$$
2. $$\text{Var}(\log(X)) = \psi_1(\alpha) - \psi_1(\alpha + \beta)$$

where $$\psi$$ and $$\psi_1$$ are the digamma and trigamma functions, respectively. I get this since the pdf of a beta distribution can be written as:

$$f(x|\alpha, \beta) = \frac{1}{\text{B}(\alpha, \beta)}\exp\left( (\alpha -1)\log(x) + (\beta -1)\log(1-x)\right)$$

so the $$t_i$$ functions are logarithms. I don't know if i am missing something, or maybe there is another parameterization of the distribution so i can get $$E(X)$$ and $$\text{Var}(X)$$ instead, so any help will be appreciated.

• The identities are for the natural statistics, which is $\log(X)$ for the Beta. – Xi'an Dec 3 '20 at 16:56