3 paths to Black-Scholes equation

• Derivation of Black-Scholes equation by three different methods;
• Use of portfolio replication and Martingale representation;
• Comments on the use of risk neutral measure;

Eh..what´s up, Docs? Undoubtedly, the Black-Scholes equation is one of the largest achievements in the field of quantitative finance. In this post, I´ll show three different paths that could be used to derive this equation.
The first path is the usual one, assuming a portfolio evolving as the money market. The second path uses replication of portfolio assuming non-arbitrage conditions. The third explores the martingale property of the risk-neutral measure. Despite the fact that these three paths provide the same result, the third one is rather more abstract and general, allowing us to obtain partial differential equation for pricing derivatives on asset with any stochastic differential model. More comments are presented in the end of this post.

Before going further, let´s take a look in the assumptions of the Black-Scholes from the original paper:

1. The short term interest rate is constant;
2. The underlying asset follows a geometric Brownian motion;
3. There is no dividends;
4. The option is a vanilla European;
5. There is no transaction costs or fees;
6. The market admits the use of short position for the underlying asset;
7. It is possible to borrow and/or invest any fraction of the price of the securities in the money market with the constant interest rate.

Additionally to these initial assumptions, others also play important roles:

1. Trades can be done continuously on time;
2. (Delta) Hedging is done continuously;
3. Bid-ask spreads for the options, interest rate and underlying assets are neglected;
4. The asset used to hedge the position in the derivative is the own underlying asset.

1st: Portfolio with derivative and its underlying asset

Consider we want to hedge a long position in a vanilla European call option with maturity $T$ by using a short position in the underlying asset. The value of this portfolio at time $t<T$:

$x(t)=C(t)-\Delta (t) S(t)$

where $\Delta(t)$ is the size of the short position in the underlying asset. Since there is no arbitrage opportunities with the portfolio (we are imposing this), the evolution of portfolio value follows the interest rate of the money market. Hence:

$dx(t)=rx(t)dt=r(C(t)-\Delta S)dt=dC(t)-\Delta dS$

From the expression above, one might ask why we have considered only the differential term $\Delta  dS$ but not $d\Delta S$, since both are time-dependent. The answer is that the size of the hedge can only be defined after the interval $t$ and $t+dt$, after which the value of the $C(t)$ assumes a different level.

As described in the assumption 2 (in the first list above), the underlying asset follows a geometric Brownian motion (see technical note in the end of the post):

$dS=rSdt+\sigma S dW$

where $W$ denotes the Brownian motion. 

In order to write $dC(t)$ one should use the Itô´s lemma. Shreve´s book provides an amazing description of the Itô´s lemma. Here, we can simply say that when a function depends on a random variable, the differential of this function requires a second order term on the random variable, similar to a second-order Tailor expansion. In the case of $dC(t)$ the random variable is the underlying asset $S(t)$. Therefore:

$dC(t)=\frac{\partial C}{\partial t}dt + \frac{\partial C}{\partial S}dS + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}dSdS$

Replacing the geometric Brownian motion of the asset in the expression for $dC(t)$ we obtain:

$dSdS=(rSdt+\sigma S dW)(rSdt+\sigma S dW)=(rS)^2dtdt+2r\sigma S^2dtdW + (\sigma S)^2 dWdW$

Here we need the following relations:

$dtdt=0$

$dtdW=0$

$dWdW=dt$

The last relation comes from a convergence relation which will not be detailed here. Replacing these relations we obtain:

$dSdS=(\sigma S)^2 dt$

Therefore:

$dC(t)=\frac{\partial C}{\partial t}dt + rS \frac{\partial C}{\partial S}dt+\sigma S \frac{\partial C}{\partial S} dW + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma S)^2 dt$

The equation of the evolution of the portfolio can now be rewritten as:

$dx(t)=r(C(t)-\Delta S)dt=\frac{\partial C}{\partial t}dt + rS \frac{\partial C}{\partial S}dt+\sigma S \frac{\partial C}{\partial S} dW + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma S)^2 dt-\Delta dS$

Thus:

$\frac{\partial C}{\partial t}dt + rS \frac{\partial C}{\partial S}dt+\sigma S \frac{\partial C}{\partial S} dW + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma S)^2 dt- r\Delta Sdt+\sigma \Delta S dW-rC(t)dt+r\Delta Sdt=0$

