Introduction
A proficiency in the change-of-measure technique is useful to the working quant. An excellent summary of the important results is the note Girsanov, Numeraires and all that, by Andrew Lesniewski. In this post, I would like to derive relevant results and then we can enjoy pricing some payoffs together!
Girsanov Theorem
Theorem 1 (Girsanov Theorem) Let \((W^{\mathbb{P}},t\geq 0)\) be a \(\mathbb{P}\) standard brownian motion on \((\Omega,\mathcal{F},\mathbb{P})\) and let \(\phi\) be any adapted process. Choose fixed \(T\) and defined the process \(L\) on \([0,T]\) by:
\[ \begin{align*} dL_t = \phi_t L_t dW^{\mathbb{P}}_t \end{align*} \tag{1}\]
\[ \begin{align*} L_0 = 1 \end{align*} \tag{2}\]
that is:
\[ \begin{align*} L_t = \exp\left(\int_0^t \phi_s dW^{\mathbb{P}}_s - \int_0^t \phi^2_s ds\right) \end{align*} \tag{3}\]
Assume that:
\[ \begin{align*} \mathbb{E}^{\mathbb{P}}[L_T] = 1 \end{align*} \tag{4}\]
and define the new probability measure \(\mathbb{Q}\) on \(\mathcal{F}_t\) by:
\[ L_T = \frac{d\mathbb{Q}}{d\mathbb{P}} \tag{5}\]
Then,
\[ dW^{\mathbb{P}}_t = dW^{\mathbb{Q}}_t + \phi_t dt \tag{6}\]
where \(dW^{\mathbb{Q}}_t\) is a \(\mathbb{Q}\)-standard brownian motion.
Proof.
Our claim is that, under the \(\mathbb{Q}\) measure, the increments \((W^{\mathbb{Q}}_t - W^{\mathbb{Q}}_s)\) are normally distributed with mean \(0\) and variance \((t-s)\). We start with the special case \(s=0\). Using moment generating functions, it is enough to show that:
It is straightforward to derive Equation 3 using Ito’s lemma. Let \(f(x) = \ln x\). Then, \(f_x = -\frac{1}{x}\), \(f_{xx} = -\frac{1}{x^2}\).
\[ \begin{align*} d(\ln L_t) &= -\frac{1}{L_t}dL_t + \frac{1}{L_t^2}(dL_t)^2 \\ &= -\frac{1}{L_t}{\phi_t L_t dW^{\mathbb{P}}_t} + \frac{1}{L_t^2}\phi_t^2 L_t^2 dt\\ &= - \phi_t dW^{\mathbb{P}}_t + \phi_t^2 dt \\ L_t &= \exp\left(-\int_0^t \phi_s dW^{\mathbb{P}}_s + \frac{1}{2}\int_0^t \phi_s^2 ds \right) \end{align*} \]
To prove our main result, we will now use the MGF of the increments. For \(n \in \mathbb{N}\) and \((t_j,j\leq n)\) a partition of \([0,T]\), with \(t_n = T\), I will show that :
\[ \begin{align*} \mathbb{E}^{\mathbb{Q}}\left([\exp\left(\sum_{j=0}^{n-1}\lambda_j (W^{\mathbb{Q}}_{t_{j+1}} - W^{\mathbb{Q}}_{t_{j}})\right)\right] = \exp\left(\lambda_j^2(t_{j+1}-t_j)\right) \end{align*} \tag{7}\]
This proves that the increments are the ones of standard brownian motion.
