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This paper is showing how KL divergence can be minimized by matching the expected values of the sufficient statistics. More precisely, For any distribution p of the exponential family with pdf: $$ p_{\theta}(x) = \frac{1} {Z(\theta)}\exp(\theta^T \phi(x))$$ the distribution $$p_{θ∗} $$ which minimises the Kullback-Leibler divergence, $$ KL (p||p_{θ∗})$$ over the exponential family with natural statistic $\phi$ is implicitly given by: $$E_{p_{\theta}*(x)} [\phi(x)] = E_{p_{(x)}} [\phi(x)]$$

I need help to show that this is true for the particular case of a normal-gamma distribution. So I need to make the same proof that is on the paper but for the special case of Normal gamma.

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We already have that the sufficient statistics are $$ T_1 = \ln T,\; T_2 = T,\; T_3 = TX,\; T_4=TX^2. $$

You should select the approximating density $q(X,T)$. Then, by Theorem 1, you simply equate the true expectations with the expectations under the approximating distribution. That is, $$ \begin{align} E_q(\ln T) &= \psi(\alpha) - \ln(\beta)\\ E_q(T) &= \frac{\alpha}{\beta},\\ E_q(TX) &= \frac{\alpha\mu}{\beta},\\ E_q(TX^2) &= \frac{\alpha\mu^2}{\beta} + \frac{1}{\lambda}. \end{align} $$ Until you specify what your approximating density is, this is all I can say for this problem. The cited note is explaining assumed density filtering where the posterior distribution is the true distribution and the approximating density $q$ is an exponential family.

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  • $\begingroup$ I don't totally understand what you explained but the density that I want to approximate is another normalgamma $\endgroup$
    – sam
    Mar 15, 2022 at 17:37
  • $\begingroup$ That would be trivial. The KL divergence will obviously be minimized when the parameters are identical between the two normal-gamma distributions. $\endgroup$
    – Daeyoung
    Mar 15, 2022 at 17:48
  • $\begingroup$ So here P and P* are normal gamma but they don't have the same parameters $\endgroup$
    – sam
    Mar 16, 2022 at 13:14
  • $\begingroup$ @SamiHadouaj They will once you’ve minimized the KL divergence. $\endgroup$
    – Daeyoung
    Mar 16, 2022 at 13:39
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    $\begingroup$ @SamiHadouaj It is impossible to "find" a distribution that minimizes the KL divergence because the class of distributions is infinite. A family of distributions must be specified and that will give you a handle on the minimization step. And it can't be another normal gamma because the KL divergence is minimized when they're the same distribution, i.e., same parameters. $\endgroup$
    – Daeyoung
    Mar 16, 2022 at 19:29

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