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I'm fitting some data with a pdf that is the mixture of two gaussians with the same mean (this is, the sum of the density functions). The weight of each gaussian after the fit is $w_{\rm core}$ and $w_{\rm tail}$,

$$ f_{\rm Global} ( \mu, \sigma_{\rm core}, \sigma_{\rm tail} ) = w_{\rm core} N(\mu, \sigma_{\rm core}) + w_{\rm tail} N(\mu, \sigma_{\rm tail}) $$

, where $N(\mu, \sigma)$ is the normalized normal distribution

Is there an analytical expression to compute the variance of the global pdf (as a function of $w_{\rm core}$, $w_{\rm tail}$, $\sigma_{\rm core}$ and $\sigma_{\rm tail}$)?

Answer

$$ \sigma_{\rm Global}^2 = w_{\rm core} \sigma_{\rm core}^2 + w_{\rm tail} \sigma_{\rm tail}^2 $$

This is derived in this post for the general case $\mu_{\rm core} \neq \mu_{\rm tail}$. I'm quoting it in case somebody searches directly for the specific case (like me) and does not find the other answer.

Albercoc
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    By "sum" do you mean *mixture* (as suggested by your formula) or a true *sum* (in the sense of random variables)? If it's the former, your question is answered at https://stats.stackexchange.com/questions/16608 and if it's the latter, the sum (as is well known and shown in many threads here) is Gaussian again and you can find its variance by adding the variances of the two random variables. – whuber Mar 11 '21 at 17:24
  • I meant mixture. Thanks for the comment, unfortunately I cannot post the answer properly. – Albercoc Mar 12 '21 at 14:28
  • I understand that one can read off the answer to this question from the answer of @whuber in the other post, but that is just a more general case. I don't think that the definition of "duplicate" in stackexchange applies for a "simplified case" of another question. – Albercoc Mar 13 '21 at 18:57
  • When all you have to do is plug numbers into a formula that has already been given and explained in another thread, that thread is a duplicate. – whuber Mar 13 '21 at 18:59

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