We have determined that we may model a primitive asset such as a stock as,
$$S_T = S_t e^{ (r \; – \frac{\sigma^2}{2} )\tau + \sigma \sqrt{\tau}Z} \tag{1}$$
Question: What exactly is $e^{\sigma \sqrt{\tau}Z}$ ? Or since we realise that it is probably a random variable, we should ask what is the expectation $\mathbb{E} (e^{\sigma \sqrt{\tau}Z})$ or more generally $\mathbb{E} (e^{\lambda Z})$?
We start by recalling the definition of the expectation (mean) of a random variable.
$$\mathbb{E} (x) = \int_{-\infty}^{\infty} x \cdot f(x) dx$$
where $f(x)$ is the probability distribution function.
As $Z_T \sim N (0 \; , 1)$, we may use the probability density function which is given for any normal random variable.
\begin{align}\mathbb{E} (e^{\lambda Z}) &= \int_{-\infty}^{\infty} e^{\lambda t} \cdot \underbrace{\frac{1}{\sqrt{2 \pi}} e^{-t^2/2}}_{\text{pdf}} dt \\
&= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{-t^2/2 \; + \; \lambda t} dt \\
\\
&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{ -\frac{1}{2} (t^2 \; – \; 2 \lambda t)} dt
\end{align}
Technique: Complete the square in the exponent
$$t^2 \; – \; 2 \lambda t ={(t – \lambda)}^2 \; – \lambda^2 $$ then
$$ \mathbb{E} (e^{\lambda Z}) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \text{exp} \Big( -\frac{1}{2}[{(t – \lambda)}^2 \; – \lambda^2 ] \Big) dt $$
Technique: Isolate the squared exponent in $t \; – \lambda$
\begin{align}
&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \text{exp} ( -\frac{1}{2}{(t – \lambda)}^2 ) \cdot \text{exp} ( \frac{1}{2}\lambda^2 )dt \\ \\
&= \frac{1}{\sqrt{2 \pi}} \; e^{\lambda^2 / 2} \int_{-\infty}^{\infty} \text{exp} \Big( -\frac{1}{2}{(t – \lambda)}^2 \Big) dt \end{align}
Observe:
\begin{align}= e^{\lambda^2 / 2} \; \times \underbrace{ \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \text{exp} \Big( -\frac{1}{2}{(t – \lambda)}^2 \Big) dt}_{\text{integral of the pdf displaced by } \lambda \; = \; \text{integral of the pdf} \; = \; 1 } \end{align}(See Integrating probability density functions)
Conclude:
$$ \mathbb{E} (e^{\lambda Z}) = e^{\lambda^2 / 2}$$
In alternative notation
$$ \mathbb{E} (e^{c Z}) = e^{c^2 / 2} \quad \text{where} \qquad \lambda \equiv c \equiv \sigma \sqrt{\tau} , \; \tau = T \; – t$$ For any real or complex number $c$.
Substituting for $\mathbb{E}e^{\sigma \sqrt{\tau}Z}$ in $(1)$,
$$\mathbb{E}S_T = S_t e^{ (r \; – \frac{\sigma^2}{2} )\tau + \frac{1}{2} \sigma^2 \tau}$$ or given $t = 0$
\begin{align}\mathbb{E}S_T &= S(0) e^{ (r \; – \frac{\sigma^2}{2} )T + \frac{1}{2} \sigma^2 T}\\ \\
\mathbb{E} S (T) &= S(0) e^{rT}\end{align}
This allows us to make a number of important claims. First, although $S_T$ is a random function, its expectation is deterministic and therefore “risk-neutral”. (Observe that the random term has disappeared).
There is only one unknown parameter required to obtain the future $T$ asset price given its current $t$ price. The unknown parameter is the volatility $\sigma$.