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Suppose we have $n$ random variables distributed normally with the same mean and different variances. Suppose we know these variances. Which will be the variance of the marginal distribution induced by a finite mixture of $n$ normal distributions with the same mean and different variances?

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    $\begingroup$ The variance of the sum the $n$ random variables? Or the variance of the marginal distribution induced by a finite mixture of $n$ normal distributions with the same mean and different variances? $\endgroup$ – Andrew M Oct 13 '16 at 21:05
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    $\begingroup$ The second one. Sorry it should be easy, but can not figure out... $\endgroup$ – German Demidov Oct 13 '16 at 21:07
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    $\begingroup$ You may want to rephrase your question to clarify exactly what it is you're asking. $\endgroup$ – dsaxton Oct 13 '16 at 21:46
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    $\begingroup$ Based on your comments, it is by no means obvious that you understand where your descriptions are not being understood. The complete calculation for the variance of a random variable that has a normal mixture density, that is, the density is of the form $$f(x) = \sum_{I=1}^n p_i f_i(x), \ \ \ p_i > 0,\ \ \ \sum_ip_i = 1,$$ is given in this answer on math.SE. Please read it and decide if that is what you wanted to know. If not, perhaps you can clarify further what you understand by "finite mixture of random variables" $\endgroup$ – Dilip Sarwate Oct 13 '16 at 22:33
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    $\begingroup$ @GermanDemidov then it seems that your problem is much more complicated then your description says. $\endgroup$ – Tim Sep 22 '17 at 9:37

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