Let \((\mathcal{F}_{t_j},j\leq n)\) be the filtrations of the Brownian motion at the time of the partition. The proof is by successively conditioning from \(t_{n-1}\) to \(t_1\). We have:
\[ \begin{aligned} \mathbb{E}^{\mathbb{Q}}\left[\exp\left(\sum _{j=0}^{n-1} \lambda _{j} (W_{t_{j+1}}^{\mathbb{Q}} -W_{t_{j}}^{\mathbb{Q}} )\right)\right] & =\mathbb{E}^{\mathbb{P}}\left[\mathbb{E}^{\mathbb{P}}\left[ M_{t_{n}}\exp\left(\sum _{j=0}^{n-1} \lambda _{j} (W_{t_{j+1}}^{\mathbb{Q}} -W_{t_{j}}^{\mathbb{Q}} )\right) |\mathcal{F}_{t_{n-1}}\right]\right]\\ & =\mathbb{E}^{\mathbb{P}}\left[\sum _{j=0}^{n-2} \lambda _{j} (W_{t_{j+1}}^{\mathbb{Q}} -W_{t_{j}}^{\mathbb{Q}} )\mathbb{E}^{\mathbb{P}}\left[ M_{t_{n}}\exp\left( \lambda _{n-1} (W_{t_{n}}^{\mathbb{Q}} -W_{t_{n-1}}^{\mathbb{Q}} )\right) |\mathcal{F}_{t_{n-1}}\right]\right]\\ & =\mathbb{E}^{\mathbb{P}}\left[\sum _{j=0}^{n-2} \lambda _{j} (W_{t_{j+1}}^{\mathbb{Q}} -W_{t_{j}}^{\mathbb{Q}} )\mathbb{E}^{\mathbb{P}}\left[\exp\left( -\int _{t_{n-1}}^{t_{n}} \theta dW_{s}^{\mathbb{Q}} +\frac{1}{2}\int _{t_{n-1}}^{t_{n}} \theta _{s}^{2} ds\right)\exp\left( \lambda _{n-1}\int _{t_{n-1}}^{t_{n}} dW_{s}^{\mathbb{Q}}\right) |\mathcal{F}_{t_{n-1}}\right]\right]\\ & =\mathbb{E}^{\mathbb{P}}\left[\sum _{j=0}^{n-2} \lambda _{j} (W_{t_{j+1}}^{\mathbb{Q}} -W_{t_{j}}^{\mathbb{Q}} )\mathbb{E}^{\mathbb{P}}\left[\exp\left(\int _{t_{n-1}}^{t_{n}}( -\theta _{s} +\lambda _{n-1}) dW_{s}^{\mathbb{Q}} +\frac{1}{2}\int _{t_{n-1}}^{t_{n}} \theta _{s}^{2} ds\right) |\mathcal{F}_{t_{n-1}}\right]\right]\\ & =\mathbb{E}^{\mathbb{P}}\left[\sum _{j=0}^{n-2} \lambda _{j} (W_{t_{j+1}}^{\mathbb{Q}} -W_{t_{j}}^{\mathbb{Q}} )\mathbb{E}^{\mathbb{P}}\left[\exp\left(\int _{t_{n-1}}^{t_{n}}( -\theta _{s} +\lambda _{n-1})\left( dW_{s}^{\mathbb{P}} +\theta ds\right) +\frac{1}{2}\int _{t_{n-1}}^{t_{n}} \theta _{s}^{2} ds\right) |\mathcal{F}_{t_{n-1}}\right]\right]\\ & =\mathbb{E}^{\mathbb{P}}\left[\sum _{j=0}^{n-2} \lambda _{j} (W_{t_{j+1}}^{\mathbb{Q}} -W_{t_{j}}^{\mathbb{Q}} )\exp\left(\frac{1}{2}\int _{t_{n-1}}^{t_{n}} \theta _{s}^{2} +\int _{t_{n-1}}^{t_{n}} \theta _{s}( -\theta _{s} +\lambda _{n-1}) ds\right) \ \exp\left(\frac{1}{2}\int _{t_{n-1}}^{t_{n}}( -\theta _{s} +\lambda _{n-1})^{2} ds\right)\right]\\ & \left\{\ \int XdW_{s}^{\mathbb{P}} \ \text{ is a }\mathcal{N}^{\mathbb{P}}\left( 0,\int \mathbb{E}\left[ X^{2}\right] ds\right) \ \text{gaussian random variable.}\right\} \ \\ & =\mathbb{E}^{\mathbb{P}}\left[\sum _{j=0}^{n-2} \lambda _{j} (W_{t_{j+1}}^{\mathbb{Q}} -W_{t_{j}}^{\mathbb{Q}} )\exp\left(\int _{t_{n-1}}^{t_{n}}\left( -\frac{1}{2} \theta _{s}^{2} +\lambda _{n-1} \theta _{s} +\frac{1}{2} \theta _{s}^{2} -\lambda _{n-1} \theta +\lambda _{n-1}^{2}\right) ds\right)\right]\\ & =\mathbb{E}^{\mathbb{P}}\left[\sum _{j=0}^{n-2} \lambda _{j} (W_{t_{j+1}}^{\mathbb{Q}} -W_{t_{j}}^{\mathbb{Q}} )\exp\left( \lambda _{n-1}\int _{t_{n-1}}^{t_{n}} ds\right)\right]\\ & =\exp( \lambda _{n-1}( t_{n} -t_{n-1})) \cdot \mathbb{E}^{\mathbb{P}}\left[\sum _{j=0}^{n-2} \lambda _{j} (W_{t_{j+1}}^{\mathbb{Q}} -W_{t_{j}}^{\mathbb{Q}} )\right] \end{aligned} \]
Here, I used the fact that \(M_{t_{n-1}}\) is \(\mathcal{F}_{t_{n-1}}\) measurable. I can now condition on \(\mathcal{F}_{t_{n-2}}\) down to \(\mathcal{F}_{t_1}\) and proceed as above to obtain the desired result.
