# Find a formula for the price of a derivative paying $\max(S_T(S_T-K),0)$

Develop a formula for the price of a derivative paying

$$\max(S_T(S_T-K))$$

in the Black Scholes model.

Apparently the trick to this question is to compute the expectation under the stock measure. So,

$$\frac{C_0}{S_0} = \mathbb{E}[\frac{S_T\max{(S_T-K,0)}}{N_T}]$$

and taking $$N_T = S_T$$. We can split this expectation into two parts,

$$\mathbb{E}_{new}[\max(S_T-K,0)] = \mathbb{E}_{new}[S_T\mathbb{I}_{S_T>K}] - \mathbb{E}_{new}[K\mathbb{I}_{S_T>K}]$$

Focusing on the second term, we can show that the final stock price is distributed in the stock measure is,

$$S_T = S_0 \exp{\{ (r+\frac{\sigma^2}{2})T +\sigma \sqrt{T} N(0,1) \}}\tag{1}$$

And then we have $$\mathbb{E}_{new}[K\mathbb{I}_{S_T>K}] = K \mathbb{P}(S_T > K) = K N(d_1)$$.

Now concentrating on $$\mathbb{E}_{new}[S_T\mathbb{I}_{S_T>K}]$$, we can rewrite the expectation as an integral,

$$\mathbb{E}_{new}[S_T\mathbb{I}_{S_T>K}] = \frac{S_0}{\sqrt{2\pi}} \int^{\inf}_l \exp{\frac{-x^2}{2}}\exp{(r+\frac{\sigma^2}{2})T+\sigma\sqrt(T) x} dx\tag{2}$$

with

$$l = \frac{\ln(k/S_0)-(r+\frac{\sigma^2}{2})T}{\sigma \sqrt{T}}$$

1. How has $$(1)$$ been derived? How do we go from the stock price distribution in the normal numeraire as $$S_t = S_0 \exp{\{ (r-\frac{\sigma^2}{2})T +\sigma \sqrt{T}W_t \}}$$ to $$(1)$$? Could this be explained in detail please as it is key to understanding how to solve these questions. I need to understand how all the moving parts bit together.

2. How has this last equality been derived? I am guessing that the $$\mathbb{P}$$ is different, but again I cannot see how to derive it. Moreover, could it be explained in detail as to how the $$d_1$$ comes into it.

3. How has this integral been derived? I cannot see where the $$\exp{\frac{-x^2}{2}}$$ come into the integral, this seems to be some distibution from somewhere.

• Please note that I have already read the books on topics such as Girsanov's theorem, change of measure, probability theory etc. I just cannot understand how all the parts fit together to answer this question Jul 6, 2020 at 17:12
• If and when you use things like change of numeraire, fundamental theorem of asset pricing, radon nikodym, please clearly state when and why you are using them Jul 6, 2020 at 17:14
• What is your understanding of a martingale? Jul 6, 2020 at 17:20
• excellent @DaneelOlivaw i have all the requisite maths, check out my math s.e. if you want to know more Jul 6, 2020 at 17:31
• To heuristically understand that change of measure with $-\sigma t$ (see the replies of Kupoc allahoui or Jan Stuller), you might know that assets divided by numéraire must be martingales. Your market, as explained by KeSchn, has actually 2 assets: the stock $S_t$ and the bank account $B_t$. You should also know that a martingale $M$ has no drift, i.e. $dM_t=0\times dt+(\dots)\times dW_t$. Now, apply Itô's Lemma to $B_t/S_t$ and ask yourself "how should I tweak $W_t$ such that $B_t/S_t$ has no drift?" Jul 6, 2020 at 19:23

I provide a solution in three steps.

• The first step carefully outlines how to split up the expectation and what new measures are used. This first step does not require any special model assumption and holds in a very general framework. I derive a formula for the option price that resembles the standard Black-Scholes formula.
• In a second step, I assume that the stock price follows a geometric Brownian motion and use Girsanov's theorem to derive a precise formula for all (probabilistic) terms involved. However, I want to present a more elegant approach which does not require to integrate the Gaussian density. That's just pointlessly tedious and makes it harder to generalise the approach to other processes.
• The third section states Girsanov's theorem, links it to numéraire changes and outlines how this change impacts the drift of the stock price.

## General Numéraire Changes

As you said, the key is a numéraire change as originally outlined by Geman et al. (1995). The standard risk-neutral measure ($$\mathbb Q$$ or $$\mathbb Q^0$$) uses the (locally) risk-free bank account, $$B_t=e^{rt}$$, as numéraire. We could easily allow for a general interest rate process $$B_t=\exp\left(\int_0^t r_s\mathrm{d}s\right)$$. We define a new probability measure, $$\mathbb Q^1\sim\mathbb Q^0$$ which uses the stock price, $$S_t$$ as numéraire. The new measure, $$\mathbb Q^1$$, is defined via

\begin{align*} \frac{\mathrm{d}\mathbb Q^1}{\mathrm d\mathbb Q^0} = \frac{S_T}{S_0}\frac{B_0}{B_T}=\frac{S_T}{S_0}e^{-rT}. \end{align*}

If the stock pays dividends at rate $$\delta$$, you use the reinvested stock price, $$S_te^{\delta t}$$, as numéraire.

