Pricing a Squared-Stock Payoff with Numéraire Changes in Black–Scholes

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


# 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

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

- 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.

- 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.

- 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.

## Answer by Kevin (score 16, accepted)

https://quant.stackexchange.com/a/55477

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 $r+\alpha\sigma^2$.

## Answer by Kupoc (score 4)

https://quant.stackexchange.com/a/55405

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: $$\mathbb{E}^{Q_{1}}[\max(S_T-K,0)] = \mathbb{E}^{Q_{1}}[S_T\mathbb{I}_{S_T>K}] - \mathbb{E}^{Q_{1}}[K\mathbb{I}_{S_T>K}]$$

## Answer by Jan Stuller (score 4)

https://quant.stackexchange.com/a/55479

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):

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

- Part 2: Radon-Nikodym Derivative:

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:

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

## Answer by alexprice (score 3)

https://quant.stackexchange.com/a/55476

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".