Framework

Two assets:

• Risky asset (stock) with a GBM dynamics $\frac{dS_t}{S_t} = \mu dt + \sigma dW_t$

• Risk-free asset with dynamics $\frac{d\beta_t}{\beta_t} = r dt$

Warning: in previous sections, $r$ represents the risk asset returns. Here, it’s the risk-free asset.

Change of measure

A change of measure is needed to be under the risk-neutral probability and thus ensure no arbitrage opportunity. Under this risk-neutral measure, the expected return of all assets is the risk-free return.

In other words, risk-neutral pricing removes the need to estimate the asset’s real-world expected return $\mu$. The physical drift $\mu$ is replaced by the risk-free rate $r$ for pricing purposes.

B-S equation (largely inspired from Hull)

Let’s now consider the price of a derivative $f$. $f$ is a function of $S_t$ and $t$ and we can apply Itô’s lemma:

Note (1): in the original formula, one would need to use $\mu S_t$ as drift and $\sigma S_t$ as volatility.

Note (2): the Wiener process is the same as the one driving the risky asset.

In discrete time:

We can now construct a portfolio containing a stock and a derivative so that the Wiener process is eliminated.

-1: derivative (short position)

+$\frac{\partial f}{\partial S_t}$: shares

Let $V_t$ be the value of the portfolio.

Using $(*)$ and $(**)$:

The equation doesn’t involve any Wiener process, hence the portfolio is riskless (only!) during time $\Delta t$. This implies that the portfolio must earn the same rate as the short-term risk-free asset:

It follows that

This is the Black-Scholes-Merton differential equation (PDE).

See these notes for the step-by-step solve of the Black-Scholes PDE.

Greeks

In the case of a portfolio containing one derivative only, we define:

Hence, the BSM differential equation becomes

When the portfolio is delta-hedged ($\Delta = 0$):

These greeks are also used to decompose the P&L of an option position, see P&L attribution.

Call price

Under Black-Scholes, the call price is:

with $d_1 = \frac{\ln\frac{S_t}{K}+\left(r+\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}} \text{ and } d_2 = d_1 - \sigma \sqrt{T-t}$ \

We note that:

• T $\uparrow$ => price $\uparrow$ (the more the maturity increases, the more chances we have to exercise)

• K $\uparrow$ => price $\downarrow$ (the more the strike increases, the less chances we have to exercise)

• $\sigma \uparrow$ => price $\uparrow$ (the more the volatility increases, the more chances we have to exercise)

• $S_t \uparrow$ => price $\uparrow$ (the more the underlying increases, the more chances we have to exercise)

Remark: $\Phi$ is a monotonic increasing function.

In theory, the B-S formula has 7 parameters:

• $S_t$: spot price

• $K$: strike

• $r$: risk-free rate

• $T$: maturity

• $\sigma$: volatility

• $q$: dividend yield

• $b$: repo rate (cost of borrowing the stock to short it when hedging)

Dividends and repo both reduce the expected growth of the stock under $\mathbb{Q}$ (the forward becomes $F = S_te^{(r-q-b)(T-t)}$), so they act like a discount on the stock leg:

with $d_1 = \frac{\ln\frac{S_t}{K}+\left(r-q-b+\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}} \text{ and } d_2 = d_1 - \sigma \sqrt{T-t}$. The simple formula above is the special case $q=b=0$.

Black formula (forward version)

The Black formula (Black, 1976) is the same model written in terms of the forward $F_t = F(t,T)$ instead of the spot $S_t$:

The call price becomes:

with $d_1 = \frac{\ln\frac{F_t}{K}+\frac{\sigma^2}{2}\left(T-t\right)}{\sigma\sqrt{T-t}} \text{ and } d_2 = d_1 - \sigma \sqrt{T-t}$

We note that the drift terms ($r$, $q$, $b$) have disappeared from $d_1$ and $d_2$: they are all embedded in the forward. Plugging $F_t = S_te^{(r-q-b)(T-t)}$ gives back exactly the Black-Scholes formula with dividends and repo above.

In practice, the Black formula is preferred when the forward is directly observable or easier to estimate than dividends and repo separately. For example, we can deduce the implied forward from market option prices using the call-put parity.

Interpretation of $d_1$ and $d_2$

$\ln\frac{S_t}{K}$ is called the log-moneyness. $\frac{S_t}{K}$ measures the distance to the strike, and we take the log on top of it to:

• get symmetry: $\ln\frac{100}{50} = -\ln\frac{50}{100}$, unlike percentages (+100% vs -50%)

• be consistent with the continuous returns assumption ($\ln\frac{S_{t+1}}{S_t}$)

Under Black-Scholes (risk-neutral measure $\mathbb{Q}$), we know that:

Since $\ln\frac{S_T}{K} = \ln\frac{S_t}{K} + \ln\frac{S_T}{S_t}$, the log-moneyness at maturity is:

Since $W_T-W_t \sim \mathcal{N}\left(0, T-t\right)$, the log-moneyness at maturity is normally distributed:

The expected log-moneyness at maturity is simply the mean of this normal distribution:

Normalizing it by its standard deviation $\sigma\sqrt{T-t}$ gives exactly $d_2$:

In other words, $d_2$ is the expected log-moneyness at maturity normalized by the number of standard deviations:

$\Phi(d_2)$ is thus the probability that the option ends ITM at maturity. Note: this is a risk-neutral probability (the drift is $r$), not the real-world one (the drift would be $\mu$).

