Talk:Harmonic mean

message: Is this true?: "in an electrical circuit you have two resistors connected in parallel, one with 40 ohms and the other with 60 ohms, then the average resistance is 48 ohms" 210.224.218.13 15:51, Nov 20, 2003 (UTC)
:The total resistance is given by the harmonic mean divided by the number of resistors, not just the harmonic mean. The resistance should then be around 24 ohms. 128.84.45.4 20:55, Jul 06, 2004 (UTC)
::Exactly. The average resistance is 48 ohms, while the equivalent resistance is 24 ohms. (I don't think "average resistance" is a term any physicist/electrician/electrical engineer would use, since it would have different meanings for resistors in parallel as for resistors in series, but that's how this article is using the term, and I think that's fine.) Ruakh 02:32, 8 Mar 2005 (UTC)
replies: text:  Shouldn't that read arithmetic mean resistance rather than average?. Also resistance should strictly be impedance Smidoid 12:18, 23 Feb 2008 (UTC)
replies: text:  See my section on "taking the average of ratios",  The ratios are voltages/currents which is the definition of resistance.  In parallel, the voltages are all the same, but the currents are different.  So the problem falls in the case of denominators unknown which implies use of the harmonic mean.  We obtain the "ratio of averages" the voltage average is fixed at V.  But what we need is the ratio of V to total current so we must multiply the average current denominator by N (the number of resistors) to get total current (same as dividing the "average of ratios" by N).   Thus we get the value of the equivalent resistance V/I(total).  Note that we can do this without assigning any numerical values to voltage and currents.David S. Lawyer 10:23, 23 November 2015 (UTC)
thread_title:  Electrical Resistance? 

message: What if the two numbers are negative? The harmonic mean of -1 and -2 is 2/(-1 + -(1/2)), which equals 2/(-1.5) or 2 * -(2/3) or -4/3, which is larger than the arithmetic mean, which is -1.5 (as compared to the geometric mean being -1.3......). However, the geometric mean would actually be the square root of 2... messed up for negative numbers, I suppose. ugen64 03:26, Dec 16, 2003 (UTC)
:The article explicitly restricts the notion of harmonic mean to the positive reals. If it didn't, then we'd have the bigger issue of finding the harmonic mean of -1 and 1 (since 2/(1/-1 + 1/1) = 2/(-1 + 1) = 2/0, which is not defined). Ruakh 02:32, 8 Mar 2005 (UTC)
thread_title:  Negative Numbers? 

message: added the definition when frequency is not unary. Nova67 16:16, 12 June 2005 (UTC)
replies: text: I don't understand. Are you saying that you added such a definition? If so, there's really no need to mention it here; the article history will show that. Are you saying that such a definition needs to be added? Please explain what you mean. Ruakh 14:27, 15 Jun 2005 (UTC)
thread_title:  stats comment 

message: I removed the following text that a non-registered user inserted into the article, as its placement and internal organization were confusing and ill-suited to the article:
:===An example===
:* An experiment yields the following data: 34,27,45,55,22,34 To get the harmonic mean
:# How many items? There are 6. Therefore n=6
:# What is the sum on the bottom of the fraction? It is 0.181719152307
:# Get the reciprocal of that sum. It is 5.50299727522
:# To get the the harmonic mean multiply that by n to get <U>33.0179836513</U>
:--
If anyone wants to rescue it, feel free. It needs to be rewritten, given lead text, and placed appropriately (perhaps in a sidebar; the "ExampleSidebar" template could be used). Ruakh 19:48, 8 September 2005 (UTC)
thread_title:  Removed example. 

message: This statement warrants a little qualification:
if for half the distance of a trip you travel at 40 miles per hour and for the other half of the distance you travel at 60 miles per hour, then your average speed for the trip is given by the harmonic mean of 40 and 60, which is 48; that is, the total amount of time for the trip is the same as if you traveled the entire trip at 48 miles per hour.
it may be true, of course, but not independent of the premises of the problem which are rather subtely expressed only. It is equally true that:
if for half the time of a trip you travel at 40 miles per hour and for the other half of the time you travel at 60 miles per hour, then your average speed for the trip is given by the arithmetic mean of 40 and 60, which is 50; that is, the total distance you covered is the same as if you traveled the entire trip at 50 miles per hour.
While this isn't central to a discussion of harmonic means, as I discovered recently people are easily lulled, by reading such concise statements, into the mistaken belief that the harmonic mean is THE correct manner of averaging independent speed measures. And this, it is not. It is A correct measure given certain premises concerning the problem. Bwechner 05:00, 25 November 2005 (UTC)
replies: text: Thank you for pointing that out.  I've changed the 50 to 40.  Dbfirs 15:47, 2 October 2019 (UTC)
thread_title:  Average speed 

