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Gordon
Gordon
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Per @SKRX's suggestion, another solution is provided below.

For simplicity, we assume that the stock price process $\{S_t \mid t \geq 0\}$ follows an SDE, under the risk-neutral measure $\mathbb{Q}$, of the form \begin{align*} \frac{dS_t}{S_t} = r dt + \sigma dW_t, \end{align*} where $r$ is the constant interest rate, $\sigma$ is the constant volatility, and $\{W_t \mid t \geq 0\}$ is a standard Brownian motion. Moreover, let $B_t = e^{rt}$ be the money-market account value at time $t$.

Note that \begin{align*} (S_T-K)^+ &= (S_T-K)\mathbb{1}_{S_T >K}\\ &= S_T\mathbb{1}_{S_T >K} - K \mathbb{1}_{S_T >K}. \end{align*} Then, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}\mathbb{Q}(S_T >K)\\ &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}N(d_2). \end{align*} To compute the expectation $\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big)$, we define the probability measure $\widetilde{\mathbb{Q}}$, so that we have the Radon-Nikodym derivative of the form \begin{align*} \frac{d\widetilde{\mathbb{Q}}}{d\mathbb{Q}}\big|_t &= \frac{S_t}{B_t S_0}\\ &=\exp\left(-\frac{\sigma^2}{2} t + \sigma W_t \right). \end{align*} By Girsanov theorem, $\{\widetilde{W}_t \mid t \geq 0\}$, where \begin{align*} \widetilde{W}_t = W_t - \sigma t, \end{align*} is a standard Brownian motion under the probability measure $\widetilde{\mathbb{Q}}$. Moreover, under $\widetilde{\mathbb{Q}}$, \begin{align*} \frac{dS_t}{S_t} = \left(r+ \sigma^2 \right) dt + \sigma d\widetilde{W}_t. \end{align*} Note also that \begin{align*} \frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_t &= \frac{B_tS_0}{S_t}. \end{align*} Therefore, \begin{align*} e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) &=e^{-rT}\mathbb{E}_{\widetilde{\mathbb{Q}}}\left(\frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_T S_T\mathbb{1}_{S_T >K}\right)\\ &=S_0 \widetilde{\mathbb{Q}}(S_T >K)\\ &=S_0 N(d_1). \end{align*} That is, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &=S_0 N(d_1) - K e^{-rT}N(d_2), \end{align*}\begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &= S_0 \widetilde{\mathbb{Q}}(S_T >K) - K e^{-rT}\mathbb{Q}(S_T >K) \\ &= S_0 N(d_1) - K e^{-rT}N(d_2), \end{align*} which is the Black-Scholes formula.

Per @SKRX's suggestion, another solution is provided below.

For simplicity, we assume that the stock price process $\{S_t \mid t \geq 0\}$ follows an SDE, under the risk-neutral measure $\mathbb{Q}$, of the form \begin{align*} \frac{dS_t}{S_t} = r dt + \sigma dW_t, \end{align*} where $r$ is the constant interest rate, $\sigma$ is the constant volatility, and $\{W_t \mid t \geq 0\}$ is a standard Brownian motion. Moreover, let $B_t = e^{rt}$ be the money-market account value at time $t$.

Note that \begin{align*} (S_T-K)^+ &= (S_T-K)\mathbb{1}_{S_T >K}\\ &= S_T\mathbb{1}_{S_T >K} - K \mathbb{1}_{S_T >K}. \end{align*} Then, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}\mathbb{Q}(S_T >K)\\ &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}N(d_2). \end{align*} To compute the expectation $\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big)$, we define the probability measure $\widetilde{\mathbb{Q}}$, so that we have the Radon-Nikodym derivative of the form \begin{align*} \frac{d\widetilde{\mathbb{Q}}}{d\mathbb{Q}}\big|_t &= \frac{S_t}{B_t S_0}\\ &=\exp\left(-\frac{\sigma^2}{2} t + \sigma W_t \right). \end{align*} By Girsanov theorem, $\{\widetilde{W}_t \mid t \geq 0\}$, where \begin{align*} \widetilde{W}_t = W_t - \sigma t, \end{align*} is a standard Brownian motion under the probability measure $\widetilde{\mathbb{Q}}$. Moreover, under $\widetilde{\mathbb{Q}}$, \begin{align*} \frac{dS_t}{S_t} = \left(r+ \sigma^2 \right) dt + \sigma d\widetilde{W}_t. \end{align*} Note also that \begin{align*} \frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_t &= \frac{B_tS_0}{S_t}. \end{align*} Therefore, \begin{align*} e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) &=e^{-rT}\mathbb{E}_{\widetilde{\mathbb{Q}}}\left(\frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_T S_T\mathbb{1}_{S_T >K}\right)\\ &=S_0 \widetilde{\mathbb{Q}}(S_T >K)\\ &=S_0 N(d_1). \end{align*} That is, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &=S_0 N(d_1) - K e^{-rT}N(d_2), \end{align*} which is the Black-Scholes formula.

