Volatility Adjustment for Compounding Normally Distributed Returns
Summary
The document asks whether normally distributed simple returns need the same volatility adjustment used when translating lognormal growth into a continuously compounded rate. It explains that a lognormal price model includes a variance term because exponentiating a normal random variable raises its expected value, then compares this with compounding simple returns. A two-period example shows that returns of 9% and 11% compound to slightly less than two returns of 10%, despite sharing the same arithmetic average.
The author reports a Monte Carlo observation: averaging annual returns arithmetically overstates the realized compound growth, with the gap increasing as volatility rises. The simulation suggests the adjustment is close to, but not exactly, half the variance. The document poses the correction as an open question and does not provide a derivation, a general formula, or details of the simulation. Its comparison is therefore useful as a prompt about arithmetic versus geometric return, but not as a settled adjustment rule.
Key ideas
- Arithmetic average returns can exceed the compound growth rate when returns vary across periods.
- The gap between arithmetic and realized compound returns grows with volatility.
- The familiar half-variance adjustment for lognormal growth may not directly apply to normally distributed simple returns.
- The reported simulation is suggestive but does not establish a general correction formula.
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Full text
# If you have normally distributed returns, shouldn't you have the same adjustment factor as lognormally distributed?
# If you have normally distributed returns, shouldn't you have the same adjustment factor as lognormally distributed?
We know that when using lognormal returns, the number you need to plug in is not the apparent return, but $\mu-\sigma^2/2$ because what you really have is, in essence, (1) a deterministic growth of $\mu$, plus (2) a zero-mean, specified variance, normally distributed random element. In other words, if
$$s=s_0e^{(\mu +N(0,\sigma^2))t}=s_0e^{\mu t}e^{N(0,t\sigma^2))} $$
Then, since $e^{anything}>0$, that second term introduces a bias. Thus, for lognormal distributions, the appropriate 'return' to plug in, if you want it to square with simple compounding, is $\mu-\sigma^2/2$, so the expected value makes the $\sigma^2/2$ wash out. (Otherwise, $E[s_t]=s_0e^{t(\mu+\sigma^2/2)})$
Now lets think about simple-minded compounding of normal returns.
If a bond delivers 10% annually, then in two years \$100 will become \$121. But if returns are binomial, with 9% and 11%, and you get two 'matched' years, even though the (arithmetic) average is 10% the real return is smaller (instead of \$121 you get \$120.99, pretty clearly). If the spread is wider you get an even lower number.
My question is this: assume simple, normal annual returns. The result is the same. If you have returns that are distributed as $N(\mu, \sigma^2)$, then as $\sigma^2$ grows, the actual return must be smaller than $\mu$. That is why two back-to-back returns of 10% don't get you as much as one period of 11% and one period of 9%.
So, is the correction term the same - i.e., $-\sigma^2/2$? If periodic returns are normally distributed as $N(\mu-\sigma^2/2, \sigma^2)$, will $E[s_t]=e^{\mu t}$???
NOTE: Since I wrote this, I have done a Monte Carlo simulation. Indeed, if one calculates the average return per year by averaging the returns of each year (the naive arithmetic approach), instead of calculating and using
$$(\frac {p_{final}}{p_{initial}})^{\frac{1}{num\_preiods}}-100\%$$
Then, with normal returns and any reasonable volatility, there is indeed a gap between what return you 'expect' to achieve and what you do. However, it does not look like the correction term is $\sigma^2/2$, though it is close to it.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.