The price of your option is then

\begin{align*} e^{-rT}\mathbb{E}^\mathbb{Q}[\max\{S_T^2-KS_T,0\}] &=e^{-rT}\mathbb{E}^{\mathbb{Q}^1}\left[\frac{\mathrm{d}\mathbb Q^0}{\mathrm d\mathbb Q^1}\max\{S_T^2-KS_T,0\}\right] \\ &= S_0\mathbb{E}^{\mathbb{Q}^1}\left[\max\{S_T-K,0\}\right] \\ &= S_0\left(\mathbb{E}^{\mathbb{Q}^1}[S_T\mathbb{1}_{\{S_T\geq K\}}] -K\mathbb{E}^{\mathbb Q^1}[\mathbb{1}_{\{S_T\geq K\}}]\right) \\ &= S_0\left(\mathbb{E}^{\mathbb{Q}^1}[S_T\mathbb{1}_{\{S_T\geq K\}}] -K\mathbb Q^1[\{S_T\geq K\}]\right). \end{align*}

To compute the first expectation, we (again) use a change of numéraire. I follow this great paper from Mark Joshi. Let $$N_{t,T}^\alpha$$ be the time-$$t$$ price of an asset (claim) paying $$S_T^\alpha$$ at time $$T$$. Because of Jensen's inequality, $$N_{t,T}^\alpha\neq S_t^\alpha$$ if $$\alpha\neq0,1$$. There is of course a restriction on the choice of $$\alpha$$. If $$\alpha$$ is too large, then $$S_t^\alpha$$ may not be integrable (in particular if your stock price model includes fat tails). So, for now we just assume that $$\alpha$$ is chosen appropriately. Then,

\begin{align*} \frac{\mathrm{d}\mathbb Q^\alpha}{\mathrm d\mathbb Q^0} = \frac{N_{T,T}^\alpha B_0}{N_{0,T}^\alpha B_T} . \end{align*}

Thus,

\begin{align*} \frac{\mathrm{d}\mathbb Q^\alpha}{\mathrm d\mathbb Q^1} =\frac{\mathrm{d}\mathbb Q^\alpha}{\mathrm d\mathbb Q^0} \frac{\mathrm{d}\mathbb Q^0}{\mathrm d\mathbb Q^1} = \frac{N_{T,T}^\alpha B_0}{N_{0,T}^\alpha B_T} \frac{S_0B_T}{S_TB_0} = \frac{S_{T}^\alpha}{N_{0,T}^\alpha } \frac{S_0}{S_T}. \end{align*}

Using $$\alpha=2$$, we obtain

\begin{align*} \mathbb E^{\mathbb Q^1}[S_T\mathbb 1_{\{S_T\geq K\}}] = \frac{N_{0,T}^2}{S_0}\mathbb E^{\mathbb Q^2}[\mathbb 1_{\{S_T\geq K\}}] =\frac{N_{0,T}^2}{S_0}\mathbb Q^2[\{S_T\geq K\}]. \end{align*}

The final option price thus reads as $$e^{-rT}\mathbb{E}^\mathbb{Q}[\max\{S_T^2-KS_T,0\}] = N_{0,T}^2\mathbb Q^2[\{S_T\geq K\}] - KS_0\mathbb Q^1[\{S_T\geq K\}],$$

which beautifully resembles the Black-Scholes formula. This also hints to how a formula for the price of a general power option looks like.

## Black-Scholes Model

To actually implement the above equation, we need to find expressions for $$\mathbb Q^\alpha[\{S_T\geq K\}]$$ and $$N_{t,T}^\alpha$$. These formulae will depend on the chosen stock price model. Here, we opt for the simplest one, the Black-Scholes setting with a log-normally distributed stock price.

Let's begin with the simpler problem: the price of a claim paying $$S_T^\alpha$$. Using standard risk-neutral pricing and the martingale property $$\mathbb{E}[e^{\sigma W_t}|\mathcal{F}_s]=e^{\frac{1}{2}\sigma^2(t-s)+\sigma W_s}$$, we obtain \begin{align*} N_{t,T}^\alpha &= e^{-r(T-t)}\mathbb{E}^{\mathbb Q}[S_T^\alpha|\mathcal{F}_t] \\ &= e^{-r(T-t)}\mathbb{E}^{\mathbb Q}\left[S_0^\alpha\exp\left(\alpha\left(r-\frac{1}{2}\sigma^2\right)T+\alpha\sigma W_T \right)\bigg|\mathcal{F}_t\right] \\ &= e^{-r(T-t)}S_0^\alpha\exp\left(\alpha\left(r-\frac{1}{2}\sigma^2\right)T+\frac{1}{2}\alpha^2\sigma^2(T-t)+\sigma\alpha W_t\right) \\ &= e^{-r(T-t)}S_t^\alpha\exp\left(\alpha\left(r-\frac{1}{2}\sigma^2\right)(T-t)+\frac{1}{2}\alpha^2\sigma^2(T-t)\right) \\ &= S_t^\alpha \exp\left((T-t)(r(\alpha-1)+0.5\sigma^2(\alpha^2-\alpha)\right) \end{align*}