It’s almost the same concept for $\Phi(d_1)$:

The expected log-moneyness is now increased by $\sigma^2(T-t)$. This can be explained looking at the initial formula of the price as an expected payoff, which can be split in 2 expectations:

Reminder: the expectation of an indicator is a probability: $\mathbb{E}\left[\mathbf{1}_A\right] = \mathbb{P}(A)$.

We notice that the second integral is weighted by $x$ (i.e. $S_T$ the stock value). These weights don’t average to 1, so we normalize them by $\mathbb{E}^{\mathbb{Q}}\left[S_T\right] = S_te^{r(T-t)}$ to get a probability.

Formally, multiplying the (lognormal) density $f$ by $x$ and dividing by $\mathbb{E}^{\mathbb{Q}}\left[S_T\right]$ gives another lognormal density, whose log has its mean shifted by $\sigma^2(T-t)$:

That’s why the expected log-moneyness is increased in $d_1$: $\Phi(d_1)$ is the probability to end ITM when scenarios are weighted by the stock value (i.e. under the measure where the stock is the numeraire).

Summary: both terms are probabilities to end ITM, but with different weights on the scenarios:

In other words, the price of a call is what we receive (the stock, with a probability to end ITM that favours the scenarios where the stock finishes high) minus what we pay (the discounted strike, times the probability to end ITM).

Note: the stock leg is also discounted, but this cancels out with its expected growth: $e^{-r(T-t)}\mathbb{E}^{\mathbb{Q}}[S_T] = S_t$ (assuming no dividends and no repo).

Proof of the Black-Scholes formula

We define:

The payoff of a European option is: $f(x) = \max (S_T-K,0)$

Using the fact that $S_te^z-K \leq 0$ => $z \leq \ln \frac{K}{S_t}$ and $\int^c_a f(x)dx = \int^b_a f(x)dx + \int^c_b f(x)dx$, we have:

Reminders:

• $X \sim \mathcal{N}\left( \mu, \sigma^2 \right)$ => $f(x) = \frac{1}{\sqrt{2\pi\sigma}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$

• $X$ standardized => $P(X \leq x) = CDF = \int_{-\infty}^{\color{red}x} f(x)dx = \Phi\left(\frac{x-\mu}{\sigma}\right)$

• $\int_a^bf(x)dx = - \int_b^af(x)dx$

Let $A = Ke^{-r(T-t)}\Phi\left(-\frac{\ln\frac{K}{s}- \left(r-\frac{\sigma^2}{2}\right)\left(T-t\right)}{\sigma\sqrt{T-t}}\right)$

Implementation

import numpy as np
from scipy.stats import norm

S0 = 100
r = .05
K = 110
sigma = 0.2
T = 1/2    # 6 months maturity
t = 0

d1 = (np.log(S0/K)+(r+.5*(sigma**2))*(T-t))/(sigma*np.sqrt(T-t))
d2 = d1 - sigma*np.sqrt(T-t)
price = S0*norm.cdf(d1) - K*np.exp(-r*T)*norm.cdf(d2)
print(round(price,2))

The above prints $2.91$. Intuitively, it means that “on average” we earn $2.91$.

With the following parameters:

$S_0 = 100$

$r = 5\%$

$K = 105$

$\sigma = 0\%$

$T = 2$

It would mean that in 2 years, the stock is certain to end at 110.25 so we would earn 5.25 exercising the option (a bit less taking the cost of opportunity into account). Now in order for NFL to hold, price $\approx$ 5 (B-S gives 4.99).

ATM price approximation

When the option is ATM, $K=S_t$, thus $d1 = \frac{1}{2}\sigma \sqrt{T}$ and $d2 = -\frac{1}{2}\sigma \sqrt{T}$. Hence the price becomes

Using Taylor series for the first order, when $x \to 0$, $\Phi(x) \approx \Phi(0) + \Phi’(0)x$.

Reminder: Taylor series when $x \to a$, $f(x) = f(a) + f’(a)(x-a) + \frac{1}{2!}f’‘(a)(x-a)^2 + \frac{1}{3!}f^{(3)}(a)(x-a)^3 + …$

Thus $\Phi(x) + \Phi(-x) = 2\Phi’(0)x = \frac{1}{\sqrt{2 \pi }}x$

Finally, when ATM,

Note: for quick calculations, $\sqrt{2 \pi} \approx \frac{5}{2}$. A good benchmark is a call option with maturity 1Y and volatility 20% -> price=8% (i.e. purchase amount = 8% x underlying price).

Note (2): this equation can be especially useful to compute the implied volatility. One could then compare the historical and implied volatility to know whether the volatility is overvalued or not. If the volatility is overvalued (implied volatility » historical volatility), premiums could be too high.

Limits

The Black-Scholes model has many unrealistic assumptions:

• Returns are assumed normally distributed, whereas in reality the kurtosis is high (= fat tails) and the skew is negative (larger losses because investors tend to react stronger when the market is down).

• The volatility is assumed constant

As a result, the BS model tends to misprice options that are deep ITM or deep OTM.

In practice, the Black-Scholes model is used to find this implied volatility rather than to compute option prices.