In this page we derive the famous Black-Scholes Model for European option pricing. I first
Assuming the stock price follows geometric brownian motion and the SDE is: \[dS_t = \mu S_t dt + \sigma S_t dW_t\] and \(g(x) = ln(x)\) Apply the Ito’s formula:
\[ \begin{aligned} dg(S_t) &= d(ln(S_t)) \\ &= \frac{1}{S_t}dS_t + \frac{1}{2} \cdot (-1) \cdot \frac{1}{S_t^2} \sigma^2 S_t^2dt \\ &= \frac{1}{S_t}(\mu S_t dt + \sigma S_t dW_t) - \frac{1}{2} \sigma^2dt \\ &= (\mu-\frac{1}{2} \sigma^2) dt + \sigma dW_t \end{aligned} \] Take the integral, then we get the stock price \(S_t\) at time \(t\):
\[S_t = S_0 \cdot e^{(\mu-\frac{1}{2} \sigma^2) t + \sigma W_t}\]
Expected value of \(S_t\):
\[ \begin{aligned} E[S_t] &= S_0 \cdot e^{(\mu-\frac{1}{2} \sigma^2) t } \cdot E[e^{\sigma W_t}] \\ &= S_0 \cdot e^{(\mu-\frac{1}{2} \sigma^2) t} \cdot e^{\frac{\sigma^2 t}{2}} \\ &= S_0 \cdot e^{(\mu-\frac{1}{2} \sigma^2) t+\frac{1}{2}\sigma^2 t} \\ &= S_0 \cdot e^{\mu t} \\ \end{aligned} \] We haven’t proved that \(E[e^{\sigma W_t}] = e^{\frac{1}{2}\sigma^2t}\) yet. Let’s derive \(E[e^{W_t}] = e^{\frac{1}{2}t}\) fisrt:
\[ \begin{aligned} E[e^{W_t}] &= \int e^x \cdot \frac{1}{\sigma_w\sqrt{2\pi}} \cdot e^{-\frac{1}{2}(\frac{x-\mu}{\sigma_w})^2} dx\\ &= \int \frac{1}{\sqrt{t}\sqrt{2\pi}} \cdot e^{x-\frac{1}{2}(\frac{x}{\sqrt{t}})^2} dx \\ &= \int \frac{1}{\sqrt{t}\sqrt{2\pi}} \cdot e^{-x-\frac{x^2}{2t}} dx\\ &= \int \frac{1}{\sqrt{t}\sqrt{2\pi}} \cdot e^{-\frac{2xt+x^2}{2t}} dx\\ &= \int \frac{1}{\sqrt{t}\sqrt{2\pi}} \cdot e^{-\frac{x^2+2xt+t^2}{2t}+\frac{t^2}{2t}} dx\\ &= \int \frac{1}{\sqrt{t}\sqrt{2\pi}} \cdot e^{-\frac{1}{2}(\frac{x-(-t)}{\sqrt{t}})^2+\frac{t}{2}} dx \\ &= e^{\frac{t}{2}} \cdot \int \frac{1}{\sqrt{t}\sqrt{2\pi}} \cdot e^{-\frac{1}{2}(\frac{x-(-t)}{\sqrt{t}})^2} dx \\ &= e^{\frac{t}{2}} \\ \end{aligned} \] We can then extend to that \(E[e^{\sigma W_t}] = e^{\frac{\sigma^2 t}{2}}\) by replacing \(\sigma_w = \sqrt{t}\) with \(\sigma_w = \sigma \sqrt{t}\)
Geometric Brownian Motion means that the log of stock return follows normal distribution.
[To be validated] The pricing of European option requires changing from phycial measure to risk neutral measure using Girsanov’s Theorem.
The price of an European call option can be expressed as:
\[C = E[e^{-\mu_cT} \cdot (S_T-K)^+]\]
Where \(\mu\) is the required rate of return for the option, \(S_T\) is the stock price at maturity date \(T\) and \(K\) is the strike price. Since the required rate of return for the option in unknown, we need to do some tranformation in order to price the option: change to risk-neutral measure.
Under Physical measure \(P\), the stock price has SDE:
\(S_T = S_0 \cdot e^{(\mu -\frac{1}{2}\sigma^2)T + \sigma W_T^P}\)
where \(\mu\) is the expected rate of return of the stock.
