What price should we pay for a derivative contract that has the characteristics of a European Call or a European Put? We know the pay off of the option at expiry:

\begin{equation*}
\text{European Call Pay-off} =
\begin{cases}
S(T) – K\ &\text{ if } S(T)>k \\
0 &\text{ otherwise }\\
\end{cases}
\end{equation*}

The payoff of the call is not a smooth linear function. It is continuous but it is not differentiable at the kink in the function at expiry. We have a closed form solution for a European option but the majority of traded options do not have nice closed-form solutions. The factors involved are as follows

Strike $K$
Expiry $T$
Current asset price $S_o$
Volatility $\sigma$

We need a model for the future asset prices. If at $t$ the asset price is $S$
and at $t+dt$ the price moves to $S+dS$, we need a model for the percentage change $\dfrac{dS}{S}$ over time $dt$

We can decompose the change into two parts, a predictable part and a random part

$\mu dt:$ predictable growth rate of the asset
$\sigma dW_t:$ random fluctuations

We can therefore compose a model for the change
$$\dfrac{dS}{S} = \mu dt + \sigma dW_t$$

The axioms of Brownian motion are used in pricing but not explicitly examined. They are however examined in numerical methods so now would be a good time to refresh your memory. The axioms of Brownian motion can be applied to the infinitessimal stock price increment at time $t$
\begin{align}dS_t = S_t( \mu dt + \sigma dW_t) \tag{1} \\ \\
dS_t = \mu S_t dt + \sigma S_t dW_t
\end{align}
it is not uncommon to see this written in a shorter form[ref]Steven E. Shreve. Stochastic Calculus for Finance II, Continuous-Time Models. Chapter 4.4.2. $a_t \equiv \mu S_t \equiv \Theta(t)$ and $b_t \equiv \sigma S_t \equiv \Delta(t)$.[/ref] as
$$dS_t = a_t dt + b_t dW_t $$
squaring the increment $(1)$
\begin{align}
(dS_t)^2 &= S_t^2( \mu dt + \sigma dW_t)^2\\ \\
&= S_t^2( \underbrace{\mu^2 dt^2}_{=0} + \underbrace{2\mu \sigma dtdW_t}_{=0} + \underbrace{\sigma^2 dW_t^2}_{=\sigma^2dt})\\ \\
(dS_t)^2 & = S_t^2 \sigma^2dt \tag{2}
\end{align}
Clearly, while $dS_t$ is random, $(dS_t)^2$ is on the contrary not random. This result leans towards the theorem known as Ito’s lemma. We can apply Ito’s rules to the taylor series expansion of the twice differentiable function $f(X_t, t)$

A quick and cavalier approach is to recall familiar differential formula with an additional second order term for $x$ but not for $t$ since higher $dt$ powers are zero by comparison.
$$df = \dfrac{\partial f}{\partial t}dt + \dfrac{\partial f}{\partial x}dx + \dfrac{\partial^2 f}{\partial x^2}(dx)^2$$
by replacing $x$ with $S_t$ can be derived more rigourously as the Ito process
\begin{align}df(S_t,t) = \partial_t f dt + \partial_s f dS_t + \dfrac{1}{2} \partial_{ss}^2 f (dS_t)^2 \tag{3}\end{align}
In our case, the function of $S_t$ and $t$ is the model in $(1)$ above. Substituting $(1)$ and $(2)$ into $(3)$
\begin{align}df(S_t,t) = \dfrac{\partial f}{\partial t} dt + \dfrac{\partial f}{\partial S} (S_t \mu dt + S_t \sigma dW_t) + \dfrac{1}{2} \dfrac{\partial^2 f}{\partial S^2} S_t^2 \sigma^2dt \end{align}
grouping by infinitesimals $dt$ and $dW_t$
\begin{align}df(S_t,t) = \Biggr( \dfrac{\partial f}{\partial t} + \mu S_t \dfrac{\partial f}{\partial S} + \dfrac{1}{2} \sigma^2 S_t^2 \dfrac{\partial^2 f}{\partial S^2}\Biggr)dt + \sigma S_t \dfrac{\partial f}{\partial S}dW_t \tag{4} \end{align}
This is Ito’s second lemma. It is not uncommon to see this written in short form as
\begin{align}df(S_t,t) = \Biggr( \dfrac{\partial f}{\partial t} + a_t \dfrac{\partial f}{\partial S} + \dfrac{1}{2} b_t^2 \dfrac{\partial^2 f}{\partial S^2}\Biggr)dt + b_t \dfrac{\partial f}{\partial S}dW_t \end{align}
As explained in Ito’s Lemma & Stochastic Differential Equations we can start with a a process such as geometric Brownian motion for the stock price and by using Ito’s second lemma for the increment arrive at the model $dX_t$. The sections below illustrate that the reverse is also possible. We have a random walk for the increment $dS_t$ which we can apply Ito’s lemma to and integrate to arrive back at the stock price process $S_t$.