The process \(\phi_t\) is called the Girsanov kernel.
What is a numeraire?
As Shreve puts it, a numeraire is the unit of account in which other assets are denominated. In practice, we tend to choose numeraires that simply the payoff expression.
Any strictly positive (non-dividend paying) price process can be chosen as a numeraire. A numeraire must be a tradable asset.
Consider a unit of stock worth \(S_t\). It can be used as numeraire, because the price process \(e^{-rt}S_t\) (assume a constant short rate) is a martingale under risk-neutral measure \(\mathbb{Q}^M\). Powers of the stock price \(S_t^\alpha\) cannot be used as numeraires, because their discounted values are not martingales under the risk-neutral measure. Clearly, set the short rate \(r = 0\), then \(\mathbb{E}^{\mathbb{Q}^M}[S_T^2] \geq (\mathbb{E}^{\mathbb{Q}^M}[S_T])^2 =S_0^2\) by the Jensen’s inequality.
The price-process \(V_t\) of a derivative contract that pays \(V_T=S_T^2\) is a martingale under \(\mathbb{Q}\) and can be used as a numeraire.
Consider the price of a contract that pays a unit sum \(1\) at maturity \(T\). This instrument is the zero-coupon bond. Its an observable and tradable asset. Its price process \(P(t,T) = \mathbb{E}^{\mathbb{Q}^M}[1/M_T]\) can be used as a numeraire. \(\mathbb{Q}^T\) is called the \(T\)-forward measure.
Abstract Bayes Formula
Theorem 2 (Abstract Bayes Formula) Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space and let \(\mathbb{Q}\) be any other probability measure on it. By the Radon-Nikodym theorem, \(\exists L = \frac{d\mathbb{Q}}{d\mathbb{P}}\), \(L \geq 0\) with \(\mathbb{E}^{\mathbb{P}}[L]=1\). Then we have:
\[ \begin{align*} \mathbb{E}^{\mathbb{Q}}[X|\mathcal{G}] = \frac{\mathbb{E}^{\mathbb{P}}[LX|\mathcal{G}]}{\mathbb{E}^{\mathbb{P}}[L|\mathcal{G}]} \end{align*} \tag{8}\]
Proof.
By the definition of conditional expectations, recall that if \(W\) is any \(\mathcal{G}\)-measurable random variable, then the conditional expectation must satisfy the relationship:
\[ \mathbb{E}[WX] = \mathbb{E}[W\mathbb{E}[X|\mathcal{G}]] \]
It is sufficient to prove that:
\[ \mathbb{E}^{\mathbb{P}}[X|\mathcal{G}]\mathbb{E}^{\mathbb{Q}}[L|\mathcal{G}] = \mathbb{E}^{\mathbb{P}}[LX|\mathcal{G}] \]
Let \(G\) be an arbitrary set in \(\mathcal{G}\). We have:
\[ \begin{align*} & \int_G \mathbb{E}^{\mathbb{Q}}[X|\mathcal{G}]\mathbb{E}^{\mathbb{P}}[L|\mathcal{G}] d\mathbb{P} \\ &= \int_G \mathbb{E}^{\mathbb{P}}[L\cdot \mathbb{E}^{\mathbb{Q}}[X|\mathcal{G}]|\mathcal{G}] d\mathbb{P} \\ & \quad \{ \mathbb{E}^{\mathbb{Q}}[X|\mathcal{G}] \text{ is }\mathcal{G}\text{-measurable } \}\\ &= \int_G L\cdot \mathbb{E}^{\mathbb{Q}}[X|\mathcal{G}] d\mathbb{P}\\ &= \int_G \frac{d\mathbb{Q}}{d\mathbb{P}}\cdot \mathbb{E}^{\mathbb{Q}}[X|\mathcal{G}] d\mathbb{P}\\ &= \int_G \mathbb{E}^{\mathbb{Q}}[X|\mathcal{G}] d\mathbb{Q} \end{align*} \]
Hence, proved.
