Expected geometric return

[nikogeorgiev]

Dear David,

In regards to calculating expected geometric return, why do we substract (sigma^2)/2 from the mean?

Regards
Niko

 

[David Harper CFA FRM]

Hi Niko,

Right, great question. As Hull/McDonald say/imply, we are not really required to define it thusly (without the variance/2 is also a valid drift) owing to the "ambiguity of expected return" (that's Hull talking). The short answer is that we are referring here to a lognormal price distribution; i.e., S(t) is lognormal under our convenient assumption that LN(St/S0) is normal. Note the lognormal is not symmetrical: its mean > median by 1/2 the variance. We can refer to an expected return that drifts toward the mean (i.e., without - variance/2) or the median (i.e., includes the "erosion of return by volatility" with - variance/2 to get the lower E[S(t)] and the one that we just happen to be referring to!) of the distribution.
… so please note, the term "mean" is very tricky (in this GBM context of expected[random lognormal variable]) b/c it can refer to a return and/or a distribution and it might be better to just refer to drift (mu) to acknowledge that there can be two drifts, one toward the lognormal median and another to the necessarily greater lognormal mean

The longer answer is given by McDonald (his Chapter 18 is the only good explain I have ever read) where I will paraphrase McDonald p 597, but I may be no better than the source (b/c maybe your next question is "why doesn't the mean add variance/2 instead of the median subtracting variance/2?"):

E(St) = S(0)*EXP(uT); let u = drift
E[ S(0) * EXP[ (u - sigma^2/2)T + sigma^2*T/2 ]; i.e., not *really* necessary but allows for:
=E[ S(0) * EXP [ (u - sigma^2/2)T * EXP [ sigma^2*T/2 ]

...now the 2nd EXP() can capture the stochastic (z)
Mean of lognormal (x) where x~N(mu, sigma^2) is given by: EXP(mu + sigma^2/2)
See http://en.wikipedia.org/wiki/Log-normal_distribution (the mean of lognormal is actually "+ variance/2"; again, the mean of lognormal > median by variance/2)

(z) is random standard normal such that:
E[EXP(x)] where mean (X) = 0, where variance (x) = sigma^2*T (i.e., if volatility = SQRT(T)*sigma, then variance = T*sigma^2) is given by: EXP(sigma^2*T/2) = E(EXP[sigma*SQRT(T)*z)

E(St) = E[EXP(u - sigma^2/2)T) * E[EXP(sigma^2/2)]
…. And since E[EXP(sigma^2/2) = EXP[sigma*sqrt(T)*z]
St = EXP(u - sigma^2/2)T * EXP[sigma*sqrt(T)*z] = EXP[(u - sigma^2/2)T + sigma*SQRT(T)*z]

Hope that sheds some light ... David

 

[nikogeorgiev]

Hi David,

You were very thorough as always. Your explanation did in fact shed some light.
Thank you very much for that

Regards
Niko

 

[nikogeorgiev]

Hi David,

Further to your excellent explanation if assume confidence of 99% we can say that the lowest future return is S(t) = S(0)exp(mu-2.33sigma) why? I can't seem to work out the mathematics here

regards
Niko

 

[David Harper CFA FRM]

Hi Niko,

I think my math above is only mean/median, so neither could I infer your equation directly.

You are showing, quite directly, geometric returns:
Arithmetic 99% worst (our typical) = worst 99% S(t) = S(0) - S(0)*[-mu + 2.33* sigma]; i.e., S(0) - absolute VaR
The Geometric analog = worst 99% S(t) = S(0)*EXP[mu - 2.33* sigma] = S(0)*(1 - EXP[mu - 2.33* sigma])

David