message: I don’t believe the pool-draining example highlights a special case, and I suspect the example might warrant clarification or removal.  As this is not my area of expertise, I submit the following for your review:
The rates provided (4 and 6) are “hours per pool”.  The harmonic mean, then, is the average number of hours it would take one pump to empty the pool (4.8 hours).  With twice as many pumps working at that average rate, it would of course take only half the time, or 2.4 hours.  —Preceding unsigned comment added by Jenburton (talk • contribs) 00:31, 25 August 2009 (UTC)
thread_title: ”Draining Pool” example

message: :I added this part and I want to thank the person who did the editing. It really flows well. 
:This example only work with the double angle tangent identity. I tried tan A = tan (3x) and the formula failed to work out. It will fail to work out for any tan(nx) where n /= 2.  
Opinionhead (talk) 20:09, 10 May 2011 (UTC)
thread_title: Trigonometric example

message: I just deleted a statement that cannot be correct, or is at least not properlyy explained/defined. DrMicro wrote that
"If all the real variables in the set are > 0 then H ≥ M2 / m where m is the minimum in the set."Jean WH (1984) The harmonic mean and other necessary conditions for stochastic dominance. J Finance 39 (2) 527
If this were the case (and M is the maximum as defined in the main article above) then the H would be greater than the maximum, meaning that it no longer qualified as a mean. If someone knows the correct statement of this inequality as stated in the paper (it's too late for me to be reading a maths paper right now) feel free to re-add and correct.
I'll message DrMicro about it too.
Cheers,
JPBrod (talk) 23:28, 21 April 2012 (UTC)
replies: text: Thank you. It must have been a typo. I have read the reference and I cannot find this inequality in this paper. Im afraid I do not have an explanation for its inclusion. Well spotted. DrMicro (talk) 11:14, 22 April 2012 (UTC)
thread_title:  Omission - Relationship with other means. 

message: Currently the subsection "Theoretical values" of the section "Statistics" reads
:Theoretical value
:The variance of the harmonic mean isZelen M (1972) Length-biased sampling and biomedical problems. In Biometric Society Meeting, Dallas, Texas
::  \operatorname{Var}\left( \frac { 1 } { x } \right) = \frac { m \left[ \operatorname{E}( 1 / x - 1 ) \right] } { n m^2 } 
:where m is the arithmetic mean of the reciprocals, x are the variates, n is the population size and E is the expectation operator. Asymptotically E(1 / x) is distributed normally.
:The mean of the sample m is also distributed normally with variance s2.
::  s^2 =  \frac  { m [ \operatorname{E}( 1 / x - 1 ) ] }{ m^2 n } ''
Unfortunately the citation for this is a paper at a 1972 conference, so there's no way to check the source to confirm that it's been copied correctly (or even that it was right in the first place since it appears not to have been refereed or published). Moreover, it's been changed in the present article from what it said earlier.
(1) This looks suspicious to me because the variance of the harmonic mean is given as the same expression as the variance of the arithmetic mean of the reciprocals.
(2) It looks suspicious because surely the original source would not have left something with m not cancelled from the numerator and denominator.
(3) It says the variance of the harmonic mean is denoted as var(1/x) where "x are the variates". That doesn't make sense to me -- if 1/x is the harmonic mean, then x is the mean of the reciprocals of the variates, not "the variates". So this expression, even if correct, may refer to the variance of the reciprocals of the variates rather than the variance of the sample harmonic mean.
Does anyone have a source for the true mean and variance of the sample harmonic mean? Duoduoduo (talk) 15:08, 5 June 2013 (UTC)
thread_title:  "Theoretical values" section is confused 

message: Anyone mind if I add a musical (harmonic ratio) example, since that's where the name of the mean comes from?  Thanks! -- Michael Scott Cuthbert (talk) 21:26, 19 June 2014 (UTC)
thread_title:  Harmonics / history / music? 

message: The article needs a new section entitled "Proof that the harmonic mean is less than the geometric mean", preferably with a source. Or, if there is enough to say, there could be a new article Inequality of harmonic and geometric means, analogous to the article Inequality of arithmetic and geometric means. Loraof (talk) 16:27, 13 December 2015 (UTC)
replies: text: Really, this is another special case of Generalized mean inequality. Boris Tsirelson (talk) 19:48, 11 June 2016 (UTC)
thread_title:  Need proof of inequality 

