# 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?

• The second link you gave is to a downloaded file in your local directory so no one else can access it – Slade Oct 14 '19 at 11:10
• Sorry, corrected. – Moronic Oct 14 '19 at 11:17

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.

• Sorry, I don't follow how this is a matter of notation? $\exp$ means $\exp$, and $exp(\mu 𝑇+\sigma 𝑊_𝑇)$ is patently different from $exp(\mu 𝑇-\frac{1}{2}\sigma^2 T+\sigma 𝑊_𝑇)$. – Moronic Oct 14 '19 at 12:40
• The stock price dynamics in hull and the actuarial question are different. So the answers are different. I can make a new stock price dynamics that say the stock price should be constant, so under those dynamics that expected future value is the same as the present value. So under different dynamics, the expected value is different. – Slade Oct 14 '19 at 12:43
• The indication of different dynamics in the actuarial exam is that they give the $\mu$ term explicitly for a log normal random variable, so the log mean does not involve $\sigma$ – Slade Oct 14 '19 at 12:44
• But the textbook gives $\sigma$ explicitly too, and defines it as "Volatility of the stock price per year". It also quotes the formula in a section (15.1) called "Lognormal property of stock prices", while the exam says "prices of both stocks are assumed to follow the lognormal model". I don't see any difference here. – Moronic Oct 14 '19 at 12:47
• the $\sigma$ term is the stock price vol. in both the Hull textbook and the exam. The difference is that the stock price dynamics in the Hull textbook includes the $-0.5 \sigma^2$ term while the exam does not. There are many types of possible log-normal models that can be used, but in this particular case, the difference is whatever I wrote above. I understand your confusion as to why the exam decided not to include it, but I think the point of the question was to just show that you know what a log-normal random variable is so as long as you can work with them, your understanding is solid. – Slade Oct 14 '19 at 12:52

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.

• It was my understanding all along that GBM and lognormal model are one and the same thing. Textbooks, including Hull, simply mention them all in one go. Is this not the case? – Moronic Oct 18 '19 at 15:05
• Also which one is "correct"? If I want to price a path-dependent option by simulating paths, which one do I discretize to do it? – Moronic Oct 18 '19 at 15:12
• I have added further details in the answer. Hope it makes sense! – Magic is in the chain Oct 18 '19 at 21:08