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I've been wondering about a simple question for few minutes, but I can't find any relevant questions or references by googling.

I want to minimize KL divergence from $p\in \Delta(A)$ where A is some discrete set to compact set $\Theta\subset \Delta(A)$ where $p \notin \Theta$. That is, I want to solve $$\min_{q\in\Theta}D(p||q)$$ Since $A$ is a discrete set, $\Theta, \Delta(A)$ is just a subset of $\mathbb{R}^{|A|}$.

Intuitively, since KL divergence measures "distance", I've conjectured that this would be minimized with $q$ closest to $p$ in Euclidean distance.

I'm not sure if I can prove it, or even if it is right at all. Does the following hold?

$$arg\min_{q\in\Theta}D(p||q)=arg\min_{q\in\Theta}d(p,q)$$ $d(p,q)$ is the standard Euclidean metric.

Any answers/counter examples/references would be helpful. Thank you!

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  • $\begingroup$ KL divergence is not even s distance. Unlike Euclidean Distance, it does not satisfy the triangle inequality, and is not symmetric with respect to its arguments. $\endgroup$ Commented Feb 13, 2022 at 16:08
  • $\begingroup$ If you define your 'Euclidean Distance' as $\sum_{i}^{} (p_i-q_i)^2$, then yes. Both achieve minimum when distributions are equal. $\endgroup$ Commented Feb 13, 2022 at 19:43

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What you are doing here is essentially an information projection. Your conjecture would thus mean that information projection is equivalent to an $\ell_2$-projection.

There is no reason why this should be the case. Take the following example:

Suppose we want to project $p= [\frac12, \frac16, \frac13]$ onto the family of distributions $\Theta = \{q_{\theta} = [\theta, \frac12-\theta, \frac12]: \theta \in (0, 1/2) \}$.

If you try to compute the corresponding projections, you will find that (solved numerically):

  • $\min_{q \in \Theta}D(p\|q) = \min_{\theta \in (0, \frac12)} D(p\| q_\theta) \approx 0.0566$ is attained at $\theta =\frac38$ (check here).
  • $\min_{q \in \Theta}\|p-q\|_2 = \min_{\theta \in (0, \frac12)} = \min_{\theta \in (0, \frac12)} \|p-q_\theta\|_2 \approx 0.2041$ is attained at $\theta = \frac{5}{12}$ (check here).
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