Is there an analytic solution/approximation to the PDF/CDF and mean of an harmonic mean of random variables? I'm wondering about beta distributions ($\beta$) or truncated exponential distributions ($E$)?
Generally, what is the PDF/CDF and mean of
$X = \dfrac{n}{\sum_{i=1}^{n}\frac{1}{\beta_i}} $
or
$Y = \dfrac{n}{\sum_{i=1}^{n}\frac{1}{E_{i}}} $
If there is no solution to $n$ distributions, I would be happy to see for $2$ and $3$…
Best Answer
A harmonic mean is the reciprocal of the mean reciprocal of data, and is used to average rates. For example, electrical capacitance of a series of $n$ connected capacitors is $\frac{1}{n}^{th}$ of the harmonic mean, and electrical resistance of $n$ parallel connected resistors is also $\frac{1}{n}^{th}$ of its harmonic mean. In other words, for resistors in parallel, the harmonic mean resistance is the mean value of each individual resistance. Here, we use $t$ for time, but the following is true for any $x$. For continuous density functions one uses a variation of the second mean value theorem for integrals, which for support on $[\alpha\geq0,\beta]$, where $\alpha<\beta$ and where $1=\int_{\alpha }^{\beta } f(t) \, dt$, the harmonic mean residence time (H-MRT) is
\begin{equation} \text{H-MRT} := \Bigg \langle \frac{1}{T} \Bigg \rangle ^{-1}=\Bigg \langle \frac{1}{T} \Bigg \rangle ^{-1}\Bigg[\int_{\alpha }^{\beta } f(t) \, dt\Bigg]^{-1}=\Bigg[\int_{\alpha }^{\beta } \frac{f(t)}{t} \, dt\Bigg]^{-1}. \end{equation} For example, the Pareto density, whose first moment is undefined (unphysical; negative) for $0<\alpha<1$, some measure other than the mean, e.g., the harmonic mean is needed when the right tail is very heavy. From the definition above, this is \begin{equation} \text{H-MRT}_{\text{Pareto}} := \Bigg[\int_{\beta }^{\infty } \frac{\alpha \beta ^{\alpha } t^{-\alpha -1}}{t} \, dt\Bigg]^{-1} = \beta \Big(1+\frac{1}{\alpha }\Big);\; \alpha,\beta>0 \end{equation}
For the case when the mean value of a Pareto distribution is undefined, three other statistical measures are defined; median, geometric mean, and the harmonic mean, all three of which are also defined when the first moment is also defined. Of those three, the harmonic mean is arguably most useful as argued in https://arxiv.org/pdf/1402.4061; "However, the harmonic mean statistic will be relatively insensitive to the value of $\alpha _{\min }$ and will tolerate an $\alpha _{\min }$ that is close to 0. That is another point in favor of using the harmonic mean instead of one of the other statistics."
For a left truncated exponential distribution, the harmonic mean is \begin{equation} \Bigg[ \int_{\alpha }^{\infty } \frac{\lambda e^{-\lambda(x-\alpha)}}{x} \, dx \Bigg]^{-1}=\frac{e^{-\alpha \lambda }}{\lambda \Gamma (0,\alpha \lambda )};\; \alpha >0,\lambda >0 \end{equation}
The left and right truncated exponential harmonic mean is
\begin{equation} \Bigg[\int_{\alpha }^{\beta } \frac{\lambda e^{\lambda (\beta -x)}}{x \left(e^{\lambda (\beta -\alpha )}-1\right)} \, dx \Bigg]^{-1} =\frac{e^{-\beta \lambda } \left(e^{\lambda (\beta -\alpha )}-1\right)}{\lambda (\text{Ei}(-\beta \lambda )-\text{Ei}(-\alpha \lambda ))};\alpha <\beta ,\alpha \neq 0, \end{equation}
where the exponential integral function $\text{Ei}(z)=-\int_{-z}^{\infty } \frac{e^{-t}}{t} \, dt$, where the principal value of the integral is taken.
For the beta distribution the harmonic mean is
\begin{equation} \Bigg[ \int_0^1 \frac{x^{\alpha -1} (1-x)^{\beta -1}}{x B(\alpha ,\beta )} \, dx\Bigg]^{-1} =\frac{a-1}{a+b-1};\;\alpha>1,\beta>0 \end{equation}