Why Log Returns Trail Arithmetic Returns by Half the Variance

# Why is logarithmic mean equal to the arithmetic expectation less one-half its variance?


# Why is logarithmic mean equal to the arithmetic expectation less one-half its variance?












I've taken it as gospel that the following equality is true:

$$\mathbb{E}[\mu_x] = m_x - \frac{1}{2}\sigma_x^2 $$

where:

$\mathbb{E}[\mu_x]$ is the expected value of the logarithmic mean of some arbitrary Numeraire process: $\mathbb{E}[\frac{dX_T}{X}dt] \to \ln (\frac{{X}_T}{{X}_{T- \Delta t}})$;

$m_x$ is the arithmetic expectation; and,

$\sigma^2_x$ is its logarithmic variance.

I know it's true because the equality can be shown to converge for random processes as sample size becomes large, but this doesn't tell me why it's true.

Does anyone know of a mathematical proof for why this is true?

Moreover, why in common derivations of expected value is $m$ used as starting point instead of $\mu$? Is it because commonly cited short rates are just given that way?

This might be elementary, and I apologize if it is, but I haven't been able to find a purely symbolic proof.

## Answer by will (score 3, accepted)

https://quant.stackexchange.com/a/33774

So i'm kinda guessing what you really mean by the logarithmic mean - i'm guessing you mean the logarithmic average of returns - where you mean geometric average.

$$ \left( \prod_{i=0}^n a_i \right)^{\frac{1}{n}} $$

where $a_i$ are our returns. We have to make an assumption here - that your underlying is described by $\mathrm{d}S = \mu S \mathrm{d}t + \sigma S \mathrm{d}W$ such that our processes evolves as $S_{t+\mathrm{d}t} = S_t e^{(\mu - \frac{1}{2} \sigma^2)\mathrm{d}t + \sigma \mathrm{d}W_t}$. i.e. our return is $e^{(\mu - \frac{1}{2} \sigma^2)\mathrm{d}t + \sigma \mathrm{d}W_t}$ - that is what we want to compare the arithmetic and eometric averages of.

First, the arithmetic average, below:

Arithmetic Average

Ultimately i think you'll just need to go through the algebra of the logarithmic mean, and then the stuff below will give you the rest of what you want.

Unless that is, you don't actually the logarithm mean as described here.

First take the following, which i presume you know $$ \begin{align} e^x &= 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \ldots\\ e^x &= \sum\limits_{k=0}^{\infty} \frac{1}{k!}x^k \\ \end{align}$$

Now, if $X$ is a normally distributed random variable, $X \sim \mathcal{N}(\mu,\sigma)$, then we can first split out the mean, such that we have $X = \mu + Y$ where $Y\sim \mathcal{N}(0,\sigma)$, and we have $Z = e^X = e^{\mu + Y} = e^\mu e^Y$

$$ \begin{align} \mathbb{E}[Z] = \mathbb{E}[e^\mu e^Y] &= e^\mu \mathbb{E}[\sum\limits_{k=0}^{\infty} \frac{1}{k!}Y^k]\\ &= e^\mu\sum\limits_{k=0}^{\infty} \frac{1}{k!}\mathbb{E}[Y^k]\\ \end{align}$$

Where each of the $\mathbb{E}[Y^k]$ are the $k^\mathrm{th}$ moments of $Y$, and since $\mu_Y=0$, they are equal to the central moments, which are given by:

$$ \mathbb{E}[Y^k] = \begin{cases} 0 & \mathrm{if\ }k \mathrm{\ is\ odd}\\ \sigma^k (k-1)!! & \mathrm{if\ }k \mathrm{\ is\ even}\\ \end{cases}$$

So now we just have some algebra to do:

$$ \begin{align} \mathbb{E}[Z] &= e^\mu\sum\limits_{k=0}^{\infty} \frac{1}{k!}\mathbb{E}[Y^k]\\ &= e^\mu\sum\limits_{k=0, \mathrm{even}}^{\infty} \frac{1}{k!}\sigma^k (k-1)!!\\ &= e^\mu\sum\limits_{k=0, \mathrm{even}}^{\infty} \frac{1}{k!!}\sigma^k\\ \end{align} $$

