B How is the harmonic mean affected by additional data points?

AI Thread Summary
The discussion centers on the impact of adding new data points on the harmonic mean (HM) of an existing dataset. Initially, a dataset of eight points has a harmonic mean of 22.86, which increases to 23.08 after adding two additional points. It is noted that the harmonic mean is influenced by the relative values of new points compared to the existing mean; points below the HM decrease it, while points above increase it. The conversation concludes that the combined harmonic mean of two datasets can be calculated using a weighted formula, confirming that the overall HM will shift based on the new data's harmonic mean. Understanding these relationships allows for predictions about how the HM will change with additional data points.
Feynstein100
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We have a collection of 8 discrete data points. They are:
10, 20, 30, 20, 30, 40, 30, 40
In increasing order:
10, 20*2, 30*3, 40*2
The harmonic mean of this data series is 22.86
I read on Wikipedia that the harmonic mean is skewed towards the smaller values i.e. smaller values will affect the HM more than larger values. So I thought that if we add 2 additional data points 20 and 30, our HM would be even smaller. And yet, when I calculated the HM of this new data series with 10 points:
10, 20*3, 30*4, 40*2
it turned out to be 23.08 i.e. higher than the previous case. Why did that happen?
One of our new points was lower than the HM whereas the other was higher. I thought the HM would be more skewed toward the lower value and thus would bring the overall mean down. Ah is it because the second datapoint was much higher than the HM?
In general, I'm interested in the question of how adding new datapoints will affect the HM of the existing data series.
We're not changing the endpoints, they remain constant. So any new point added will lie somewhere inside the bounds of the data series. In our example, that's 10 and 40.
So I think the answer is quite simple. If New point < HM, it lowers the HM. If New point > HM, it increases the HM.
It seems quite straightforward for adding one datapoint but what if we add multiple? In essence appending another data series to the existing one. Can we predict in advance if the new HM will be higher or lower?
 
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It looks like a simple quantitative question. One point at a time will give a predictable result. More than one - results?
 
Feynstein100 said:
It seems quite straightforward for adding one datapoint but what if we add multiple? In essence appending another data series to the existing one. Can we predict in advance if the new HM will be higher or lower?
Yes. If the harmonic mean of the new points is lower, the HM of all the points will be lower, If it is higher the new HM will be higher. Easy to see from HM of \frac {1}{\sum {\frac{1}{a_i}}} and \frac {1}{\sum {\frac{1}{b_i}}} is \frac {1}{ \sum {\frac{1}{a_i}} + \sum {\frac{1}{b_i}} }
 
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mathman said:
It looks like a simple quantitative question. One point at a time will give a predictable result. More than one - results?
That's............kind of what I'm asking 😂
 
willem2 said:
Yes. If the harmonic mean of the new points is lower, the HM of all the points will be lower, If it is higher the new HM will be higher. Easy to see from HM of \frac {1}{\sum {\frac{1}{a_i}}} and \frac {1}{\sum {\frac{1}{b_i}}} is \frac {1}{ \sum {\frac{1}{a_i}} + \sum {\frac{1}{b_i}} }
Thanks for the reply. I worked it out myself and turns out, the combined harmonic mean Hc of two harmonic means H1 with m items and H2 with n items will be
Hc =(m + n)/(m/H1 + n/H2)
i.e. the weighted harmonic mean of H1 and H2. And by the general property of all means, Hc will be somewhere between H1 and H2. Thus, if the new HM is lower, the overall HM will be lower as well. And if the new HM is higher, the overall HM will be higher as well.

Btw your third formula has a mistake. It should be 2/(sum of inverses), not 1/(sum of inverses) and that's a special case of when both series have the same number of items. The general formula is the weighted harmonic mean.
 
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