Arithmetic mean of log returns vs geometric mean

[hunden]

Hello,



The relation between the geometric mean and arithmetic mean of a time series is that geometric mean is always lower than or equal to the arithmetic mean, and equal if and only if the data points in the time series are equal, right?



I wonder if there is an equal relation between the geometric mean of a time series and the arithmetic mean of the logarithms of that same time series? Is there a way to establish a relation between those two ways of computing the mean?

[pj]

well,

[img]/User%20Files/126/Latex-Equation-10959.gif[/img]

so I gather it's equality if you

take a log of geometric mean Chew

[silverside]

EDIT reread question - I answered the wrong question - oops!

[aaron]

In addition to pj's mathematical answer, I would add the economic intuition that if you keep a fixed investment amount, you earn the arithmetic average return, while if you invest an initial amount and never add or subtract, you earn the geometric average return. That is:



A: Keeps $100 in the stock market, adjusting at the end of every month to get up to or down to $100. If the monthly returns are r_1, r_2, . . ., r_n, her total profit will be $100 times the sum of the r's and her average monthly profit will be $100 times the average of the r's.



B: Puts $100 in the stock market and forgets about it (and pays no taxes or fees, and reinvests dividends). His total profit is [(1+r_1)*(1+r_2)*. . .*(1+r_n) - 1]*$100. His compound average monthly growth rate is the geometric mean of the (1 + r's) minus 1.



Who will have more profit? B's first order profit is equal to A's, $100 times the sum of the r's. B's second order profit is the sum of all pairwise products of the r's, that is, r_i * r_j. If the r's are all positive or all negative, that is a positive amount, and B makes more money than A. But if some r's are positive and some are negative, it can be a negative amount. Say n=2, r_1 = 0.5 and r_2 = -0.4. A makes $10 and B loses $10.

[pj]

> I would add the economic intuition

A nice example.

[intradaybill]

Aaron: "In addition to pj's mathematical answer, I would add the economic intuition that if you keep a fixed investment amount, you earn the arithmetic average return, while if you invest an initial amount and never add or subtract, you earn the geometric average return..."



Aaron, thanks for the example. You are a good teacher and your students (if you are teaching) are very lucky to have you.



I remember in the mid 1990s, funds used to report arithmetic returns based on fixed inital capital because they assumed that the investor would like to withdraw some money every year for income and keep the inital capital in. Now they have switched to compounded returns. Which is more appropriated do you think? (you might have answered that already).

[aaron]

As you say, the obvious answer is arithmetic return for an investor who withdraws gains and tops up losses (or in general, who runs a balanced portfolio according to some algorithm); CAGR for the investor who never withdraws.



It's kind of like the choice between Sharpe Ratio and Treynor Ratio. If the fund represents most of the risk of your portfolio, then you probably care about geometric returns and Sharpe ratio. But if it's a small part of a diversified portfolio, arithmetic returns and Treynor ratio make more sense.



Arithmetic returns emphasize portfolio selection skills, does the manager make good bets? Geometric add in risk management skills, does the manager size bets well? If you size the bets yourself by changing the amount invested, you care more about arithmetic returns (although you still would like the manager to size his relative bets correctly). If you rely on the manager to size the bets, especially if you're locked up, geometric matters.



Another point is whether to use time-weighted or dollar-weighted returns. Do you care what return stream the manager produced in theory, or how many dollars he made or lost for investors.



By the way, you can score at geek parties by saying, "The difference between geometric and arithmetic returns is Ito's lemma."

[pj]

> By the way, you can score at geek parties



Remind me NEVER get invited at your parties.

Cool