# 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 '20 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 '20 at 17:14
• What is your understanding of a martingale? Jul 6 '20 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 '20 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 '20 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 '20 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 '20 at 10:51
• This method is attractive though Jul 11 '20 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 '20 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 '20 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 '20 at 17:05
• $W_{T}^{'}$ how do you know this is $N(0,1)$? Jul 3 '20 at 17:09
• $\mathbb{P}(S_T > K) = N(d_1)$ how has this been derived? Jul 3 '20 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 '20 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 '20 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 '20 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 '20 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 '20 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 '20 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 '20 at 6:59