For that last term, let's expand it out and see if we can see anything:

$$ \begin{align} \sum\limits_{k=0, \mathrm{even}}^{\infty} \frac{1}{k!!}\sigma^k &= 1 + \frac{1}{2}\sigma^2 + \frac{1}{8}\sigma^4 + \frac{1}{48}\sigma^6 + \ldots\\ &= \frac{1}{0!}\left(\frac{\sigma^2}{2}\right)^0 + \frac{1}{1!}\left(\frac{\sigma^2}{2}\right)^1 + \frac{1}{2!}\left(\frac{\sigma^2}{2}\right)^2 + \frac{1}{3!}\left(\frac{\sigma^2}{2}\right)^3 + \ldots \\ &= e^{\frac{\sigma^2}{2}} \end{align} $$

So we get back

$$ \begin{align} \mathbb{E}[Z] &= e^\mu\sum\limits_{k=0}^{\infty} \frac{1}{k!}\mathbb{E}[Y^k]\\ &= e^\mu e^{\frac{\sigma^2}{2}}\\ &= e^{\mu +\frac{\sigma^2}{2}}\\ \end{align} $$

Geometric Average

In the below, the sum is product/sums are over the whole time period, which then cancels out with the reciprocal power.

$$ \begin{align} \mathbb{E} \left[ \left( \prod_{i=0}^n a_i \right)^{\frac{1}{n}} \right] =& \mathbb{E} \left[ \left( \prod_{i=0}^n e^{(\mu -\frac{1}{2}\sigma^2)\mathrm{d}t + \sigma \mathrm{d}W} \right)^{\frac{1}{n}} \right] \\ =& \mathbb{E} \left[ \left( e^{\sum_{i=0}^n (\mu -\frac{1}{2}\sigma^2)\mathrm{d}t + \sigma \mathrm{d}W } \right)^{\frac{1}{n}} \right]\\ =& e^{-\frac{1}{2}\sigma^2} \mathbb{E} \left[ \left( e^{ \mu + \sigma \tilde{X}} \right) \right]\\ \end{align} $$ Which is essentialyl just the above, but with the extra negative of half the variance - i.e the difference you're looking for.

It happens because the mean you're looking at is the geometric mean of the expected returns, which are essentially the arithmetic means of the stochastic process.

Is that what you were after? Or you want it more fleshed out?

## Answer by David Addison (score 0)

https://quant.stackexchange.com/a/33976

@Chris-Degnen made me aware of the following work through which uses Itô's lemma from page 54 of Computational Financial Mathematics using Mathematica. I adopt it to answer the question as follows:

Assume stock prices evolve according to following SDE derived from a standard Wiener Process:

$dS_t = a S_t \,dt + \sigma S_t dB_t$

where: $B_t$ is a Wiener process.

Since movements of $S_t$ should be proportional to the current price, we investigate the proporties of the natural logarithm of price. Let:

$S_0 = p$; and, $z = \ln(S)$

Noticing that:

$(dS)^2 =\sigma^2 S^2\,dt$

we use Itô's lemma to derive:

$dz = d \ln(S) = \frac{1}{S}\,dS+ \frac{1}{2} \frac{-1}{S^2}(dS^2)$

$= a\, dt + \sigma \, dB - \frac{-1}{2S^2}\sigma^2S^2\,dt = (a-\frac{1}{2}\sigma^2)dt +\sigma \,dB$

Since $\sigma$ is constant, the above SDE for $z$ can be solved explicitly by applying stochastic integration, yielding:

$z_t = t (a - \frac{\sigma^2}{2})+ \sigma (B_t-B_0)+z_0 = t (a -\frac{\sigma^2}{2}) +\sigma B_t + \ln(p)$

This in turn gives the solution of the original stock price SDE. By taking its exponent, we now have:

$\Large{S_t = e^{z_t} = e^{(a-\frac{\sigma^2}{2})t + \sigma B_t + \ln(p) } = S_0 e^{(a-\frac{\sigma^2}{2})t + \sigma B_t }}$

Since "the population geometric mean is the population median for the lognormal distribution", and the geometric mean is interchangeable with the logarithmic expectation (through simple transformation), the logarithmic expectation is satisfied by the expected value of the median curve for $S_t$.

The derivation using Itô's lemma is essentially a stochastic chain rule for expressing Jensen's inequality which states that the median and the mean differ from a convex function.

The median for $\frac{dS_t}{S_t}$ is given by:

$\mathbb{E}[\ln(\frac{S_t}{S_0})] = (a-\frac{\sigma^2}{2})t$,

(or: $\mathbb{E}[\frac{dS_t}{S_t}] = (a-\frac{\sigma^2}{2})\,dt$)

while the mean value is given by:

$\ln(\frac{S_t}{S_0}) ={a\,t }$

(or: $ \frac{S_t+dS\,dt}{S_t} ={a }$).

Therefore, the expectation is equal to the mean value less one-half its variance.