Derivation of the model of the asset $S$

If we choose $f(S_t) = \text{log}(S)$ in Ito’s second lemma $(4)$, the partials evaluate as follows
$$\dfrac{\partial f}{\partial t} = 0 \qquad \dfrac{\partial f}{\partial S} = \dfrac{1}{S} \qquad \dfrac{\partial^2 f}{\partial S^2} = -\dfrac{1}{S^2}$$
$(4)$ then becomes
$$d\text{log}(S_t) =\Biggl( \mu – \dfrac{\sigma^2}{2} \Biggr)dt + \sigma dW_t$$in integral form,
$$\int_t^{t+\tau} d\text{log}(S_t) = \int_t^{t+\tau} \Biggl( \mu – \dfrac{\sigma^2}{2} \Biggr)dt + \sigma \int_t^{t+\tau} dW_t$$or equivalently,
\begin{align}
\text{log}\dfrac{S(t+\tau)}{S(t)} &= \Biggl( \mu – \dfrac{\sigma^2}{2} \Biggr)\tau + \sigma \Delta W_t \\ \\
&=\Biggl( \mu – \dfrac{\sigma^2}{2} \Biggr)\tau + \sigma \sqrt{\tau} Z_t
\end{align}
taking exponentials
\begin{align}
\dfrac{S(t+\tau)}{S(t)} &= e^{\Biggl( \mu – \dfrac{\sigma^2}{2} \Biggr)\tau + \sigma \sqrt{\tau} Z_t}\\ \\
\end{align}

Modelling Derivatives $f(S_t, t)$ of an Asset $S_t$ in 4 Key Steps

Consider a derivative asset $C(S_t, t)$

A.

Apply Ito’s Lemma to determine the infinitesimal increment in derivative value w.r.t the underlying asset $S_t$ and $t$. Factorise into a time-trending $dt$ infinitesimal and a risky $dW_t$ infinitesimal.
\begin{align}dC(S_t,t) = \Biggr( \dfrac{\partial C}{\partial t} + \mu S_t \dfrac{\partial C}{\partial S} + \dfrac{1}{2} \sigma^2 S_t^2 \dfrac{\partial^2 C}{\partial S^2}\Biggr)dt + \sigma S_t \dfrac{\partial C}{\partial S}dW_t \tag{5} \end{align}

B.

Build a hedged portfolio $\Pi$ of the derivative asset comprised of

  $\; \; 1 \quad$ European Call $C$
$-\Delta \quad $ amount of the underlying stock $S$. (Short sell)

Such that at $t$,
$$\Pi_t = C \; – \; \Delta \cdot S$$
over the time interval $dt$,
$$d\Pi = dC \; – \; \Delta \cdot dS $$
$dC$ and $dS$ are given by equations $(5)$ and $(1)$ respectively above. Quick tip: in $dC(S_t,t)$, subtract $\Delta$ from $\dfrac{\partial C}{\partial S}$ as it multiplies the terms common to $dS$ and $dC(S_t,t)$.

\begin{align}d\Pi(S_t,t) = \Biggr( \dfrac{\partial C}{\partial t} + \mu S_t \Bigl[ \dfrac{\partial C}{\partial S} – \Delta \Bigr] + \dfrac{1}{2} \sigma^2 S_t^2 \dfrac{\partial^2 C}{\partial S^2}\Biggr)dt + \sigma S_t \Bigl[ \dfrac{\partial C}{\partial S} – \Delta \Bigr]dW_t \tag{6} \end{align}

C.

Compute the quantity $\Delta$ of $S$ with respect to, or, in ‘units of’ the derivative, that will make the risky term $dW_t$ vanish. Obtain infnitessimal incremental value $d\Pi$ of portfolio. Clearly for the risky term $dW_t$ to vanish,
$$\Delta = \dfrac{\partial C}{\partial S}$$
\begin{align}
\therefore \quad d\Pi(S_t,t) = \Biggr( \dfrac{\partial C}{\partial t} + \dfrac{1}{2} \sigma^2 S_t^2 \dfrac{\partial^2 C}{\partial S^2}\Biggr)dt \tag{7}\end{align}
This is now a risk free protfolio.

D.

Determine the conditions necessary for the portfolio to be arbitrage-free over the small interval $dt$. For the risk free portfolio to be arbitrage free over the infinitesimal interval $dt$, it must be equal to the portfolio growth rate of a risk free bank account over $dt$. Let the growth rate be $r$
\begin{align}d\Pi &= r\Pi dt\\ \\
& = r(C- \Delta \cdot S) dt\\ \\
& = r(C- \dfrac{\partial C}{\partial S} \cdot S) dt \tag{8} \end{align}
equating $(7)$ and $(8)$ gives
\begin{align}
\quad d\Pi(S_t,t) = \dfrac{\partial C}{\partial t} + \dfrac{1}{2} \sigma^2 S_t^2 \dfrac{\partial^2 C}{\partial S^2} + rS\dfrac{\partial C}{\partial S} – rC = 0 \tag{9}\end{align}

This is the famous Black-Scholes partial differential equation

As $\mu$ does not appear in the equation, the Black-Schoes is said to give the risk-neutral price of the derivative. This suggests $\mu$ can be replaced with $r$ in $(1)$ above. Therefore

\begin{align}
S(t+\tau) &= S(t) e^{\Biggl(r – \dfrac{\sigma^2}{2} \Biggr)\tau + \sigma \sqrt{\tau} Z_t}
\end{align}

The Black-Scholes equation can be solved by a number of ways. One way is to transform it into the heat equation. The solution (given) is
$$C(S_t) = S \cdot \Phi(d_1) – Ke^{-r(T-t)} \Phi (d_2)$$
where
$$N(x) = \dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-y^2/2} dy$$