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It is known that mixture of Gaussians are dense in the set of all distribution functions. A 1-dimensional Gaussian has the following density: $$ \frac{1}{\sqrt{2\pi \sigma^2}} e^{-\frac{(\omega-\beta)^2}{2\sigma^2}} $$ Now if we just apply a Taylor linearization ($e^a \approx 1+a$) to the Gaussian density we will have $$ \frac{1}{\sqrt{2\pi}}\frac{2\sigma}{2\sigma^2+(w-\beta)^2} $$ which can be considered as another density (if I am not mistaken one needs to normalize by $\sqrt{\pi}$) If we plot them together (for mean 3.3 and std 1.4) they look like following: enter image description here

It is obvious that quadratic denominator distribution is heavier-tailed. Now my question is the following: Can we claim a similar denseness as in the case of mixture of Gaussians? i.e., can I approximate any density arbitrarily well using mixtures of "quadratic denominators"?

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  • $\begingroup$ Your "quadratic denominator" distribution is actually called a Cauchy distribution. [Note that you got the normalization factor wrong.] Can you explain the sense in which you're measuring how well the density is approximated? $\endgroup$ – Glen_b Mar 14 '16 at 8:38
  • $\begingroup$ @Glen_b thanks for your prompt response. I am not sure about the approximation error measure. The denseness argument is quoted from this paper which indeed quotes Kostantinos, N. Gaussian mixtures and their applications to signal processing. $\endgroup$ – YBE Mar 14 '16 at 8:41
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    $\begingroup$ You might like to consider that any mixture of Cauchy distributions will not have finite mean... it might make it unsuitable for many purposes even if the approximation of the density is adequate by some measure. $\endgroup$ – Glen_b Mar 14 '16 at 12:15

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