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calculating arithmetic mean of a continuous distribution function is straight forward.

$AM(a,b)=\int^b_a{f(x)}dx/(b-a)$

But for GM it's harder.

Note: AM is not just for a and b ie not (a+b)/2,but all the continuous values inbetween,that's what I mean by continuous distribution, i.e. not discrete.

Mini kute
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  • Posible duplicate: https://math.stackexchange.com/questions/18710/what-is-to-geometric-mean-as-integration-is-to-arithmetic-mean – user326159 Aug 21 '21 at 19:10
  • That is for generalized means, but he got without explaining...The geometric mean of yi is nothing but exp of the arithmetic mean of logyi, and this generalizes in the straightforward way to integration: exp(∫x1x0logf(x)dx∫x1x0dx). - This is the formula's proof i need – Mini kute Aug 21 '21 at 20:36
  • It is not straightforward to me as the answerer says. – Mini kute Aug 21 '21 at 20:36
  • Although its close but not a duplicate as my question is just assumed, and not discussed – Mini kute Aug 21 '21 at 20:39
  • The questioner says... "Similarly, we can find the geometric mean of f(x) by taking n samples. "?? but how is my question. – Mini kute Aug 21 '21 at 20:40
  • The problem is products are involved and integration becomes harder – Mini kute Aug 21 '21 at 20:42
  • i did it this way, i had only one side and used $e^{log a\epsilon}=1+loga\epsilon$ and i get infinte products so $(dx)^2 $ or higher powers appear now i cant integrate products. – Mini kute Aug 21 '21 at 20:53

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