Per @SKRX's suggestion, another solution is provided below.

For simplicity, we assume that the stock price process $\{S_t \mid t \geq 0\}$ follows an SDE, under the risk-neutral measure $\mathbb{Q}$, of the form \begin{align*} \frac{dS_t}{S_t} = r dt + \sigma dW_t, \end{align*} where $r$ is the constant interest rate, $\sigma$ is the constant volatility, and $\{W_t \mid t \geq 0\}$ is a standard Brownian motion. Moreover, let $B_t = e^{rt}$ be the money-market account value at time $t$.

Note that \begin{align*} (S_T-K)^+ &= (S_T-K)\mathbb{1}_{S_T >K}\\ &= S_T\mathbb{1}_{S_T >K} - K \mathbb{1}_{S_T >K}. \end{align*} Then, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}\mathbb{Q}(S_T >K)\\ &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}N(d_2). \end{align*} To compute the expectation $\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big)$, we define the probability measure $\widetilde{\mathbb{Q}}$, so that we have the Radon-Nikodym derivative of the form \begin{align*} \frac{d\widetilde{\mathbb{Q}}}{d\mathbb{Q}}\big|_t &= \frac{S_t}{B_t S_0}\\ &=\exp\left(-\frac{\sigma^2}{2} t + \sigma W_t \right). \end{align*} By Girsanov theorem, $\{\widetilde{W}_t \mid t \geq 0\}$, where \begin{align*} \widetilde{W}_t = W_t - \sigma t, \end{align*} is a standard Brownian motion under the probability measure $\widetilde{\mathbb{Q}}$. Moreover, under $\widetilde{\mathbb{Q}}$, \begin{align*} \frac{dS_t}{S_t} = \left(r+ \sigma^2 \right) dt + \sigma d\widetilde{W}_t. \end{align*} Note also that \begin{align*} \frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_t &= \frac{B_tS_0}{S_t}. \end{align*} Therefore, \begin{align*} e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) &=e^{-rT}\mathbb{E}_{\widetilde{\mathbb{Q}}}\left(\frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_T S_T\mathbb{1}_{S_T >K}\right)\\ &=S_0 \widetilde{\mathbb{Q}}(S_T >K)\\ &=S_0 N(d_1). \end{align*} That is, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &= S_0 \widetilde{\mathbb{Q}}(S_T >K) - K e^{-rT}\mathbb{Q}(S_T >K) \\ &= S_0 N(d_1) - K e^{-rT}N(d_2), \end{align*} which is the Black-Scholes formula.

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Gordon
Gordon
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Per @SKRX's suggestion, another solution is provided below.

For simplicity, we assume that the stock price process $\{S_t \mid t \geq 0\}$ follows an SDE, under the risk-neutral measure $\mathbb{Q}$, of the form \begin{align*} \frac{dS_t}{S_t} = r dt + \sigma dW_t, \end{align*} where $r$ is the constant interest rate, $\sigma$ is the constant volatility, and $\{W_t \mid t \geq 0\}$ is a standard Brownian motion. Moreover, let $B_t = e^{rt}$ be the money-market account value at time $t$.