Note that, the filtration \(\mathcal{G}\) is the same irrespective of what probability measure we construct on \(\Omega\).
Martingale property
Proposition 1 Assume that there exists a numeraire \(M\) and a probability measure \(\mathbb{Q}^M\), such that the price of any traded asset \(X\) (without intermediate payments) relative to \(M\) is a martingale under \(\mathbb{Q}^M\).That is:
\[ \frac{X_t}{M_t} = \mathbb{E}^{\mathbb{Q}^M} \left\{\frac{X_T}{M_T}|\mathcal{F}_t\right\} \]
Let \(N_t\) be an arbitrary numeraire. Then, there exists a probability measure \(\mathbb{Q}^N\) such that the price of \(X\) normalized by \(N\) is a martingale under \(\mathbb{Q}^N\).
\[ \frac{X_t}{N_t} = \mathbb{E}^{\mathbb{Q}^N} \left\{\frac{X_T}{N_T}|\mathcal{F}_t\right\} \]
Moreover, the Radon-Nikodym derivative defining the measure \(\mathbb{Q}^N\) is given by:
\[ \frac{d\mathbb{Q}^N}{d\mathbb{Q}^M} = \frac{N_T/N_0}{M_T/M_0} \]
Proof.
We have:
\[ X_0 = M_0 \mathbb{E}^{\mathbb{Q}^M}\left[\frac{X_T}{M_T}\right] \]
Imposing the simple fact that, the price of the derivative contract should be the same, even if we switch numeraires from \(M\) to \(N\), we should have:
\[ X_0 = N_0 \mathbb{E}^{\mathbb{Q}^N}\left[\frac{X_T}{N_T}\right] \]
Thus,
\[ \begin{aligned} N_{0}\mathbb{E}^{\mathbb{Q}^{N}}\left[\frac{X_{T}}{N_{T}}\right] & =M_{0}\mathbb{E}^{\mathbb{Q}^{M}}\left[\frac{X_{T}}{M_{T}}\right]\\ \frac{N_{T}}{N_{0}} \times N_{0}\mathbb{E}^{\mathbb{Q}^{N}}\left[\frac{X_{T}}{N_{T}}\right] & =\frac{N_{T}}{N_{0}} \times M_{0} \ \mathbb{E}^{\mathbb{Q}^{M}}\left[\frac{X_{T}}{M_{T}}\right] \quad \left\{\text{Multiplying both sides by }\frac{N_{T}}{N_{0}}\right\}\\ \Longrightarrow \mathbb{E}^{\mathbb{Q}^{N}}[ X_{T}] & =\mathbb{E}^{\mathbb{Q}^{M}}\left[\frac{N_{T} /N_{0}}{M_{T} /M_{0}} X_{T}\right] \end{aligned} \]
But, we know that:
\[ \mathbb{E}^{\mathbb{Q}^N}[X_T] = \mathbb{E}^{\mathbb{Q}^M}\left[\frac{d\mathbb{Q}^N}{d\mathbb{Q}^M}X_T\right] \]
Consequently, our candidate for the Radon-Nikodym derivative should be:
\[ L_T = \frac{d\mathbb{Q}^N}{d\mathbb{Q}^M} = \frac{N_T/N_0}{M_T/M_0} \]
Further \((X_t/N_t)\) is a martingale under \(\mathbb{Q}^N\). Its easy to see that:
\[ \begin{aligned} \mathbb{E}^{\mathbb{Q}^{N}}\left[\frac{X_{T}}{N_{T}} |\mathcal{F}_{t}\right] & =\frac{\mathbb{E}^{\mathbb{Q}^{M}}\left[ L_{T} \cdot \frac{X_{T}}{N_{T}} |\mathcal{F}_{t}\right]}{\mathbb{E}^{\mathbb{Q}^{M}}[ L_{T} |\mathcal{F}_{t}]} \quad \left\{\text{ Abstract bayes formula }\right\}\\ & =\frac{\mathbb{E}^{\mathbb{Q}^{M}}\left[ L_{T} \cdot \frac{X_{T}}{N_{T}} |\mathcal{F}_{t}\right]}{L_{t}}\\ & =\frac{\mathbb{E}^{\mathbb{Q}^{M}}\left[\frac{N_{T}}{N_{0}} \cdot \frac{M_{0}}{M_{T}} \cdot \frac{X_{T}}{N_{T}} |\mathcal{F}_{t}\right]}{\frac{N_{t}}{N_{0}} \cdot \frac{M_{0}}{M_{t}}}\\ & =\frac{\mathbb{E}^{\mathbb{Q}^{M}}\left[\frac{N_{T}}{N_{0}} \cdot \frac{M_{0}}{M_{T}} \cdot \frac{X_{T}}{N_{T}} |\mathcal{F}_{t}\right]}{\frac{N_{t}}{N_{0}} \cdot \frac{M_{0}}{M_{t}}}\\ & =\frac{M_{t}}{N_{t}} \cdot \mathbb{E}^{\mathbb{Q}^{M}}\left[\frac{X_{T}}{M_{T}}\right]\\ & =\frac{M_{t}}{N_{t}} \cdot \frac{X_{t}}{M_{t}}\\ & =\frac{X_{t}}{N_{t}} \end{aligned} \]
Since we determined the relevant likelihood process, it is easy to find the Girsanov Kernel.
Drift transformation under change of numeraire
Suppose we are interested in the dynamics of the stochastic process \((X_t,t\geq 0)\). Under \(\mathbb{Q}^M\) measure, its dynamics reads:
\[ \begin{aligned} dX(t) = \mu_X^{\mathbb{Q}^M}(t) dt + c_X(t)dW^{\mathbb{Q}^M}_t \end{aligned} \tag{9}\]
I supressed \(\mu_X^{\mathbb{Q}^M}(t,X_t)\) as \(\mu_X^{\mathbb{Q}^M}(t)\) for brevity.
Under the \(\mathbb{Q}^N\) measure, its dynamics reads:
\[ \begin{aligned} dX(t) = \mu_X^{\mathbb{Q}^N}(t) dt + c_X(t)dW^{\mathbb{Q}^N}_t \end{aligned} \tag{10}\]
Remember that the diffusion coefficients in these equations are unaffected by the change of measure! We assume that \(\mathbb{Q}^M\) is associated with the numeraire \(M(t)\) whose dynamics is given by:
\[ dM(t) = \mu_M(t)dt + c_M(t)dW^{\mathbb{Q}^M} \]
and that the numeraire \(N\) has \(\mathbb{Q}^M\) dynamics:
\[ dN(t) = \mu_N(t)dt + c_N(t)dW^{\mathbb{Q}^N} \]
According to the Girsanov theorem, the likelihood process \(L(t)\) accompanying this change of measure is a martingale under the measure \(\mathbb{Q}^M\) measure and satisfies the stochastic differential equation:
\[ dL_t = L(t)\theta(t)dW^{\mathbb{Q}^M}_t \]
Explicitly, the likelihood process \(L(t)\) is given by the stochastic exponential of the martingale \(\int_0^t \theta_s dW^{\mathbb{Q}^M}_s\):
\[ L(t) = \exp\left(\int_0^t \theta_s dW^{\mathbb{Q}^M}_s - \frac{1}{2}\int_0^t \theta^2_s ds \right) \]
On the other hand, from Proposition 1, we have:
\[ L_t = \frac{N_t / N_0}{M_t / M_0} \]
Differentiating using Ito’s lemma, we have:
\[ \begin{aligned} dL_{t} & =\frac{M_{0}}{N_{0}} d\left(\frac{N_{t}}{M_{t}}\right)\\ & =\frac{M_{0}}{N_{0}}\left( -\frac{N_{t}}{M_{t}^{2}} dM_{t} +\frac{1}{M_{t}} dN_{t} +\frac{1}{2} \cdot \frac{2N_{t}}{M_{t}^{3}}( dM_{t})^{2} -\frac{1}{M_{t}^{2}}( dM_{t} \cdot dN_{t})\right)\\ & \begin{array}{l} =\frac{M_{0}}{N_{0}}( -\frac{N_{t}}{M_{t}^{2}}\left( \mu _{M}( t) dt+c_{M}( t) dW_{t}^{\mathbb{Q}^{M}}\right) +\frac{1}{M_{t}}\left( \mu _{N}( t) dt+c_{N}( t) dW_{t}^{\mathbb{Q}^{M}}\right)\\ +\frac{N_{t}}{M_{t}^{3}} c_{M}^{2}( t) dt-\frac{1}{M_{t}^{2}} c_{M}( t) c_{N}( t) dt \end{array} \end{aligned} \]
