B Is there a relationship between the harmonic mean and the standard deviation?

AI Thread Summary
The discussion explores the relationship between the harmonic mean (HM) and the arithmetic mean (AM) in a sequence of numbers, highlighting that while the AM remains constant, the HM decreases as the values become more extreme. It emphasizes that the HM is more sensitive to lower values, which skews its calculation downward. The conversation also delves into the potential connection between HM and standard deviation (σ), suggesting that increased fluctuations in values could lead to a lower HM. Participants seek a rigorous mathematical expression for HM in relation to σ, indicating that the expectation of the distribution plays a role in this relationship. The thread concludes with a clarification on the distinction between HM and the expectation in the context of deriving a formula.
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I was doing a thought experiment on the harmonic mean. Let's say we have a sequence of 4 numbers:

100, 110, 90, 100

The arithmetic mean (AM) of this sequence is 100 but the harmonic mean (HM) is 99.4975.
Let's imagine that the initial and final values remain constant but the middle ones get more and more extreme.

100, 120, 80, 100

The AM is still 100 but the HM is now 97.9592

100, 130, 70, 100

AM = 100, HM = 95.2880

and so on

I think I understand why this happens. The AM is symmetric, if you will. So an equal movement above the mean will cancel out an equal movement below it. However, the HM is more skewed toward the lowest value in the sequence. As such, it only takes one low value to bring the HM down. And with each step, we have a lower minimum and thus a lower HM. This is basically a crude verbal proof. Is there a more rigorous, mathematical one?

I'm also interested to know if there's a relationship between the HM and the fluctuation (I guess the arithmetic standard deviation is a good way to represent this). I mean, there has to be, as higher fluctuations mean higher likelihood of lower values and thus a lower HM, right?

Basically, HM = f(σ)?
 
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Imagine a variable x with a distribution N(μ,σ2), i.e. a Normal distribution with mean μ and variance σ2. Think of x as μ + δ where δ = N(0,σ2), and (for simplicity) σ << μ. Can you derive an (approximate) expression for HM, where HM = 1/E(1/x)?
 
mjc123 said:
Imagine a variable x with a distribution N(μ,σ2), i.e. a Normal distribution with mean μ and variance σ2. Think of x as μ + δ where δ = N(0,σ2), and (for simplicity) σ << μ. Can you derive an (approximate) expression for HM, where HM = 1/E(1/x)?
Sounds interesting. But what exactly is E in this case? 😅
 
The expectation (mean of the distribution).
 
mjc123 said:
The expectation (mean of the distribution).
Ah okay but isn't that literally the definition of the harmonic mean? Not sure what we'd get by proving it. I mean, I'm trying to find a relationship between HM and σ, not HM and E.

Also, in the example, the E remained constant whereas the HM kept decreasing, exactly illustrating my point. Although, oh wait. That was the AM, not E(1/x). My bad 😅
 
I wasn't asking you to prove HM = 1/E(1/x), but to use that definition to derive an expression for HM in terms of μ and σ.
 
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