Equation above can scare a little bit... let´s organize the mess:

$\{\frac{\partial C}{\partial t} + rS \frac{\partial C}{\partial S} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma S)^2 -rC(t)\}dt+\{\sigma S \frac{\partial C}{\partial S} +\sigma \Delta S\} dW=0$

Each one the terms $\{...\}dt$ and $\{...\}dW$ should be zero independently. The term with the differential of the Brownian motion allows us to obtain the size of the hedge position:

$\sigma S \frac{\partial C}{\partial S} +\sigma \Delta S=0 \to \Delta = \frac{\partial C}{\partial S}$

while the other term with $dt$ provides us the Black-Scholes equation:

$\frac{\partial C}{\partial t} + rS \frac{\partial C}{\partial S} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma S)^2 -rC(t)=0$

2nd: Replication of portfolio

In some sense, the first method above and the second method presented hereafter are quite similar. In the first path described above, the key point was the hedge of the long position using a short position with the underlying asset. In this second method, we will replicate the value of the option $C(t)$ using a portfolio $x(t)$ with a position in the underlying asset and a position in the money market. This method uses the following idea: suppose one assumes a short position with a vanilla European option, with the money obtained from this position one buys a position in the underlying asset and the rest is invested in the money market.
These portfolios presents the same discounted value evolving in time:
$d(e^{-rt}C(t))=d(e^{-rt}x(t))$
where:
$d(e^{-rt}C(t))=-re^{-rt}C(t)dt+e^{-rt}dC(t)$
and:
$d(e^{-rt}x(t))=-re^{-rt}x(t)dt+e^{-rt}dx(t)=-re^{-rt}x(t)dt+e^{-rt}\{ \Delta dS + r(x(t)-\Delta S)dt\}$
Using the hypothesis of the geometric Brownian motion for the underlying asset (see the technical note in the end of the post):
$d(e^{-rt}x(t))= e^{-rt} \sigma \Delta  SdW$
Equating the relations for $C(t)$ and $x(t)$ we obtain:
$-rC(t)dt+dC(t)=\sigma \Delta  SdW$
Replacing the value of $dC(t)$ obtained with the Itô´s lemma already presented in the 1st path above, we obtain:
$-rC(t)dt+\frac{\partial C}{\partial t}dt + rS \frac{\partial C}{\partial S}dt+\sigma S \frac{\partial C}{\partial S} dW + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma S)^2 dt=\sigma \Delta  SdW$
Once again we´ve obtained a quite ugly equation... let´s clean up this mess:
$\{-rC(t)+\frac{\partial C}{\partial t} + rS \frac{\partial C}{\partial S} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma S)^2\} dt=\{ \sigma S \frac{\partial C}{\partial S} +\sigma \Delta  S \} dW$
In order to be satisfied, each side of the relation above must be zero. Taking first the right-hand side, we have:

$\sigma \Delta S -\sigma S \frac{\partial C}{\partial S}=0 \to \Delta=\frac{\partial C}{\partial S}$

and from the left hand side we obtain the Black-Scholes equation:

$\frac{\partial C}{\partial t} + rS \frac{\partial C}{\partial S} + \frac{1}{2}(\sigma S)^2 \frac{\partial^2 C}{\partial S^2}-rC(t)=0$

