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I am looking at question 1b of the following notes: http://www.gatsby.ucl.ac.uk/teaching/courses/ml1/asst1.pdf

In 1a, I have shown that the Beta distribution has a density that can be written in the form $g(\boldsymbol{\theta}) f(\textbf{x}) e^{\phi(\boldsymbol{\theta})^T T(\textbf{x})}$ where:

$g(\boldsymbol{\theta}) = \frac{1}{B(\alpha,\beta)}$

$f(x) = \frac{1}{x(1-x)}$

$\phi(\boldsymbol{\theta}) = \begin{pmatrix} \alpha \\ \beta \end{pmatrix}$

$T(x) = \begin{pmatrix} \log{x} \\ \log{(1-x)} \end{pmatrix}$

Now, according to this thread, I can calculate the expectation of the sufficient statistic, $E[T(x)]$, by rewriting the density as

$f(\textbf{x}) e^{\phi(\boldsymbol{\theta})^T T(\textbf{x})} e^{-A(\boldsymbol{\theta})}$

and then using $E[T(x)]_i = \frac{A'(\boldsymbol{\theta})_i}{\phi'(\boldsymbol{\theta})_i}$

To do this, we define $A(\theta) = - \log{g(\theta)} = \log{B(\alpha,\beta)}$ but then I become uncertain of what to do. My attempt so far looks like:

$A'(\theta) = \frac{1}{B(\alpha,\beta)} \begin{pmatrix} B_\alpha \\ B_\beta \end{pmatrix}$ where $B_\alpha = \frac{\partial B}{\partial \alpha}$ etc. I also find $\phi'(\theta) = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

I'm pretty sure I've made a mistake with the calculation of $phi'(\theta)$ since the vector of ones seems wrong, and yet $\phi'(\theta) = \frac{\partial}{\partial \theta_i} \phi(\theta) = \begin{pmatrix} \frac{\partial}{\partial \alpha} \\ \frac{\partial}{\partial \beta} \end{pmatrix} \phi(\alpha, \beta)$. Is this going to become a 2x2 matrix?

I have managed to get this technique to work for distributions such as binomial and Poisson where the sufficient statistic is a scalar. I am getting confused in this case where we have a vector of sufficient statistics.

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    $\begingroup$ Please make the question self-contained rather than expecting the reader to follow a web link. $\endgroup$ – Xi'an May 8 at 8:14
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The thread you provide only operates for one-dimensional exponential families. In general, when the exponential family is written wrt its natural parameter: $$f(x) e^{\theta^T T(x)} e^{-A(\boldsymbol{\theta})}$$ the mean of the associated (sufficient) statistic is $$\Bbb E_\boldsymbol\theta[T(X)] = \nabla A(\boldsymbol{\theta})$$ Since this is the case for the Beta distribution, namely $\phi( \boldsymbol\theta)=\boldsymbol\theta$, $$\nabla A(\boldsymbol{\theta})=\frac{1}{B(\alpha,\beta)} \begin{pmatrix} \frac{\partial B(\alpha,\beta)}{\partial\alpha} \\ \frac{\partial B(\alpha,\beta)}{\partial\beta} \end{pmatrix}$$

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  • $\begingroup$ It seems that we cannot always have $\phi(\theta) =\theta$ however. For example, with the gamma distribution, I find $\phi(\theta) =( \alpha - 1, - \beta) ^T$. Under these circumstances, if I follow through the algebra, we would have $(\nabla \phi) E[T(x)] = \nabla A$. Is that correct? $\endgroup$ – user11128 May 26 at 9:14
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    $\begingroup$ You can always reparameterise $\phi(\theta)$ as the default $\theta$. $\endgroup$ – Xi'an May 26 at 10:22

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