Why Lognormal Stock Price Expectations Depend on the Drift Convention

# What is the stock price expectation?


# What is the stock price expectation?












The Hull textbook (and accompanying technical note) says that the expected stock price $\mathbb{E}[S_T]=S_0 \exp(\mu T)$. However, the answers to a British actuarial examination (Q4 for September 2018) require it to be $S_0 \exp (\mu T +\frac{\sigma^2}{2}T)$. Why don't these values agree?

## Answer by Slade (score 5)

https://quant.stackexchange.com/a/49199

In Hull's textbook, the stock price dynamics is lognormal: $S_T = S_0 \exp(\mu T - \frac{1}{2}\sigma^2T + \sigma W_T)$, where $W_t$ is a standard brownian motion. And so the mean of this is the mean of a lognormal random variable with the log mean as $\ln S_0 + \mu T - \frac{1}{2}\sigma^2T$ and the log standard deviation as $\sigma \sqrt{T}$, and so the mean is (link to info on log-normal mean), $S_0\exp(\mu T)$.

In the exam, the stock price dynamics is also lognormal, but now there is no mean term involving $\sigma$, so $S_T = S_0 \exp(\mu T + \sigma W_T)$. This is since the question is saying that there is, according to the wikipedia (and also standard) notation, a $\mu$ parameter and a $\sigma$ parameter, and so the $\mu$ value is as given. So the mean of this log normal random variable is $S_0\exp(\mu T+ \frac{1}{2}\sigma^2T)$.

So the two results actually do not contradict. They are different answers but neither is 'wrong', although the Hull result is the typical one you will see all over finance textbooks. The actuarial exam result is just a particular case of a log-normal dynamics for the stock price that they decided to use.

## Answer by Magic is in the chain (score 1)

https://quant.stackexchange.com/a/49213

Let's focus on the technical note, but let's change the symbols for easy comparison. The note has two parts:

Firstly, if we assume that the stock price is log normal with the following parameters:

$S_t \sim \mathrm{LN}\left(\ln S_0+\mu t,\sigma^2 t\right)$

then by definition, its log is normally distributed:

$\ln S_t \sim \mathrm{N}\left(\ln S_0+\mu t,\sigma^2 t\right)$

We can write it in terms of standard normal:

$\ln S_t=\ln S_0+\mu t+\sigma \sqrt{t} Z$

which means:

$S_t=S_0 e^{\mu t+\sigma \sqrt{t} Z}$

The mean of the above variable is indeed:

$E\left[S_t \right]=S_0 e^{E \left[ {\mu t+\sigma \sqrt{t} Z}\right]+ 0.5 V\left[ {\mu t+\sigma \sqrt{t} Z}\right]}=S_0 e^{\mu t+0.5\sigma^2 t}$

This is the equivalent of $e^{m+s^2/2}$ in the Hull's technical note (just after equation 1).

On the other hand, if we we assume that the stock price follows GBM:

$\frac{dS_t}{S_t}=\mu dt + \sigma dW_t$

Then by applying ito's lemma, we know that this means:

$d \ln S_t=\mu dt+\sigma dW_t-\frac{1}{2}\sigma^2 dt$

$\ln S_t=\ln S_0+\mu t+\sigma W_t-\frac{1}{2}\sigma^2 t$

or in terms of standard normal (after rearrangement):

$\ln S_t=\ln S_0+\left(\mu-\frac{1}{2}\sigma^2\right) t+\sigma \sqrt{t}Z$

$S_t=S_0 e^{\left(\mu-\frac{1}{2}\sigma^2\right) t+\sigma \sqrt{t}Z}$

So it is log normal but with mean $\ln S_0+\left(\mu-\frac{1}{2}\sigma^2\right) t$ and variance $\sigma^2t$. And the Hull's technical note gives these formulae immediately after the equation $e^{m+s^2/2}$ that I referenced above. The note references the text, which I believe is equation 13.3 in Hull's textbook (7th Edition). We can thus calculate its expected value as follows:

$E\left[S_t \right]=S_0 e^{E \left[ {\left(\mu-0.5 \sigma^2\right) t+\sigma \sqrt{t} Z}\right]+ 0.5 V\left[ {\left(\mu-0.5\sigma^2\right) t+\sigma \sqrt{t} Z}\right]}=S_0e^{\left(\mu-0.5\sigma^2\right) t+0.5\sigma^2 t}=S_0 e^{\mu t}$

Now to the exam terminology: It seems like the exam is using log normal model with drift $\mu$ to mean lognormal distribution with mean $\mu$.

Re-comment, the 1D-marginal distribution of the GBM process is indeed log normal, but the tricky part is the connection between the drift, and the mean. If we represent the GBM with drift $\mu t$ and variance $\sigma^2 t$ by $\mathrm{GBM} \left(\mu t, \sigma^2 t\right)$,then the process value at time t, $S_t$ given its current value, $S_0=1$, has $\mathrm{LN} \left(\mu t-\frac{1}{2}\sigma^2t, \sigma^2 t \right)$.

For path-dependent options, if you were to use the probability distribution approach, you will need the joint distribution of the process values at different times. The joint distribution for most pay-offs will not be available in tractable form. Though for some path-dependent options, e.g., geometric average, the joint distribution is easy to use because the joint distribution is again log-normal. So the best would be to use a numerical method: simulate the process values under the risk neutral measure, or use PDE based approach.