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.


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