# Questions on continuously compounded return vs long term expected return

I have reading a paper from Oliver Grandville on long term expected return. I am trying to reconcile what I am reading in that paper vs what I see under "Application to Stock Market" in Kelly criterion page on Wikipedia.

https://en.wikipedia.org/wiki/Kelly_criterion

In his paper, he define the following:

1. $R_{t-1,t} = \frac{S_t-S_{t-1}}{S_{t-1}}$ as the yearly rate of return compounded once per year
2. $X_{t-1,t} = 1 + R_{t-1,t} = \frac{S_t}{S_{t-1}}$ as the yearly dollar return
3. $log \left( X_{t-1,t} \right) = log \left(\frac{S_t}{S_{t-1}}\right)$ as the yearly continuously compounded rate of return
4. $E(X_{t-1,t}) = E(1+R_{t-1,t})$ as the expected value of the yearly dollar return and $V(X_{t-1,t})$ as its variance

He derives that given $log \left( X_{t-1,t} \right) = log \left(\frac{S_t}{S_{t-1}}\right) = \mu + \sigma N(0,1)$ (ie log returns follow normal distribution of mean $\mu$ and variance $\sigma^2)$, we must have that $E(X_{t-1,t}) = exp \left( \mu + \frac{\sigma^2}{2}\right)$.

My questions are:

1. For the wiki article vs this article, the wiki one says expected log return is $R_s = \left( \mu - \frac{\sigma^2}{2}\right)t$. This seems similar to the Grandville article in that if you take $log(E(X_{t-1,t})) = \mu + \frac{\sigma^2}{2}$, but its not exactly the same. Is the because the assumption the wiki article makes is different (ie the stock price moves like Brownian motion?)
2. In the derivation of the max of the expected value in wiki page. Shouldnt it be $G(f) = fE(stock) + (1-f)E(bond)=f*(\mu - \frac{\sigma^2}{2}) + (1-f)*r$, but the vol term is different ($f^2$ vs $f$). Why is that?
3. Finally, the last line of the wiki article mentions that "Remember that $\mu$ is different from the asset log return $R_s$. Confusing this is a common mistake made by websites and articles talking about the Kelly Criterion." My understanding is that $\mu$ is the expected of the log returns $E(log(X_{t-1,t}))$ whereas $R_s$ is the log of the expected returns $log(E(X_{t-1,t}))$ and that this distinction is important because $E(log(X))$ and $log(E(X))$ are not always equal. Is this understanding correct?
• I think Granville says $(\mu+\frac{\sigma^2}{2})$ is the annual expected arithmetic return, the Wiki says $(\mu-\frac{\sigma^2}{2})$ is the annual expected logarithmic return. Arithmetic return is $\frac{S_t-S_{t-1}}{S_{t-1}}$, logarithmic return is $\log \frac{S_t}{S_{t-1}}$. But I find the notation in both articles and your post extremely confusing and possibly inconsistent, so I am not sure. – Alex C Jul 8 '17 at 19:00
• Isnt $\mu$, not $\left( \mu - \frac{\sigma^2}{2} \right)$, the annual expected log returns? – qwer Jul 9 '17 at 18:57
• Also for 2) why is it $f^2$ vs $f$ for the vol term – qwer Jul 9 '17 at 18:57

There are two communities that parametrize the random variables of the returns and levels differently. And unfortunately, they both use the same notation. One community uses $\mu$ to denote the mean of the logarithm of the level. The other uses $\tilde{\mu}$ to denote the logarithm of the mean of the level.

Yes, the order in which you apply transformations to random variables and take their expectation is very important. Only in the case of affine/linear transformations does the order not matter. In your case, the logarithm is a nonlinear function, and so interchanging the order matters very much.

In the statistics community, where it is more common to deal with single random variables and not continuous time stochastic processes, we would probably say $X$ follows a Lognormal distribution with parameters $\mu$ and $\sigma^2$. This is written $X \sim \text{Lognormal}(\mu, \sigma^2)$. However, these parameters do not denote the mean and variance of the random variable $X$. Instead, the denote the mean and variance of $R = \log X$, which is a Normally distributed single period return. For us the expected level is $E[X] = \exp[ \mu + \sigma^2/2].$

When people talk about SDEs, which I'm less familiar with, they mention that the expected level after one period is $e^{\tilde{\mu}}$. Here they are using the parameter to denote the logarithm of the mean of the level. The relationship between the two parametrizations is as follows: $$\left[ \begin{array}{c} \tilde{\mu} \\ \tilde{\sigma}^2 \end{array}\right] = \left[ \begin{array}{c} \mu + \frac{\sigma^2}{2} \\ \sigma^2 \end{array}\right],$$ where the left hand side is the set of parameters used by the SDE community.

Make the substitutions and you can see that both communities are saying the same thing about returns: $$E[R] = E[\log X] = \mu = \tilde{\mu} - \frac{\tilde{\sigma}^2}{2}.$$

If you make the substitutions again, you can see they are also saying the same thing about the levels: $$E[X] = \exp(\mu + \sigma^2/2) = \exp(\tilde{\mu}).$$

This is about the second question only.

In the theory of SDEs the parameter $\tilde{\mu}$ is sometimes called the instantaneous rate of return, that is the rate of return over an interval of time so small that the randomness can be neglected (i.e. in the limit $\Delta t\to 0$) . However, over a finite interval of time this rate of return does not apply, the randomness in the process causes the expected rate to be less than this, and over one time unit (taken to be a year for illustration) the rate of return is $\tilde{\mu}- \frac{1}{2}\sigma^2$

Now turning to the Wikipedia article. What is the yearly expected return for an investment that combines Stock in the proportion $f$ and Tbills in the proportion $(1-f)$?

The Stock has instantaneous rate of return $\tilde{\mu}$ and variance $\sigma^2$

The Tbill has rate of return $r$ and variance 0.

The combination of the two has instantaneous rate of return $f \tilde{\mu} + (1-f) r$ and has variance $f^2 \sigma^2$.

Applying the formula above we find that the yearly (or one time unit) rate of return for the blended investment is $f\tilde{\mu} + (1-f) r-\frac{1}{2}f^2 \sigma^2$. Which is the expression given in the Wikipedia article.

The mysterious "minus half the variance" correction term that makes its appearance is called the "Ito correction" and really requires a knowledge of stochastic calculus to explain.