Note that \begin{align*} (S_T-K)^+ &= (S_T-K)\mathbb{1}_{S_T >K}\\ &= S_T\mathbb{1}_{S_T >K} - K \mathbb{1}_{S_T >K}. \end{align*} Then, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}\mathbb{Q}(S_T >K)\\ &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}N(d_2). \end{align*} To compute the expectation $\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big)$, we define the probability measure $\widetilde{\mathbb{Q}}$, so that we have the Radon-Nikodym derivative of the form \begin{align*} \frac{d\widetilde{\mathbb{Q}}}{d\mathbb{Q}}\big|_t &= \frac{S_t}{B_t S_0}\\ &=\exp\left(-\frac{\sigma^2}{2} t + \sigma W_t \right). \end{align*} By Girsanov theorem, $\{\widetilde{W}_t \mid t \geq 0\}$, where \begin{align*} \widetilde{W}_t = W_t - \sigma t, \end{align*} is a standard Brownian motion under the probability measure $\widetilde{\mathbb{Q}}$. Moreover, under $\widetilde{\mathbb{Q}}$, \begin{align*} \frac{dS_t}{S_t} = \left(r+ \sigma^2 \right) dt + \sigma d\widetilde{W}_t. \end{align*} Note also that \begin{align*} \frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_t &= \frac{B_tS_0}{S_t}. \end{align*} Therefore, \begin{align*} e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) &=e^{-rT}\mathbb{E}_{\widetilde{\mathbb{Q}}}\left(\frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_T S_T\mathbb{1}_{S_T >K}\right)\\ &=S_0 \widetilde{\mathbb{Q}}(S_T >K)\\ &=S_0 N(d_1). \end{align*} That is, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &=S_0 N(d_1) - K e^{-rT}N(d_2), \end{align*} which is the Black-Scholes formula.

Per @SKRX's suggestion, another solution is provided below.

For simplicity, we assume that stock price process $\{S_t \mid t \geq 0\}$ follows an SDE, under the risk-neutral measure $\mathbb{Q}$, of the form \begin{align*} \frac{dS_t}{S_t} = r dt + \sigma dW_t, \end{align*} where $r$ is the constant interest rate, $\sigma$ is the constant volatility, and $\{W_t \mid t \geq 0\}$ is a standard Brownian motion. Moreover, let $B_t = e^{rt}$ be the money-market account value at time $t$.

Note that \begin{align*} (S_T-K)^+ &= (S_T-K)\mathbb{1}_{S_T >K}\\ &= S_T\mathbb{1}_{S_T >K} - K \mathbb{1}_{S_T >K}. \end{align*} Then, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}\mathbb{Q}(S_T >K)\\ &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}N(d_2). \end{align*} To compute the expectation $\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big)$, we define the probability measure $\widetilde{\mathbb{Q}}$, so that we have the Radon-Nikodym derivative of the form \begin{align*} \frac{d\widetilde{\mathbb{Q}}}{d\mathbb{Q}}\big|_t &= \frac{S_t}{B_t S_0}\\ &=\exp\left(-\frac{\sigma^2}{2} t + \sigma W_t \right). \end{align*} By Girsanov theorem, $\{\widetilde{W}_t \mid t \geq 0\}$, where \begin{align*} \widetilde{W}_t = W_t - \sigma t, \end{align*} is a standard Brownian motion under the probability measure $\widetilde{\mathbb{Q}}$. Moreover, under $\widetilde{\mathbb{Q}}$, \begin{align*} \frac{dS_t}{S_t} = \left(r+ \sigma^2 \right) dt + \sigma d\widetilde{W}_t. \end{align*} Note also that \begin{align*} \frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_t &= \frac{B_tS_0}{S_t}. \end{align*} Therefore, \begin{align*} e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) &=e^{-rT}\mathbb{E}_{\widetilde{\mathbb{Q}}}\left(\frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_T S_T\mathbb{1}_{S_T >K}\right)\\ &=S_0 \widetilde{\mathbb{Q}}(S_T >K)\\ &=S_0 N(d_1). \end{align*} That is, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &=S_0 N(d_1) - K e^{-rT}N(d_2), \end{align*} which is the Black-Scholes formula.

Per @SKRX's suggestion, another solution is provided below.

For simplicity, we assume that the stock price process $\{S_t \mid t \geq 0\}$ follows an SDE, under the risk-neutral measure $\mathbb{Q}$, of the form \begin{align*} \frac{dS_t}{S_t} = r dt + \sigma dW_t, \end{align*} where $r$ is the constant interest rate, $\sigma$ is the constant volatility, and $\{W_t \mid t \geq 0\}$ is a standard Brownian motion. Moreover, let $B_t = e^{rt}$ be the money-market account value at time $t$.