Of course, the price $$N_{t,T}^\alpha$$ is log-normally distributed. By the way, using Itô's Lemma, we obtain $$\mathrm{d}N_{t,T}^\alpha=rN_{t,T}^\alpha\mathrm{d}t+\alpha\sigma N_{t,T}^\alpha\mathrm{d}W_t$$.

To conclude, we need to compute the exercise probability $$\mathbb{Q}^\alpha[\{S_T\geq K\}]$$. Under $$\mathbb{Q}$$, the stock price has drift $$r$$ and under $$\mathbb Q^1$$, the stock price has drift $$r+\sigma^2$$, see this excellent answer and this question for an intuitive explanation. Under $$\mathbb Q^\alpha$$, the stock price has drift $$r+\alpha\sigma^2$$. I explain this in detail in the third section of this answer.

For now, let's accept the above drift changes. Let $$S_T$$ be a geometric Brownian motion under any arbitrary probability measure $$\mathcal{P}$$ (this could be the real world measure $$\mathbb P$$, the risk-neutral measure $$\mathbb Q$$ or a stock measure $$\mathbb Q^\alpha$$). Then, $$S_T=S_0\exp\left(\left(\mu-\frac{1}{2}\sigma^2\right)T+\sigma W_T\right)$$, where $$\mu$$ is the drift under the respective measure $$\mathcal{P}$$. Thus, using that $$W_T\sim N(0,T)$$, \begin{align*} \mathcal{P}[\{S_T\geq K\}] &= \mathcal{P}[\{\ln(S_T)\geq\ln(K)\}] \\ &=\mathcal{P}\left[\left\{\left(\mu-\frac{1}{2}\sigma^2\right)T+\sigma W_T \geq -\ln\left(\frac{S_0}{K}\right)\right\}\right] \\ &=\mathcal{P}\left[\left\{ Z \geq -\frac{\ln\left(\frac{S_0}{K}\right)+ \left(\mu-\frac{1}{2}\sigma^2\right)T }{\sigma \sqrt{T}}\right\}\right] \\ &=1-\Phi\left(-\frac{\ln\left(\frac{S_0}{K}\right)+ \left(\mu-\frac{1}{2}\sigma^2\right)T }{\sigma \sqrt{T}}\right)\\ &=\Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+ \left(\mu-\frac{1}{2}\sigma^2\right)T }{\sigma \sqrt{T}}\right), \end{align*} where $$Z\sim N(0,1)$$. I used the property $$\Phi(x)=1-\Phi(-x)$$.

Depending on which measure we use for $$\mathcal{P}$$, we merely need to the right drift. For example, under $$\mathbb{Q}^\alpha$$, we use $$r+\alpha\sigma^2$$ as drift ($$\mu$$) of the stock price. Thus, \begin{align*} \mathbb{Q}^\alpha[\{S_T\geq K\}] = \Phi\left(\frac{\ln\left(\frac{S_0}{K}\right)+\left(r+\left(\alpha-\frac{1}{2}\right)\sigma^2\right)T}{\sigma\sqrt{T}}\right). \end{align*}

We recover the special cases $$\mathbb Q^1[\{S_T\geq K\}]=\Phi(d_1)$$ and $$\mathbb Q^0[\{S_T\geq K\}]=\Phi(d_2)$$.

I thoroughly recommend reading Joshi's paper which contains more details and applications of numéraire changes, including an introductory section on the Black-Scholes model!

## Girsanov's Theorem

I will first state Girsanov's theorem and use the change of numeraire formula to show you how to switch between two risk-neutral probability measures. Then, I'll describe how this change affects the drift of the stock price.

I cite (the one-dimensional) Girsanov theorem from Björk's book, Theorem 12.3. As an alternative, see Shreve or any other textbook on stochastic calculus.