Girsanov’s Theorem:
Define: \[\frac{dQ}{dP} = Z_t = e^{\int_0^t \varphi_sdW_s^P-\frac{1}{2}\int_0^t \varphi_s^2ds}\] want Q as a probability measure: \(Z_t \geq 0\) and \(Q[\Omega] = 1\)
\[W_t^Q = W_t^P - \int_0^t \varphi_sds\] move it around we get:
\[ \begin{aligned} W_t^P &= W_t^Q + \int_0^t \varphi_sds \\ &= W_t^P - \int_0^t (-\varphi_s)ds \end{aligned} \]
Hence
\[ \begin{aligned} \frac{dP}{dQ} &= e^{\int_0^t (-\varphi_s)dW_s^Q-\frac{1}{2}\int_0^t (-\varphi_s)^2ds} \\ &= e^{-\int_0^t \varphi_sdW_s^Q-\frac{1}{2}\int_0^t \varphi_s^2ds} \end{aligned} \]
The choice \(\varphi_s = \frac{r-\mu}{\sigma}\) (constant). So:
\[ \begin{aligned} \frac{dQ}{dP} &= Z_t\\ &= e^{\int_0^t \frac{r=\mu}{\sigma}dW_s^P -\frac{1}{2}\int_0^t (\frac{r-\mu}{\sigma})^2ds} \\ &= e^{\frac{r-\mu}{\sigma}W_t^P-\frac{1}{2}(\frac{r-\mu}{\sigma})^2t} \end{aligned} \]
and
\[W_t^Q = W_t^P - \int_0^t \frac{r-\mu}{\sigma}ds = W_t^P-\frac{r-\mu}{\sigma}t\]
Now, we can price the option under the risk-neutral measure \(Q\) by taking the expected value \(E^Q\)
\[ \begin{aligned} C &= E^Q[e^{-rT}(S_T-K)^+] \\ &= E^Q[e^{-rT}S_T \cdot I_{\{S_T-K\}}] - E_Q[e^{-rT}K \cdot I_{\{S_T-K\}}] \\ &= \text{1st term} - \text{2nd term} \end{aligned} \]
Under the risk-neutral measure all the assets earn risk-free rate, so both \(\mu_c\) and \(\mu\) are replaced with \(r\). The stock price at \(T\) is therefore \(S_T = S_0 \cdot e^{(r-\frac{\sigma^2}{2})T + \sigma W_T^Q}\). We will tackle the 1st term and 2nd term individually.
A good analogy I read on stackexchange post said that “it’s like pricing an umbrella, the price depends on how likely it is to rain, and how much people hate rain”. In risk-neutral pricing, we change the require rate of return \(mu\) to risk-free rate \(r\) (change how people hate rain), we must, at the same time, also change the probability to under risk-neutral measure by Girsanov’s theorem (change how likely it is to rain) so that the pricing is the same as under physical measure.
\[ \begin{aligned} E^Q[e^{-rT}S_T \cdot I_{\{S_T-K\}}] &= E^Q[e^{-rT}S_0 \cdot e^{(r-\frac{1}{2}\sigma^2)T+\sigma W_T^Q}\cdot I_{\{S_T-K\}}]\\ &= S_0 \cdot E_Q[e^{-\frac{1}{2}\sigma T+\sigma W_T^Q} I_{\{S_T-K\}}]\\ \end{aligned} \] Now we have to change the measure again using the Girsanov’s Theorem . Define:
\[\frac{dQ^s}{dQ} = Z_t = e^{-\frac{1}{2}\sigma T+\sigma W_T^Q}\] Then we can continue as \[ \begin{aligned} S_0 \cdot E^Q[e^{-\frac{1}{2}\sigma T+\sigma W_T} I_{\{S_T-K\}}] &= S_0 \cdot E^Q[\frac{dQ^s}{dQ} \cdot I_{\{S_T-K\}}]\\ &= S_0 \cdot E^{Q^s}[I_{\{S_T-K\}}] \\ &= S_0 \cdot Q^s[S_T>K] \\ &= S_0 \cdot Q^s[S_0 \cdot e^{(r-\frac{1}{2}\sigma^2)T+\sigma W_T^Q}>K] \\ &= S_0 \cdot Q^s[S_0 \cdot e^{(r-\frac{1}{2}\sigma^2)T+\sigma(W_T^{Q^s}+\sigma T)}>K] \\ &= S_0 \cdot Q^s[S_0 \cdot e^{(r+\frac{1}{2}\sigma^2)T+\sigma W_T^{Q^s}}>K] \\ &= S_0 \cdot Q^s[W_T^{Q^s}> \frac{ln(K/S_0)-(r+\frac{1}{2}\sigma^2)T}{\sigma}] \\ &= S_0 \cdot Q^s[\epsilon<\frac{ln(S_0/K)+(r+\frac{1}{2}\sigma^2)T}{\sigma \sqrt{T}}] \\ &= S_0 \cdot \Phi(d_1) \end{aligned} \]
Where
\[d_1 = \frac{ln(S_0/K)+(r+\frac{1}{2}\sigma^2)T}{\sigma \sqrt{T}}\]
THe derivation of the 2nd term is always under \(Q\) measure. i.e. \(E[x] = E^Q[x]\) and \(W_T = W^Q_T\)
\[ \begin{aligned} E[e^{-rT}K \cdot I_{\{S_T-K\}}] &= e^{-rT} \cdot K \cdot Q[S_T>K]\\ &= e^{-rT} \cdot Q[S_0 \cdot e^{(r-\frac{1}{2}\sigma^2)T + \sigma W_T}>K] \\ &= e^{-rT} \cdot Q[\epsilon>\frac{ln(K/S_0)-(r-\frac{1}{2}\sigma^2)T}{\sigma \sqrt{T}}] \\ &= e^{-rT} \cdot Q[\epsilon<\frac{ln(S_0/K)+(r-\frac{1}{2}\sigma^2)T}{\sigma \sqrt{T}}] \\ &= e^{-rT} \cdot K \cdot \Phi(d_2) \end{aligned} \]
Where \[d_2 = \frac{ln(S_0/K)+(r-\frac{1}{2}\sigma^2)T}{\sigma \sqrt{T}}\]