But since \(L_t\) is driftless, we can ignore the \(dt\) terms (whatever they are, they are bound to cancel out) and only look at the diffusion coefficient. So, we can write:
\[ \begin{aligned} dL_{t} & =\frac{M_{0}}{N_{0}}\left( -\frac{N_{t}}{M_{t}^{2}} c_{M}( t) +\frac{1}{M_{t}} c_{N}( t)\right) dW{_{t}^{\mathbb{Q}}}^{M}\\ & =\frac{N_{t} /N_{0}}{M_{t} /M_{0}}\left(\frac{c_{N}( t)}{N_{t}} -\frac{c_{M}( t)}{M_{t}}\right) dW{_{t}^{\mathbb{Q}}}^{M}\\ & =L_{t}\left(\frac{c_{N}( t)}{N_{t}} -\frac{c_{M}( t)}{M_{t}}\right) dW{_{t}^{\mathbb{Q}}}^{M} \end{aligned} \]
Comparing this, we can infer that:
\[ \theta_t = \frac{c_N(t)}{N_t} - \frac{c_M(t)}{M_t} \]
Pricing the payoff \(V(T) = S_T(S_T - K)^{+}\)
Under the risk-neutral measure \(\mathbb{Q}^M\) associated with the money-market account numeraire \(M(t)\), the stock price \(S(t)\) evolves according to:
\[ \begin{align*} dS_t = r S_t dt + \sigma S_t dW^{\mathbb{Q}^M} \end{align*} \tag{11}\]
which has the solution:
\[ \begin{align*} S_t = S_0\exp\left[\left(r - \frac{\sigma^2}{2}\right)t + \sigma W^{\mathbb{Q}^M}(t)\right] \end{align*} \tag{12}\]
By the risk-neutral pricing formula, we have:
\[ \begin{align*} V(0) &= \mathbb{E}^{\mathbb{Q}^M}\left[\frac{V(T)}{M(T)}\right] \\ &= \mathbb{E}^{\mathbb{Q}^M}\left[\frac{S_T}{M_T}(S_T - K)^{+}\right] \end{align*} \tag{13}\]
By the change-of-measure formula, if \(N\) is any other numeraire with associated probability measure \(\mathbb{Q}^N\), we know that:
\[ \begin{align*} \mathbb{E}^{\mathbb{Q}^N}[V(T)]=\mathbb{E}^{\mathbb{Q}^M}\left[\frac{d\mathbb{Q}^N}{d\mathbb{Q}^M}V(T)\right] \end{align*} \tag{14}\]
We switch to the stock numeraire \(S_t\). The Radon-Nikodym derivative \(L_T = d\mathbb{Q}^S/d\mathbb{Q}^M\) is simply:
\[ \begin{align*} L_T = \frac{d\mathbb{Q}^S}{d\mathbb{Q}^M} = \frac{S(T)/S(0)}{M(T)/M(0)} = \frac{1}{S(0)}\cdot \frac{S(T)}{M(T)} \end{align*} \tag{15}\]
The dynamics of \(L_t\) under the \(\mathbb{Q}_M\) measure is given by:
\[ dL_t = \phi_t L_t dW^{\mathbb{Q}^M} \]
where the Girsanov kernel \(\phi_t = \sigma\). Then, the Girsanov transformation is:
\[ W^{\mathbb{Q}^M}_T = W^{\mathbb{Q}^S}_T + \sigma T \tag{16}\]
So, the \(\mathbb{Q}^S\) dynamics of the stock price is:
\[ \begin{align*} S_T &= S_0 \exp\left[\left(r - \frac{\sigma^2}{2}\right)T + \sigma (W^{\mathbb{Q}^S}_T + \sigma T)\right]\\ &= S_0 \exp\left[\left(r + \frac{\sigma^2}{2}\right)T + \sigma (W^{\mathbb{Q}^S}_T)\right] \end{align*} \tag{17}\]