Let $$(\Omega,\mathcal{F},(\mathcal{F}_t),\mathbb{P})$$ be a filtered probability space carrying a standard Brownian motion $$W_T^\mathbb{P}$$. Let $$\varphi_t$$ be an adapted process (pricing kernel''). Define $$\mathrm{d}L_t=\varphi_tL_t\mathrm{d}W_t^\mathbb{P}$$ with $$L_0=1$$ such that $$L_t=\exp\left(\int_0^t \varphi_s\mathrm{d}W_s^\mathbb{P}-\frac{1}{2}\int_0^t \varphi_s^2\mathrm{d}s\right)=\mathcal{E}\left(\int_0^t \varphi_s\mathrm{d}W_s^\mathbb{P}\right)$$. Assume that $$\mathbb{E}^\mathbb{P}[L_T]=1$$. We define a new probability measure $$\mathbb{Q}$$ on $$\mathcal{F}_T$$ via $$\frac{\mathrm{d}\mathbb{Q}}{\mathrm d\mathbb{P}}=L_T$$. Then, $$\mathrm{d}W_t^\mathbb{P}=\varphi_t\mathrm{d}t+\mathrm{d}W_t^\mathbb{Q}$$ where $$W^\mathbb{Q}$$ is a $$\mathbb{Q}$$-Brownian motion.

Here $$\mathcal{E}$$ is the Doléans-Dade exponential. For the sake of completeness, I repeat the change of numéraire formula. Let $$B_t$$ be the price of our standard numéraire (bank account) with probability measure $$\mathbb Q=\mathbb Q^0$$. Let $$N_t$$ be the price process of a new numéraire. The corresponding martingale measure $$\mathbb{Q}^N$$ is defined via $$\frac{\mathrm d\mathbb{Q}^N}{\mathrm d \mathbb{Q}} = \frac{N_TB_0}{N_0B_T}.$$

Example 1: let $$B_t=e^{rt}$$ and $$N_t=S_t$$. This means we switch from the standard risk-neutral measure $$\mathbb Q=\mathbb Q^0$$ to the stock measure $$\mathbb Q^1$$. Thus, $$\frac{\mathrm{d}\mathbb{Q}^1}{\mathrm{d}\mathbb{Q}^0} = \frac{S_T}{S_0e^{rT}} =e^{-\frac{1}{2}\sigma^2T+\sigma W_T^{\mathbb Q^0}}=\mathcal{E}(\sigma W_T^{\mathbb Q^0})$$. I use a superscript to highlight that $$W_t^{\mathbb Q^0}$$ is a standard Brownian motion with respect to the risk-neutral measure $$\mathbb{Q}^0$$. In the sense of Girsanov's theorem, $$\varphi_t \equiv\sigma$$. Thus, $$\mathrm{d}W_t^{\mathbb Q^0}=\sigma \mathrm{d}t+\mathrm{d}W_t^{\mathbb Q^1}$$. This agrees with what Gordon derived here (he called the new Brownian motion $$\hat{W_t}$$ instead of $$W_t^{\mathbb Q^1}$$).

Example 2: let $$B_t=e^{rt}$$ and the new numéraire is $$N_{t,T}^\alpha$$, the time-$$t$$ price of an asset paying $$S_T^\alpha$$ at time $$T$$. Thus, $$\frac{\mathrm{d}\mathbb{Q}^\alpha}{\mathrm{d}\mathbb{Q}^0} = \frac{S_T^\alpha}{S_0^\alpha e^{rT}} =e^{-\frac{1}{2}\alpha^2\sigma^2 T+\alpha\sigma W_T^{\mathbb Q^0}}=\mathcal{E}(\alpha\sigma W_T^{\mathbb Q^0})$$. In the sense of Girsanov's theorem, $$\varphi_T \equiv\alpha\sigma$$. Thus, $$\mathrm{d}W_t^{\mathbb Q^0}=\alpha\sigma \mathrm{d}t+\mathrm{d}W_t^{\mathbb Q^\alpha}$$.

Okay, starting with the numéraire change, we could use Girsanov's theorem to change a Brownian motion between the two probability measures. How does now the drift of the stock change?

Well, under the risk-neutral measure $$\mathbb Q^0$$, we have $$\mathrm{d}S_t=rS_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t^{\mathbb{Q}^0}$$. And we are now able to express $$\mathrm{d}W_t^{\mathbb{Q}^0}$$ under the new measure $$\mathbb{Q}^1$$. Thus, \begin{align*} \mathrm{d}S_t&=rS_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t^{\mathbb{Q}^0} \\ &=rS_t\mathrm{d}t+\sigma S_t\left( \sigma \mathrm{d}t+\mathrm{d}W_t^{\mathbb Q^1}\right) \\ &=(r+\sigma^2)S_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t^{\mathbb Q^1}. \end{align*}

Similarly, \begin{align*} \mathrm{d}S_t&=rS_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t^{\mathbb{Q}^0} \\ &=rS_t\mathrm{d}t+\sigma S_t\left( \alpha\sigma \mathrm{d}t+\mathrm{d}W_t^{\mathbb Q^\alpha}\right) \\ &=(r+\alpha\sigma^2)S_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t^{\mathbb Q^\alpha}. \end{align*}

Here we go. The drift of the stock price under the standard risk-neutral measure is $$r$$ and under a stock measure, $$\mathbb Q^\alpha$$, this drift changes to $$r+\alpha\sigma^2$$.

