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Can you find the geometric mean of a continuous variable, in a similar why to find the mean of a continuous variable.

To find the regular mean of between to points on a function you do $\frac{\int_{a}^{b} f(x) \,dx}{b-a} $

If there is 1 point that lands in $0$ it would all become all zero so it might not make any sense to take a geometric mean of a continuous variable.

Mean $$\frac{\sum_{n=1}^{k} a_k}{n}$$ Geometric Mean $$(\prod_{n=1}^{k} a_k)^{\frac{1}{n}}$$

  • @saulspatz I think so sorry couldn't find it earler –  Apr 25 '21 at 15:13
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    The mean of a continuous random variable $X\sim f$ is $$E(X)=\int_{\mathbb R}xf(x)\mathrm{d}x$$ What you wrote, namely $\frac{1}{b-a}\int_a^bf(x)\mathrm{d}x$, is the average value of a function $f$, which is something different. I think a natural definition for the geometric mean of a continuous random variable is $$\exp\Big[{E\big(\ln(X)\big)}\Big]$$ Of course, this assume $X$ is supported on $(0,\infty)$ – Matthew H. Apr 25 '21 at 15:16

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