Why N(d1) Is the In-the-Money Probability Under the Stock Measure

# Understanding $N(d_1)$ and how to use the stock itself as the numeraire?


# Understanding $N(d_1)$ and how to use the stock itself as the numeraire?












Assume the stock price follows a geometric Brownian motion Then in Black-Scholes pricing model, $N(d_2)$ is the risk-neutral probability that the option expires in-the-money. However, it is said that $N(d_1)$ is also the probability that the option expires in-the-money under the measure that uses the stock itself as the numeraire.

I understand that risk-neutral measure uses discounted stock price $\frac{S}{B}$ as the numeraire, but how do you use the stock itself as the numeraire?

## Answer by Gordon (score 8)

https://quant.stackexchange.com/a/21486

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

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

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

## Answer by user9403 (score 5)

https://quant.stackexchange.com/a/19053

How to use the stock as Numeraire:

$$\mathbb{\tilde{E}}[e^{-rT}(S_T-K)^+]=\mathbb{\tilde{E}}\left[e^{-rT}S_T\left(1-\frac{K}{S_T}\right)^+\right]$$ $$=S_0\mathbb{\tilde{E}}\left[\frac{e^{-rT}S_T}{S_0} \left(1-\frac{K}{S_T}\right)^+\right]$$ $$=S_0\mathbb{\hat{E}}\left[\left(1-\frac{K}{S_T}\right)^+\right]$$ Where under $\mathbb{\hat{P}}$ the stock follows $dS=(r+\sigma^2)Sdt+\sigma S d\hat{W}_t$. The rest is straightforward computation.

How $\mathcal{N}(d_1)$ is the probability that the option expires in the money under the stock measure:

The delta of a call option is $\frac{\partial C}{\partial S}$. Writing the value of a call options as $\mathbb{\tilde{E}}[e^{-rT}(S_T-K)^+]=\mathbb{\tilde{E}}[e^{-rT}(S_0e^{(r-\frac{\sigma^2}{2})T+\sigma W_T}-K)^+]$ and taking the derivative with respect to $S_0$, $$\frac{\partial C}{\partial S}=\mathbb{\tilde{E}}\left[e^{-rT}(e^{(r-\frac{\sigma^2}{2})T+\sigma W_T})\mathbb{I}_{S_T>K}\right]$$ Where $\mathbb{I}$ is the indicator function. $$=\mathbb{\tilde{E}}\left[e^{\sigma W_T-\frac{T\sigma^2}{2}}\mathbb{I}_{S_T>K}\right]$$ $$=\mathbb{\hat{E}}\left[\mathbb{I}_{S_T>K}\right]=\mathbb{\hat{P}}(S_T>K)$$

## Answer by user16891 (score 0)

https://quant.stackexchange.com/a/19039

If $\{N_t\}_t$ be a numerair,and $S_t$ any asset price process, then there exists a measure $Q^S$ which is equivalent to the risk-neutral measure $Q$ such that process $$\frac{S_t}{N_t}$$ is a martingale under $Q^S$ and Derivative pricing under the new measure $Q^S$ is similar as risk-neutral pricing. In particular, the time-t price of a derivative that pays $X$ at maturity $T$ is given by $$V(S_t,t,T)=N_t\mathbb{E^{Q^S}}[N_T^{-1}X|\mathcal{F}_t]$$. Now, if $S_t$ be a numerair then $Q=Q^S$ and change measure is Meaningless.

## Answer by SmallChess (score 0)

https://quant.stackexchange.com/a/19046

Risk-neutral is just one of the many possible measures. It's the most common because we can discount an asset by the risk-free rate under this measure. Of course, we can use any other measure, such as pricing under the stock measure. The mathematics will be very similar, you'll still try to form a martingale under the measure.

In @Farahvartish's answer, you can basically replace N with anything.