# Harmonic mean of random variables

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 $3$...

• Is your question a bit confused perhaps? I just took it to ask about what the continuous variable harmonic mean is for those distributions. – Carl Mar 31 '17 at 23:25
• Such problems usually need to be solved on a case by case basis. The question is too general to be answered. But if I were trying for a given random variable $V$, I would first find the pdf of $Z =1/V$ , then try to find the pdf of the sample mean for $Z$, and then the pdf of the inverse of the latter. – wolfies Apr 1 '17 at 15:20
• @wolfies U R correct in that not every $f(x)$ has a harmonic mean, but, that is easy to define, i.e., when $\int_{\alpha }^{\beta } \frac{f(x)}{x} \, dx$ is not integrable. – Carl Apr 2 '17 at 20:50
• researchgate.net/publication/… – kjetil b halvorsen Sep 30 '17 at 23:25

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

$$\text{H-MRT} := \Bigg \langle \frac{1}{T} \Bigg \rangle ^{-1}=\Bigg \langle \frac{1}{T} \Bigg \rangle ^{-1}\Bigg[\int_{\alpha }^{\beta } f(t) \, dt\Bigg]^{-1}=\Bigg[\int_{\alpha }^{\beta } \frac{f(t)}{t} \, dt\Bigg]^{-1}.$$ 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 $$\text{H-MRT}_{\text{Pareto}} := \Bigg[\int_{\beta }^{\infty } \frac{\alpha \beta ^{\alpha } t^{-\alpha -1}}{t} \, dt\Bigg]^{-1} = \beta \Big(1+\frac{1}{\alpha }\Big);\; \alpha,\beta>0$$

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 $$\Bigg[ \int_{\alpha }^{\infty } \frac{\lambda e^{-\lambda(x-\alpha)}}{x} \, dx \Bigg]^{-1}=\frac{e^{-\alpha \lambda }}{\lambda \Gamma (0,\alpha \lambda )};\; \alpha >0,\lambda >0$$

The left and right truncated exponential harmonic mean is

$$\Bigg[\int_{\alpha }^{\beta } \frac{\lambda e^{\lambda (\beta -x)}}{x \left(e^{\lambda (\beta -\alpha )}-1\right)} \, dx \Bigg]^{-1} =\frac{e^{-\beta \lambda } \left(e^{\lambda (\beta -\alpha )}-1\right)}{\lambda (\text{Ei}(-\beta \lambda )-\text{Ei}(-\alpha \lambda ))};\alpha <\beta ,\alpha \neq 0,$$

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

$$\Bigg[ \int_0^1 \frac{x^{\alpha -1} (1-x)^{\beta -1}}{x B(\alpha ,\beta )} \, dx\Bigg]^{-1} =\frac{a-1}{a+b-1};\;\alpha>1,\beta>0$$

• I'm not sure what you mean by resistors adding by harmonic mean. If anything, they add as the reciprocal of the sum of reciprocals. The harmonic mean off by a factor of $n$. – Neil G Apr 6 '17 at 2:45
• @NeilG Thanks for noticing, making the change now. – Carl Apr 6 '17 at 4:31
• @NeilG Well, stated in opposite direction, resistance of $n$ resistors in parallel is $\frac{1}{n}^{th}$ of the harmonic mean, which latter already has a name. I think the important point is that when the measurement system is reciprocal one has to use reciprocation. For example, the total electrical conductance (in siemens or mhos) of parallel resistors is the sum of the conductances of each resistor. But the total resistance of series resistors is the sum of the resistances (in ohms). – Carl Apr 6 '17 at 5:21
• Sure, but it's annoying that you have to say "$n$ times the harmonic mean" of the resistances. It would be nicer to say "the harmonic sum" of the resistances, in my opinion. – Neil G Apr 6 '17 at 5:32
• that's the harmonic series. Anyway, you can define anything you like. – Neil G Apr 4 '18 at 11:55