Usually, I find the units of the mean and the standard deviation of a distribution to be (quite obviously) the same.

Can anyone come up with a really simple explanation (for MBA students, some of whom are essentially “poets”, taught the absolute minimum of statistics), of the seeming paradox specifically as regards the units involved, of the subtraction of half the square of SD in calculating the geometric rate of return:


I should add that I have already checked the origin of this expression, via the application of Ito’s Lemma, and its relationship to the difference between the arithmetic and geometric mean in relation to the lognormal distribution, including a variety of Wikipedia entries – and even asked a couple of experts to explain, but none has been able to make clear the answer to this apparently simple question about the units involved.

Answer What are the units of the variables appearing in a standard stochastic differential equation for a Wiener process? comes close, but doesn't quite answer it sufficiently for my target audience.

The best source on this seem to be http://www.timworrall.com/eco-30004/bscholes.pdf in which Tim Worrall makes clear on p. 17 that this correction factor is in fact an approximation

  geometric mean ≈ arithmetic mean – 0.5 variance

But I'd rather not give the students an unsatisfying “hand-waving” answer that "It's a small number and the seeming difference in the units doesn’t really matter at the end of the day."

Help greatly appreciated.

  • 3
    $\begingroup$ You are right mean and standard deviation share the same units. But remember that, (1) you are here looking at returns i.e. price ratios, hence dimensionless quantities. (2) Using the notations you provide, the mean of the (arithmetic) return over the period $\Delta t$ indeed writes $r\Delta t$ (dimensionless) while the standard deviation is $\sigma \sqrt{\Delta t}$ (dimensionless). Hence the units of $r$ are $1/\Delta t$ and the units of $\sigma=1/\sqrt{\Delta t}$. This shows that the units of $r$ are indeed the same as that of $\sigma^2$. $\endgroup$ – Quantuple Mar 11 '17 at 15:29
  • $\begingroup$ The rest I can make sense of (as I do in tedious detail below), but how do you conclude that the standard deviation is σ√Δt (dimensionless)? $\endgroup$ – iSeeker Mar 11 '17 at 19:55
  • $\begingroup$ The term $r-0.5\sigma^2$ appears when transforming the SDE for arithmetic returns: $dS_t/S_t = r dt + \sigma dW_t $ into the SDE for geometric (or log-) returns $d\ln(S_t) = (r-0 5\sigma^2)dt + \sigma dW_t$. Should you disrectise the fist SDE you would have $\Delta S_t/S_t = r \Delta t + \sigma \Delta W_t$. Yet by definition of a Wiener process the increment $\Delta W_t \sim N(0, \Delta t)$. Hence the result for the mean and variance of the arithmetic return $\Delta S_t/S_t = (S_t-S_{t-\Delta t})/S_t$ $\endgroup$ – Quantuple Mar 12 '17 at 0:22
  • $\begingroup$ Thanks again, Quantuple, for focusing me on the point where it arises - I'll try to translate that into a form that's convincing for our "poets" and be able to convince them that the units of both terms in r−0.5σ^2 are the same. $\endgroup$ – iSeeker Mar 12 '17 at 11:37
  • $\begingroup$ My pleasure. Good luck :) $\endgroup$ – Quantuple Mar 12 '17 at 11:39

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