Note that \begin{align*} (S_T-K)^+ &= (S_T-K)\mathbb{1}_{S_T >K}\\ &= S_T\mathbb{1}_{S_T >K} - K \mathbb{1}_{S_T >K}. \end{align*} Then, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}\mathbb{Q}(S_T >K)\\ &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}N(d_2). \end{align*} To compute the expectation $\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big)$, we define the probability measure $\widetilde{\mathbb{Q}}$, so that we have the Radon-Nikodym derivative of the form \begin{align*} \frac{d\widetilde{\mathbb{Q}}}{d\mathbb{Q}}\big|_t &= \frac{S_t}{B_t S_0}\\ &=\exp\left(-\frac{\sigma^2}{2} t + \sigma W_t \right). \end{align*} By Girsanov theorem, $\{\widetilde{W}_t \mid t \geq 0\}$, where \begin{align*} \widetilde{W}_t = W_t - \sigma t, \end{align*} is a standard Brownian motion under the probability measure $\widetilde{\mathbb{Q}}$. Moreover, under $\widetilde{\mathbb{Q}}$, \begin{align*} \frac{dS_t}{S_t} = \left(r+ \sigma^2 \right) dt + \sigma d\widetilde{W}_t. \end{align*} Note also that \begin{align*} \frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_t &= \frac{B_tS_0}{S_t}. \end{align*} Therefore, \begin{align*} e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) &=e^{-rT}\mathbb{E}_{\widetilde{\mathbb{Q}}}\left(\frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_T S_T\mathbb{1}_{S_T >K}\right)\\ &=S_0 \widetilde{\mathbb{Q}}(S_T >K)\\ &=S_0 N(d_1). \end{align*} That is, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &=S_0 N(d_1) - K e^{-rT}N(d_2), \end{align*} which is the Black-Scholes formula.

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Gordon
Gordon
  • 21.3k
  • 1
  • 38
  • 83

Per @SKRX's suggestion, another solution is provided below.

For simplicity, we assume that stock price process $\{S_t \mid t \geq 0\}$ follows an SDE, under the risk-neutral measure $\mathbb{Q}$, of the form \begin{align*} \frac{dS_t}{S_t} = r dt + \sigma dW_t, \end{align*} where $r$ is the constant interest rate, $\sigma$ is the constant volatility, and $\{W_t \mid t \geq 0\}$ is a standard Brownian motion. Moreover, let $B_t = e^{rt}$ be the money-market account value at time $t$.

Note that \begin{align*} (S_T-K)^+ &= (S_T-K)\mathbb{1}_{S_T >K}\\ &= S_T\mathbb{1}_{S_T >K} - K \mathbb{1}_{S_T >K}. \end{align*} Then, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}\mathbb{Q}(S_T >K)\\ &=e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) - K e^{-rT}N(d_2). \end{align*} To compute the expectation $\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big)$, we define the probability measure $\widetilde{\mathbb{Q}}$, so that we have the Radon-Nikodym derivative of the form \begin{align*} \frac{d\widetilde{\mathbb{Q}}}{d\mathbb{Q}}\big|_t &= \frac{S_t}{B_t S_0}\\ &=\exp\left(-\frac{\sigma^2}{2} t + \sigma W_t \right). \end{align*} By Girsanov theorem, $\{\widetilde{W}_t \mid t \geq 0\}$, where \begin{align*} \widetilde{W}_t = W_t - \sigma t, \end{align*} is a standard Brownian motion under the probability measure $\widetilde{\mathbb{Q}}$. Moreover, under $\widetilde{\mathbb{Q}}$, \begin{align*} \frac{dS_t}{S_t} = \left(r+ \sigma^2 \right) dt + \sigma d\widetilde{W}_t. \end{align*} Note also that \begin{align*} \frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_t &= \frac{B_tS_0}{S_t}. \end{align*} Therefore, \begin{align*} e^{-rT}\mathbb{E}_{\mathbb{Q}}\big(S_T\mathbb{1}_{S_T >K}\big) &=e^{-rT}\mathbb{E}_{\widetilde{\mathbb{Q}}}\left(\frac{d\mathbb{Q}}{d\widetilde{\mathbb{Q}}}\big|_T S_T\mathbb{1}_{S_T >K}\right)\\ &=S_0 \widetilde{\mathbb{Q}}(S_T >K)\\ &=S_0 N(d_1). \end{align*} That is, \begin{align*} e^{-rT} \mathbb{E}_{\mathbb{Q}}\big((S_T-K)^+ \big) &=S_0 N(d_1) - K e^{-rT}N(d_2), \end{align*} which is the Black-Scholes formula.