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If $X\sim\mathcal{LN}({\mu,\sigma^2})$, then $\mathrm{E}[X]=e^{\mu+\sigma^2/2}$. My question is: what right do we have to add a mean and variance together? If $X$ has physical dimensions, then the expression $\mu + \sigma^2/2$ is incoherent. So what gives?

kjetil b halvorsen
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wvoq
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    $\mu$ is a *logarithm*: what units of measure do you suppose it has? – whuber Nov 26 '13 at 16:57
  • What "right"? This is math, not civics or law! The "right" we have is that when you work through the integral, that's the right answer! 2) The $\mu$ and $\sigma^2$ are the mean and variance of the log of $X$. Think about what that implies about their units of measure. – jbowman Nov 26 '13 at 16:58
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    Here is a puff for Finney, D.J. 1977. Dimensions of statistics. _Applied Statistics_ 26: 285-9, which may be accessible to you at http://www.jstor.org/stable/2346969 or otherwise. That is an excellent little tutorial paper. – Nick Cox Nov 26 '13 at 19:06
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    +1 It's sensible to be concerned about units, but your characterization of the meaning of the parameters of the lognormal is incorrect. – Glen_b Nov 26 '13 at 21:36

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Tongue in cheek: this sum is allowed only in free countries where this is actually considered as a basic human right.

Seriously: you are not adding a mean and a variance since $\mu$ is not the mean and $\sigma^2$ is not the variance of a lognormal variate. The mean is, as you said $\mathrm{E}[X]=e^{\mu+\sigma^2/2}$, and the variance is $\mathrm{Var}[X]=(e^{\sigma^2}-1)e^{2\mu+\sigma^2}$. Then $\mathrm{E}[X]$ and $\sqrt{\mathrm{Var}[X]}$ have the same units. You can read more on this in the following link:

http://en.wikipedia.org/wiki/Variance#Units_of_measurement

Teco
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    I appreciate your answer, but I'm not claiming that $\mu$ and $\sigma^2$ are the mean and variance of $X$; they're the mean and variance of $\log X$, the associated normal random variable. If $\mu$ and $\sigma^2$ have dimensions, which they do if $X$ has dimensions, then the formula for $$ is not dimensionally sound, which raises my eyebrows. And maybe that's 'just the way it is', but it at least seemed worth asking about. – wvoq Dec 13 '13 at 21:58
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I believe that $\mu$ and $\sigma$ don't have the same units in this case, but $\mu$ and $\sigma^2$ do.

Take for example the stochastic differential equation of a geometric brownian motion used to model stock prices in the Black–Scholes model:

$ dS_t = \mu S_t\,dt + \sigma S_t\,dW_t $

The stock price $S_t$ follows a log-normally distributed random variable.

Call $D_i$ the unit used to measure the variable $i$. Dimensional homogeneity implies that $D_\mu D_t = 1$, so if your measuring the time in days, the unit of $\mu$ is $1/day$.

Dimensional homogeneity also implies that $D_\sigma D_{W_t} = 1$. $W_t$ is a Wiener process with variance $t$, so $D_{W_t}^2=D_t$. Therefore, $D_\sigma=1/\sqrt{D_t}$. If you are measuring the time in days, the unit of $\sigma$ is $1/\sqrt{day}$.

Note that this interpretation is consistent with the formula used to convert daily volatility $\sigma_{daily}$ to annualized volatility $\sigma_{annual}$, assuming that a year has 252 trading days: $\sigma_{annual} = \sigma_{daily}\sqrt{252}$.

toliveira
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