Log-Return Drift and the Itô Correction in Price Models

# Modelling returns in the real world measure with or without drift


# Modelling returns in the real world measure with or without drift












What I would like to discuss is the following. I don't think that this is a pure duplicate, so I would be happy about comments:

On one hand it is reasonable to model log-returns as Gaussian: $$ \log(S_{t+\Delta}/{S_t}) = \sigma B_{\Delta t} \tag{1} $$ with a Gaussian random variable $B_{\Delta t} \sim N(0,\Delta t)$.

On the other hand as e.g. in the calculations of the equivalent Gaussian volatility for PRIIPS we model $$ S_{t+\Delta} = S_t \exp \left( - \sigma^2/2 \Delta + \sigma \left( B_{t+ \Delta t} - B_{t } \right) \right), $$ and thus $$ \log(S_{t+\Delta}/{S_t}) = - \sigma^2/2 \Delta + \sigma \left( B_{t+ \Delta t} - B_{t } \right), \tag{2} $$ which leads to a non-centered Gaussian.

I know that $(2)$ is the natural model if we want to use the SDE $$ dS_t = \sigma S_t dB_t, $$ whose discretized version is $$ S_{t+ \Delta t} - S_{t } \approx \sigma S_t \left( B_{t+ \Delta t} - B_{t } \right), $$ which can be reformulated as $$ \frac{S_{t+ \Delta t} - S_{t }}{S_t} \approx \sigma \left( B_{t+ \Delta t} - B_{t } \right). $$

So how does all this fit together? In risk management we often assume that log-returns are Gaussian $(1)$ and the regulator of PRIIPS assumes that arithmetic returns are approximately Gaussian? How can we interpret the correction term intuitively in $(2)$?

EDIT: Hopefully doing the right maths:

In setting (A) which gives us equation (1) we have the following stochastic model: $$ S_{t + \Delta t} = S_t \exp \left( \sigma (B_{t + \Delta t} - B_t) \right) $$ then for the log return $R_t$ we have $$ R_t = \log\left(S_{t + \Delta t}/S_t \right) = \sigma (B_{t + \Delta t} - B_t). $$ Then $R_t$ has a Gaussian distribution with expectation $0$ and variance $\sigma^2 \Delta t$.

Setting (B): $$ S_{t + \Delta} = S_t \exp \left( -\frac{\sigma^2}{2} \Delta t + \sigma (B_{t + \Delta t} - B_t) \right) $$ and get for $\log(S_{t + \Delta}/S_t)$ again something Gaussian with expectation $-\frac{\sigma^2}{2} \Delta t$ and variance $\sigma^2 \Delta t$.

For $\Delta t$ small (one or just a couple of days) the difference is negligible but for longer terms (e.g. recommended holding periods) we have to model many $\Delta t$ steps leading to a larger term there. So there is a difference on the long run.

## Answer by Richi Wa (score 2)

https://quant.stackexchange.com/a/34132

After editing my queston several times I decided to write an answer.

In setting (B) we have the Ito correction term. As Gordon mentions this makes the expected value vanish.

In setting (A) we introduce a positive drift of the size $\sigma^2 \Delta t/2$ even if we stay in discrete time. It is there.

The following code illustrates this in R. The setting `mu=0` corresponds to (A) and `mu = -sigma^2/2`to (B). In A after 10 years and using 1000 paths we have an average of 12% gain which is what we can expect if we have the drift of $0.15^2/2$ ($15\%$ vol in the example below) for 10 years which equals $11.25 \%$ (!)

In B we have -3% which I think is just a sampling error and it should be zero. In a second run I got the 0 expected growth.

Both processes reach similar maxima and do not explode.

Bottom line: on the long run not accounting for the drift correction will lead to overestimates of your performance ... which is a basic point in stochastic analysis.

```
S_0 = 100
nr_paths = 1000
delta_t = 1/250 # nr steps per year
nr_years = 10
sigma = 0.15

mu = 0 
#mu = -sigma^2/2

paths = matrix(S_0, nrow = nr_paths, ncol = nr_years/delta_t)
dBt = matrix(rnorm(nr_paths*(nr_years/delta_t-1), mean = 0, sd = 1), nrow = nr_paths, ncol = nr_years/delta_t-1)

for (i in 2: (nr_years/delta_t)) {
  paths[,i] = paths[,i-1]*exp(mu*delta_t + sigma*sqrt(delta_t)*dBt[,i-1]) 
}
plot( paths[1,], type="n", ylim = range(paths), main = paste("Drift =", toString(mu))) 

for(i in 1:nr_paths) {
  lines( paths[i,], col = rgb(0,0,1,0.05) )
}

lines( apply(paths, 2, mean), col ="red")
tail(apply(paths, 2, mean))
```

EDIT: one more additional thougt:

Setting (A) (no drift correction) leads to $$ E[\log(S_t/S_u)|\mathcal{F_u}] = 0 $$ thus the expectation of the (conditional) logreturn is 0, while Setting (B) drift correction as in PRIIPS leads to $$ E[\frac{S_t-S_u}{S_u}|\mathcal{F_u}] = 0 $$ thus the expectation of the (conditional) arithmetic return is zero. In this setting the price $S_t$ is a martingale.