We can develop the expression in Equation 13 to be:
\[ \begin{align*} V(0) &= S_0\mathbb{E}^{\mathbb{Q}^M}\left[\frac{1}{S_0}\frac{S_T}{M_T}(S_T - K)^{+}\right] \end{align*} \tag{18}\]
Applying the change-of-measure (abstract Baye’s formula),we have:
\[ \begin{align*} V(0) &= S_0\mathbb{E}^{\mathbb{Q}^S}\left[(S_T - K)^{+}\right] \end{align*} \tag{19}\]
Now, we have:
\[ \begin{align*} \mathbb{Q}^S(S_T > K) &= \mathbb{Q}^S\left(S_0 \exp\left[\left(r + \frac{\sigma^2}{2}\right)T + \sigma (W^{\mathbb{Q}^S}_T)\right] > K\right)\\ &= \mathbb{Q}^S\left(S_0 \exp\left[\left(r + \frac{\sigma^2}{2}\right)T + \sigma (-\sqrt{T}Z)\right] > K\right)\\ &= \mathbb{Q}^S\left(\log \frac{S_0}{K} + \left(r + \frac{\sigma^2}{2}\right)T > \sigma\sqrt{T}Z\right)\\ \end{align*} \tag{20}\]
where we subbed \(-\sqrt{T}Z=W^{\mathbb{Q}^S}_T\). Recall, the standard normals \(Z\) and \(-Z\) have the same distribution by symmetry.
Define
\[ d_{-} = \frac{\log \frac{S_0}{K} + \left(r + \frac{\sigma^2}{2}\right)T}{\sigma\sqrt{T}} \tag{21}\]
Then, since \(Z \sim \mathcal{N}^{\mathbb{Q}^S}(0,1)\):
\[ \mathbb{Q}^S(S_T > K) = \mathbb{Q}^S(Z < d_{-}) = \Phi(d_{-}) \tag{22}\]
So, we can expand the expectation in Equation 19. The indicator random variable \(1_{\{S_T > K\}}\) is \(1\) for all points \(Z < d_{-}\). So, the limits of integration will be \(-\infty\) to \(d_{-}\).
\[ \begin{align*} V(0) &= S_0\left[\int_{-\infty}^{d_{-}} S_T d\mathbb{Q}^S - K\int_{-\infty}^{d_{-}} dQ^S\right]\\ &= S_0\left[\int_{-\infty}^{d_{-}} S_T f_Z^{\mathbb{Q}^S}(z)dz - K\int_{-\infty}^{d_{-}} f_Z^{\mathbb{Q}^S}(z)dz\right]\\ &= S_0\left[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_{-}} S_0 \exp\left[\left(r + \frac{\sigma^2}{2}\right)T - \sigma \sqrt{T}z\right] \exp(-\frac{z^2}{2})dz - K\Phi(d_{-})\right] \end{align*} \]
Completing the square, we have:
\[ \begin{align*} V(0) &= S_0\left[\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_{-}} S_0 \exp\left[-\frac{1}{2}(z^2 + 2\sigma\sqrt{T}z + (\sigma \sqrt{T})^2)\right] \exp\left[\left(r + \sigma^2\right)T\right]dz - K\Phi(d_{-})\right]\\ &=S_0 \left[S_0\frac{e^{(r + \sigma^2) T}}{\sqrt{2\pi}}\int_{-\infty}^{d_{-}} \exp\left(-\frac{1}{2}(z + \sigma\sqrt{T})^2\right) dz - K\Phi(d_{-})\right]\\ \end{align*} \]
Let \(u = z + \sigma\sqrt{T}\). And define:
\[ d_{+} = d_{-} + \sigma \sqrt{T} = \frac{\log \frac{S_0}{K} + \left(r + \frac{3}{2}\sigma^2\right)}{\sigma \sqrt{T}} \]
We have the closed-form formula:
\[ V(0) = S_0^2 e^{(r+\sigma^2)T} \Phi(d_{+}) - KS_0 \Phi(d_{-}) \]