message: The [reference to "Fairness Opinions", on page 217] presents an example with two equal-sized positions. This is rarely the case in real life and the weighted average should be recommended instead.
Suppose you have a $100-billion firm with a P/E of 20 ($5 billion in earnings).
Average this with a $1-billion firm with a P/E of 1000 ($1 million in earnings).
The aggregate price/earnings of this combination should be dominated by the large firm.
=== Incorrect: unweighted harmonic mean ===
::  P/E =  \frac  {1}{1/20 + 1/1000} \approx 19.6 
=== Correct: ratio of aggregates, equivalent to market-cap-weighted harmonic mean ===
::  P/E =  \frac  { \$100\ billion + \$1\ billion }{\$5\ billion + \$1\ million} \approx 20.2
11:50, 20 April 2016‎ 198.180.162.5 (talk)‎
:::: The finance section is correct, but under the assumption that it is an equally weighted portfolio; otherwise, you need to use the weighted harmonic mean. So you are right, I have made the change to clarify that. Also note that in your example of the calculation of the unweigthed harmonic mean, you forgot to divide the denominator by 2, which of course leads to a wrong P/E.7804j (talk) 09:28, 29 September 2016 (UTC)
thread_title:  Finance example is [[Not even wrong]] and subsection should be removed or heavily revised 

message: ::  P/E =  \frac  { \$100\ billion + \$1\ billion }{\$5\ billion + \$1\ million} \approx 20.2
11:50, 20 April 2016‎ 198.180.162.5 (talk)‎
:::: The finance section is correct, but under the assumption that it is an equally weighted portfolio; otherwise, you need to use the weighted harmonic mean. So you are right, I have made the change to clarify that. Also note that in your example of the calculation of the unweigthed harmonic mean, you forgot to divide the denominator by 2, which of course leads to a wrong P/E.7804j (talk) 09:28, 29 September 2016 (UTC)
thread_title:  Correct: ratio of aggregates, equivalent to market-cap-weighted harmonic mean 

message: This image is from the article "Harmonic Mean:"
It appears to me, that the length of the purple line H is actually TWICE the harmonic mean for the following reason:
H = (G*G) / A = (a*b) / ( (a+b) / 2 ) = 2 * ( 1 / ( (1/a) + (1/b) ) )
Ergo: 
H is TWICE as big as the harmonic mean. 
Is this true? Please correct me, if I am wrong.
73.4.14.51 (talk) 14:19, 6 May 2016 (UTC)
replies: text: Yes, you are wrong, just at the end of your calculation: H=\frac{2}{(1/a)+(1/b)}, that is, 1/H is the arithmetic mean (rather than the sum!) of 1/a and 1/b. Boris Tsirelson (talk) 15:53, 6 May 2016 (UTC)
thread_title:  Image label incorrect? 

message: The section "In trigonometry" says
::In the case of the double-angle tangent identity, if the tangent of an angle A is given as , then the tangent of 2A is the product of (1) the harmonic mean of the numerator and denominator of tan A and (2) the reciprocal of (the denominator minus the numerator of tan A).
::In general, the double-angle formula can be written as
:::\tan(2A) = \frac{2ab}{a+b}\cdot\frac1{b-a}
and then gives a numerical example. I find this pointless—all it says is that the double angle formula does not use the harmonic mean, but can be expressed by starting with the harmonic mean and then adjusting it for the fact that the harmonic mean doesn't give the right answer. So I'm deleting it. Loraof (talk) 16:31, 12 June 2016 (UTC)
thread_title:  "In trigonometry" section 

message: Dr. Anderson has reviewed this Wikipedia page, and provided us with the following comments to improve its quality:
We hope Wikipedians on this talk page can take advantage of these comments and improve the quality of the article accordingly.
Dr. Anderson has published scholarly research which seems to be relevant to this Wikipedia article:
*Reference : Anderson, James E & Neary, J Peter, 2014. "Revenue Tariff Reform," CEPR Discussion Papers 9838, C.E.P.R. Discussion Papers.
ExpertIdeasBot (talk) 17:56, 27 June 2016 (UTC)
replies: text:  The weighted harmonic mean seems to be the correct way to average P/E to me, but I agree that the section was misleading because it was referring to the unweighted harmonic mean. I have clarified the section and added an example. The weighted arithmetic average is definitely not the right way to calculate the P/E of the portfolio though, so I think you are wrong on that point. 7804j (talk) 14:23, 29 September 2016 (UTC)
thread_title:  Dr. Anderson's comment on this article 

message: The statistics section mentions, without introduction, the harmonic means of several continuous probability distributions. There should be a definition of the continuous case somewhere. I suppose it would be something like integral(f(x))/integral(f(x)/x), where f(x) is a PDF that supplies the weights and the integrals are both taken over the whole domain. Elias (talk) 12:05, 27 November 2020 (UTC)
thread_title:  Continuous definition?