• Im struggling a bit with this. Joshi's papers has too many unexplained equations, with the detail glossed over. Jul 11, 2020 at 10:50
• $\mathbb{E}[e^{\sigma W_t}|\mathcal{F}_s]=e^{\frac{1}{2}\sigma^2(t-s)+\sigma W_s}$ where does this come from? Jul 11, 2020 at 10:51
• This method is attractive though Jul 11, 2020 at 10:52
• @Permian It comes from the fact that $X_t=e^{\sigma W_t-\frac{1}{2}\sigma^2t}$ is a martingale, i.e. $\mathbb{E}[X_t|\mathcal{F}_s]=X_s$ Jul 11, 2020 at 11:03
• @PontusHultkrantz great question. The problem is that $e^{-rt}S_t^2$ is not a martingale (Jensen’s inequality). The price (value) process, $N_{t,T}^\alpha$, is however (by construction). Remember that a numéraire is defined to be the price of an asset (which is the unit for the price of all other assets). Nov 10, 2020 at 14:27

It's just Girsanov's theorem. I suppose that under the risk neutral measure Q

$$dS_{t}= r S_{t} dt + \sigma S_{t}dW_{t},$$ $$S_{t} = S_{0}\exp\left((r-\frac{\sigma^{2}}{2})T + \sigma W_{T}\right)$$ By multiplying by $$e^{-rT}$$ I have $$e^{-rT}S_{T}$$ which is a martingale so that I can change my measure under $$Q$$ to some equivalent probabilty $$Q_{1}$$ under which $$W_{t}^{'} = W_{t} - \int_{0}^{t} \sigma_{s}ds = W_{t}-\sigma t$$ is a $$Q_{1}$$ Brownian motion from Girsanov's theorem, now $$S_{T}$$ writes: $$S_0 \exp\left((r-\frac{\sigma^{2}}{2})T + \sigma W_{T}^{'} + \sigma^{2} T\right) = S_0 \exp\left((r+\frac{\sigma ^{2}}{2})T + \sigma W_{T}^{'}\right)$$

So, $$\frac{C_{0}}{S_{0}} = E^{Q^{1}}[\max(S_{T}-K,0)]$$ and you have: $$\mathbb{E}^{Q_{1}}[\max(S_T-K,0)] = \mathbb{E}^{Q_{1}}[S_T\mathbb{I}_{S_T>K}] - \mathbb{E}^{Q_{1}}[K\mathbb{I}_{S_T>K}]$$

• $e^{-rT}S_{T}$ how do you know this is a martingale? Jul 3, 2020 at 17:05
• $W_{T}^{'}$ how do you know this is $N(0,1)$? Jul 3, 2020 at 17:09
• $\mathbb{P}(S_T > K) = N(d_1)$ how has this been derived? Jul 3, 2020 at 17:09
• @Permian $e^{-rt}S_t$ is a $Q$-martingale by definition. $W_T'$ is a Brownian motion and by definition normally distributed (but not standard normal!!). $\mathbb P(S_T>K)=N(d_1)$ is also wrong, we have $Q(S_T>K)=N(d_2)$ and $Q^1(S_T>K)=N(d_1)$, both follow from simple integration. Finally, $W_t'=W_t+\int_0^t\sigma_sds$ comes from Girsanov's theorem...
– Alex
Jul 3, 2020 at 19:07
• You know that $S_{t} = S_{0}\exp\left((r-\frac{\sigma^{2}}{2})T + \sigma W_{T}\right)$ which is the same in law as $S_{t} = S_{0}\exp\left((r-\frac{\sigma^{2}}{2})T + \sigma \sqrt(T) \mathcal{N}(0,1)}\right)$ so you can modify the event $\{ S_{T} > K \}$ as something of the form $\{ N(0,1) > d_[1} \{$ and after that it's just your typical normal CDF. :o , i don't know your maths level but these are some very basic questions when introducing stochastic calculus . You may want to check out some courses online before going into quantitative finance. Jul 4, 2020 at 13:32

Black scholes formula based on $$S_t$$ measure , theory, and formulas you mention are derived in detail in "Steven Shreve: Stochastic Calculus and Finance" draft pdf from 1997 , page 328 "stock price as numeraire".

Question 1 is answered in parts 1 through to 6: the idea is that each part slowly builds the tools required to derive the process equation for $$S_t$$ under the $$S_t$$ Numeraire.

Question 2 & Question 3 are then answered in part 7.

• Part 1: Expectation of a function of a Random variable:

Let $$X(t)$$ be some generic Random Variable with probability density function given by $$f_{X_t}(h)$$, where $$h$$ is a "dummy" variable. Let $$g(X_t)$$ be some (well-behaved) function of $$X_t$$. Then (I am stating the below without proof):

$$\mathbb{E}[g(X_t)]=\int_{-\infty}^{\infty}g(X_t)f_{X_t}(h)dh$$

Let $$\mathbb{P^1}$$ be a Probability measure defined via the Probability Density Function of some random variable $$X_t$$:

$$\mathbb{P^1}(A):=\int_{-\infty}^{a}f_{X_t}(h)dh$$

For all events $$\{A: X_t \leq a\}$$.