3rd: Martingale property of the risk-neutral measure

In the risk neutral measure, the discounted value of the option with European claim is a martingale, which allows us to write:
$e^{-rt}C(t)=E[e^{-rT}C(T)|F(t)]$
where $C(t)$ is the value of a vanilla call European option with maturity T, $E[...]$ is the expected value under the risk neutral measure, and $F(t)$ is a filtration at time $t$.
Taking the differential of the discounted price of the option:
$d(e^{-rt}C(t)) = -re^{-rt}C(t)dt +e^{-rt}dC(t)$
Replacing the value of $dC(t)$ obtained with the Itô´s lemma and with the assumption of geometric Brownian motion for the underlying asset, one has:
$dC(t)=\frac{\partial C}{\partial t}dt + rS \frac{\partial C}{\partial S}dt+\sigma S \frac{\partial C}{\partial S} dW + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma S)^2 dt$
and then:
$d(e^{-rt}C(t)) = -re^{-rt}C(t)dt +e^{-rt} \{ \frac{\partial C}{\partial t}dt + rS \frac{\partial C}{\partial S}dt+\sigma S \frac{\partial C}{\partial S} dW + \frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma S)^2 dt \}$
reorganizing the terms:
$d(e^{-rt}C(t)) = e^{-rt}\{-rC(t) +\frac{\partial C}{\partial t}+rS \frac{\partial C}{\partial S}+\frac{1}{2}\frac{\partial^2 C}{\partial S^2}(\sigma S)^2 \}dt    +e^{-rt} \sigma S \frac{\partial C}{\partial S} dW$
Using the martingale representation, we know that the discounted value of the option price should not present a term that goes with $dt$. As a consequence, the Black-Scholes equation naturally comes from the expression above and we have:
$\frac{\partial C}{\partial t}+rS \frac{\partial C}{\partial S}+\frac{1}{2}(\sigma S)^2 \frac{\partial^2 C}{\partial S^2}=rC(t)$

It is worth mentioning how this third method is powerful in the sense that one can obtain a partial differential equation for any derivative with European claims and any stochastic model for the underlying. In fact, the idea behind this Feynman-Kac theorem is to connect a stochastic differential model that describes an underlying asset to a partial differential equation. In the following, I´ll present an (informal) description of this theorem, exemplifying how far we can go. Supposing a generalized stochastic differential model for determined asset (or index):
$dX(t)= \beta (X,t)dt + \Gamma (X,t)dW$
Let $h(X(T))$ be a payoff of a derivative at maturity T. The discounted price function satisfies a Martingale process:
$f(X,t) = e^{-r(T-t)}E[h(X(T))|F(t)]$
Then, the function $f(X,t)$ satisfies the following partial differential equation:
$\frac{\partial f(X,t)}{\partial t}+ \beta \frac{\partial f(X,t)}{\partial X}+\frac{1}{2} {\Gamma}^2 \frac{\partial^2 f(X,t)}{\partial X^2}=rf(X,t)$

Technical note - Geometric Brownian motion in the risk neutral measure:

The geometric Brownian motion can be written as:

$dS=\mu Sdt + \sigma SdZ$

where $\mu$ is the average return of the asset and $dZ$ is a Brownian motion under the real world measure. In order to determine prices without arbitrage opportunities, we should rewrite this stochastic differential equation in the risk-neutral measure. This can be done by redefining the Brownian motion in a measure such that the discounted value of the asset is a Martingale.
By using the Martingale representation:
$M(t)=M(0)+ \int_{0}^{t} \Gamma (u) dW(u)$
we can notice that the differential form of a Martingale does not present a diffusive term (the one that goes with $dt$). Hence:
$d(e^{-rt}S)=-re^{-rt}Sdt + e^{-rt}dS$
$d(e^{-rt}S)=-re^{-rt}Sdt + e^{-rt} \{\mu Sdt + \sigma SdZ \}$
We will redefine the Brownian motion by using the Girsanov´s theorem, using $dZ=dW-\theta dt$.
Therefore:
$d(e^{-rt}S)=-re^{-rt}Sdt + e^{-rt} \mu Sdt +e^{-rt} \sigma S(dW-\theta dt) $
$d(e^{-rt}S)=e^{-rt} \{-r S + \mu S - \sigma \theta S\}dt  +e^{-rt} \sigma SdW $
Now we must find the value of $\theta$ that makes $\{...\}dt =0$, which in this case is:
$\theta = \frac{\mu - r}{\sigma}$
Under this definition of $\theta$, $dW$ is the a Brownian motion under the risk neutral measure. Replacing the relation of the Girsanov´s theorem in the Geometric Brownian motion, one obtains:
$dS=\mu Sdt + \sigma SdZ \to dS=\mu Sdt + \sigma S(dW-\theta dt)$
Then, finally:
$dS=rSdt + \sigma SdW$

which is the geometric Brownian motion under the risk neutral measure.

That´s all folks! May the force be with you.

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Diogo de Moura Pedroso

LinkedIn: www.linkedin.com/in/diogomourapedroso