# Return to Answer

 2 added 1 character in body edited Jun 20 '16 at 23:52 Ira Nirenberg 111 bronze badge I think a simple way to answer the question would be: If the mathematical structure is xy = k (an inverse relationship between variables) and you're looking for an average, then you need to use the harmonic mean--which amounts to a weighted arithmetic mean--consider Harmonic average = 2ab/(a+b) = a(b/a+b) + b(a/(a+b) For example: dollar cost averaging falls into this category because the amount of money you're investing (A) stays fixed, but the price per share (P) and number of shares (N) vary (A = PN). In fact, if you think of an arithmetic average as a number equally centered between two numbers, the harmonic average is also a number equally centered between totwo numbers but (and this is nice) the "center" is where the percentages (ratios) are equal. That is: (x - a)/a = (b -x)/b, where x is the harmonic average. If the mathematical structure is a direct variation y = kx, you use the arithmetic mean--which is what the harmonic mean reduces to in this case. I think a simple way to answer the question would be: If the mathematical structure is xy = k (an inverse relationship between variables) and you're looking for an average, then you need to use the harmonic mean--which amounts to a weighted arithmetic mean--consider Harmonic average = 2ab/(a+b) = a(b/a+b) + b(a/(a+b) For example: dollar cost averaging falls into this category because the amount of money you're investing (A) stays fixed, but the price per share (P) and number of shares (N) vary (A = PN). In fact, if you think of an arithmetic average as a number equally centered between two numbers, the harmonic average is also a number equally centered between to numbers but (and this is nice) the "center" is where the percentages (ratios) are equal. That is: (x - a)/a = (b -x)/b, where x is the harmonic average. If the mathematical structure is a direct variation y = kx, you use the arithmetic mean--which is what the harmonic mean reduces to in this case. I think a simple way to answer the question would be: If the mathematical structure is xy = k (an inverse relationship between variables) and you're looking for an average, then you need to use the harmonic mean--which amounts to a weighted arithmetic mean--consider Harmonic average = 2ab/(a+b) = a(b/a+b) + b(a/(a+b) For example: dollar cost averaging falls into this category because the amount of money you're investing (A) stays fixed, but the price per share (P) and number of shares (N) vary (A = PN). In fact, if you think of an arithmetic average as a number equally centered between two numbers, the harmonic average is also a number equally centered between two numbers but (and this is nice) the "center" is where the percentages (ratios) are equal. That is: (x - a)/a = (b -x)/b, where x is the harmonic average. If the mathematical structure is a direct variation y = kx, you use the arithmetic mean--which is what the harmonic mean reduces to in this case. 1 answered Jun 20 '16 at 22:45 Ira Nirenberg 111 bronze badge I think a simple way to answer the question would be: If the mathematical structure is xy = k (an inverse relationship between variables) and you're looking for an average, then you need to use the harmonic mean--which amounts to a weighted arithmetic mean--consider Harmonic average = 2ab/(a+b) = a(b/a+b) + b(a/(a+b) For example: dollar cost averaging falls into this category because the amount of money you're investing (A) stays fixed, but the price per share (P) and number of shares (N) vary (A = PN). In fact, if you think of an arithmetic average as a number equally centered between two numbers, the harmonic average is also a number equally centered between to numbers but (and this is nice) the "center" is where the percentages (ratios) are equal. That is: (x - a)/a = (b -x)/b, where x is the harmonic average. If the mathematical structure is a direct variation y = kx, you use the arithmetic mean--which is what the harmonic mean reduces to in this case.