Radon-Nikodym derivative is implicitly defined as some Random-Variable (let's call it $$Y_t$$) that satisfies the following:

$$\mathbb{P^2}(A) = \mathbb{E^{P^1}}[Y_t \mathbb{I_{\{ A\}}}]$$.

The above definition becomes more intuitive with a specific example: let $$X_t$$ be a standard Brownian Motion, i.e. $$X_t:=W_t$$, and let $$Y_t:=e^{-0.5\sigma^2t+\sigma W_t}$$. Basically $$Y_t=g(W_t)$$, where $$g()$$ is a well-behaved function: so we can make use of the result in part 1, specifically:

$$\mathbb{E^{P^1}}[Y_t \mathbb{I_{\{ A\}}}] = \mathbb{E^{P^1}}[g(W_t) \mathbb{I_{\{ A\}}}] = \\ = \int_{-\infty}^{\infty}g(X_t)f_{X_t}(h) \mathbb{I_{ \{ W_t \leq a \}}}dh = \\ = \int_{-\infty}^{a}g(X_t)f_{X_t}(h)dh = \\ = \int_{-\infty}^{a}e^{-0.5\sigma^2t+\sigma h}\frac{1}{\sqrt{2\pi}}e^{\frac{-h^2}{2t}}dh = \\ =\int_{h=-\infty}^{h=k}\frac{1}{\sqrt{2\pi}}e^{\frac{-(h^2-\sigma t)}{2t}}dh$$

(To go from the penultimate line to the last line, we just need to complete the square).

The main point: by applying the definition $$\mathbb{P^2}(A) = \mathbb{E^{P^1}}[Y_t \mathbb{I_{\{ A\}}}]$$, we can see how $$Y_t$$ "creates" a new probability measure: under $$\mathbb{P^2}$$, the same event, specifically $$A: W_t \leq a$$ has an altered probability, compared to the same event under $$\mathbb{P^1}$$.

By inspecting the probability $$\mathbb{P^2}(A)=\mathbb{P^2}(W_t \leq a) = \int_{h=-\infty}^{h=k}\frac{1}{\sqrt{2\pi}}e^{\frac{-(h^2-\sigma t)}{2t}}dh$$, we ca see that what was standard Brownian motion under $$\mathbb{P^1}$$ now has a probability distribution of a Brownian motion with a drift: so under $$\mathbb{P^2}$$, $$W_t$$ is no longer a standard Brownian motion, but a Brownian motion with drift $$\sigma t$$.

• Part 3: Cameron-Martin-Girsanov Theorem:

The theorem states that:

If $$W_t$$ is standard Brownian motion under some $$\mathbb{P^1}$$, then there exists some $$\mathbb{P^2}$$ under which $$W_t$$ is a Brownian motion with drift $$\mu t$$. The Radon-Nikodym derivative to get us from $$\mathbb{P^1}$$ to $$\mathbb{P^2}$$ is:

$$\frac{d \mathbb{P^2}}{d \mathbb{P^1}}(t)= e^{-0.5\mu^2t+\mu W_t}$$

If $$\tilde{W_t}:=W_t + \mu t$$ is a Brownian motion with some drift $$\mu t$$ under some $$\mathbb{P^1}$$, then there exists some $$\mathbb{P^2}$$ under which $$\tilde{W_t}$$ is a standard Brownian motion (i.e. no drift). The Radon-Nikodym derivative to get us from $$\mathbb{P^1}$$ to $$\mathbb{P^2}$$ is:

$$\frac{d \mathbb{P^2}}{d \mathbb{P^1}}(t)= e^{+0.5\mu^2t-\mu W_t}$$

We basically "proved" the C-M-G theorem in part 2 above.

• Part 4: Numeraire and Probability Measures

Under the risk-neutral measure, with deterministic money market as Numeraire, the stock price process is: $$S_t=S_0exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]$$. The only source of randomness in this process is $$W_t$$, which is a standard Brownian motion under $$\mathbb{P^Q}$$ associated with the Numeraire $$N_t:=e^{rt}$$.

Since $$W_t$$ is the only source of randomness, this gives us an idea of how a change of probability measure will work for the process $$S_t$$: the change of measure will be driven via a Radon-Nikodym derivative applied to $$W_t$$. If we can somehow get a Radon-Nikodym derivative that resembles the one from the C-M-G Theorem, then we're in for an easy change of measure: we could apply the CMG theorem directly to $$W_t$$ in the process equation for $$S_t$$!!

• Part 5: Change of Numeraire formula

Without proof, if we want to change numeraire from $$N_t$$ to some $$N^{2}_t$$, the Radon-Nikodym derivative we need to use is:

$$\frac{dN^{2}_t}{dN_t}:= \frac{N(t_0)N_2(t)}{N(t)N_2(t_0)}$$

(The proof of the above formula can be found here: Change of Numeraire formula)

• Part 6: Choosing $$S_t$$ as Numeraire

Applying the formula from part 5 above, we get:

$$\frac{dN^{S_t}_t}{dN_t}:= \frac{N(t_0)N^{S_t}(t)}{N(t)N^{S_t}(t_0)} = \\= \frac{1*S_t}{e^{rt}S_0}= \\ = \frac{S_0\exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]}{e^{rt}S_0}= e^{-0.5\sigma^2t+\sigma W_t}$$

The above result is great news, because we can use part 3 directly and apply $$e^{-0.5\sigma^2t+\sigma W_t}$$ as Radon-Nikodym derivative to $$W_t$$: we know this will introduce the drift $$\sigma t$$ under the probability measure defined through $$\frac{dN^{S_t}_t}{dN_t}=e^{-0.5\sigma^2t+\sigma W_t}$$.

Let $$\tilde{W_t}:=W_t-\sigma t$$ be a Brownian motion with a drift equal to $$-\sigma t$$ under $$\mathbb{P^Q}$$. Inserting $$\tilde{W_t}$$ into the process equation for $$S_t$$ under $$\mathbb{P^Q}$$, we get (pure algebraic manipulation, no tricks here):

$$S_t=S_0\exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]= \\ = S_0\exp\left[ (r-0.5 \sigma^2)t+\sigma (\tilde{W}(t)+\sigma t) \right] = \\ = S_0\exp\left[ (r-0.5 \sigma^2)t+\sigma^2 t + \tilde{W}(t) \right] = \\ = S_0\exp\left[ (r+0.5 \sigma^2)t+ \tilde{W}(t) \right]$$

The above equation is not particularly useful in any way. But we can now do the following: we can apply the Cameron-Martin-Girsanov theorem to $$\tilde{W}_t$$, which is very convenient: taking the Radon-Nikodym drivative $$\frac{dN^{S_t}_t}{dN_t}=e^{-0.5\sigma^2t+\sigma W_t}$$ and applying it to $$\tilde{W_t}$$ will add the drift $$\sigma t$$. But $$\tilde{W_t}$$ has negative drift equal to $$-\sigma t$$. Therefore, the Radon-Nikodym derivative $$\frac{dN^{S_t}_t}{dN_t}$$ will "kill" the drift of $$\tilde{W_t}$$. Consequently, under the probability measure associated with $$S_t$$ as Numeraire, $$\tilde{W_t}$$ becomes a standard Brownian motion with no drift.

That's why under the Stock numeraire, the process for the stock price becomes (with $$\tilde{W}_t$$ being a standard Brownian motion):

$$S_t=S_0\exp\left[ (r+0.5 \sigma^2)t+ \tilde{W}(t) \right]$$

It is worth noting that often people use "lazy" notation and don't put the 'tilde' sign on the Brownian motion under the new measure: but I prefer to do it to emphasize that it's a different process to the plain Brownian motion $$W_t$$ under the risk-neutral measure.

Part 7: evaluating $$\mathbb{E^{N_{S}}}[S_t\mathbb{I_{\{S_t > k\}}}]$$:

I think there are multiple ways the expectation can be evaluated. The method that uses the least advanced mathematics but involves the most labor is direct evaluation via an integral:

$$\mathbb{E^{N_{S}}}[S_t\mathbb{I_{\{S_t > k\}}}] = \int_{S_t=k}^{\infty} S_t f_{S_t}(S_t)dS_t = \int_{h=k}^{\infty} h f_{S_t}(h)dh$$

We know that $$S_t$$ is log-normally distributed, so we know the density of $$S_t$$ (https://en.wikipedia.org/wiki/Log-normal_distribution):

$$f_{S_t}(h)= \frac{1} {h \sqrt{t}\sigma \sqrt{2\pi}} e^{-\frac{(ln(h/S_0)-(r-0.5\sigma^2)t)^2}{2\sigma^2t}}$$

Plugging this into the integral results in the cancellation of the $$h$$ in the first denominator:

$$\int_{h=k}^{\infty} \frac{1} {\sqrt{t}\sigma \sqrt{2\pi}} e^{-\frac{(ln(h/S_0)-(r-0.5\sigma^2)t)^2}{2\sigma^2t}}dh$$

I am gonna do the following substitutions: $$y:=ln(h/S_0)$$, so that $$h=S_0e^e$$, $$dh=S_0e^ydy$$, and when $$h=K$$, we get $$y=ln\left( \frac{K}{S_0} \right)$$.

Integrating via substitution then yields:

$$\int_{y=ln(K/S_0)}^{\infty} \frac{1}{\sigma \sqrt{t}} \frac{1}{\sqrt{2 \pi}} e^{\frac{(y-(r-0.5\sigma^2)t)^2}{2\sigma^2t}}S_0 e^y dy$$

I am now gonna simplify the notation further with: $$\tilde{\mu}:=(r-0.5\sigma^2)t$$ and $$\tilde{\sigma}:=\sigma \sqrt{t}$$, so the integral becomes:

$$\int_{y=ln(K/S_0)}^{\infty} \frac{1}{\tilde{\sigma}} \frac{1}{\sqrt{2 \pi}} e^{\frac{(y-\tilde{\mu})^2}{2\tilde{\sigma}^2}}S_0 e^y dy$$

Completing the square between $$e^y$$ and $$e^{\frac{(y-\tilde{\mu})^2}{2\tilde{\sigma}^2}}$$ gives:

$$\exp(y) \exp\left(\frac{(y-\tilde{\mu})^2}{2\tilde{\sigma}^2}\right) = \\ = \exp \left(\frac{(y-(\tilde{\mu}+\tilde{\sigma}))^2}{2\tilde{\sigma}^2}\right)*\exp\left(\tilde{\mu}+0.5\tilde{\sigma}^2\right) = \\ =\exp \left(\frac{(y-(\tilde{\mu}+\tilde{\sigma}))^2}{2\tilde{\sigma}^2}\right)*\exp\left(rt\right)$$

The last line uses the fact that $$\tilde{\mu}+0.5\tilde{\sigma}^2=(rt-0.5\sigma^2t)+0.5\sigma^2t=rt$$.

Plugging back into the integral gives:

$$S_0e^{rt}\int_{y=ln(K/S_0)}^{\infty} \frac{1}{\tilde{\sigma}} \frac{1}{\sqrt{2 \pi}} \exp \left(\frac{(y-(\tilde{\mu}+\tilde{\sigma}))^2}{2\tilde{\sigma}^2}\right)dy$$

Finally, one last substitution: I will take $$z:=\frac{y-(\tilde{\mu}+\tilde{\sigma}^2)}{\sqrt{t}\sigma}$$, which gives $$dy=\sqrt{t}\sigma dz$$. Furthermore, when $$y=ln\left( \frac{K}{S_0} \right)$$, we get:

$$z=\frac{ln\left( \frac{K}{S_0} \right)-(\tilde{\mu}+\tilde{\sigma}^2)}{\sqrt{t}\sigma}=\frac{ln\left( \frac{K}{S_0} \right)-(rt+0.5 \sigma^2t)}{\sqrt{t}\sigma} = \\ = (-1) \frac{ln\left( \frac{S_0}{K} \right)+rt+0.5 \sigma^2t}{\sqrt{t}\sigma} = -d_1$$

So plugging this last substitution for $$y$$ into the integral gives:

$$S_0e^{rt}\int_{y=ln(K/S_0)}^{\infty} \frac{1}{\tilde{\sigma}} \frac{1}{\sqrt{2 \pi}} exp \left(\frac{(y-(\tilde{\mu}+\tilde{\sigma}))^2}{2\tilde{\sigma}^2}\right)dy= \\ = S_0e^{rt}\int_{z=-d_1}^{\infty} \frac{1}{\sqrt{2 \pi}} exp \left(\frac{z^2}{2} \right)dz= \\ =S_0e^{rt}\mathbb{P}(Z>-d_1)=S_0e^{rt}\mathbb{P}(Z \leq d_1) = S_0e^{rt} N(d_1)$$

• This is essentially what I was looking for. Will try to get full understanding by this evening Jul 7, 2020 at 10:14
• I think there are more elegant ways to evaluate $\mathbb{E^{N_{S}}}[S_t\mathbb{I_{\{S_t > k\}}}]$, but the above requires the most basic mathematics. The expense paid is the lengthy algebra. Jul 7, 2020 at 11:30
• I'm not bothered by the maths per se, it's more how to fit the pieces together in a finance setting Jul 7, 2020 at 12:04
• $\frac{d \mathbb{P^2}}{d \mathbb{P^1}}(t)= e^{-0.5\mu^2t+\mu W_t}$ (first equation of part 3). I cant see why this isnt $-\mu W_t$. I have the general form of girsanovs theorem from here, math.csi.cuny.edu/~tobias/Class416/Lecture07.pdf page 4 Jul 7, 2020 at 14:16
• @Permian: I derived it via substituting $\tilde{W}_t:=W_t-\sigma t$ in the paragraph above. An alternative way: take the equation for $S_t$ under $N^{Q}$: $S_t=S_0\exp\left[ (r-0.5 \sigma^2)t+\sigma W(t) \right]$. Apply Radon-Nikdym to $W_t$: that moves you from $N^Q$ to $N^{S_t}$. Under the new measure $N^{S_t}$, $W_t$ is no longer a Standard Brownian motion, but a Brownian with a drift. So it can be rewritten as $Z_t + \sigma t$, where $Z_t$ is Standard Brownian under the new measure $N^{S_t}$. So we get: $S_t=S_0\exp\left[ (r-0.5 \sigma^2)t+\sigma (Z(t) + \sigma t) \right]$. Jul 8, 2020 at 6:59