Most derivations of the Black-Scholes formula end up with the following dynamics of some (hedged) portfolio:

$$ \int_{t=0}^{T} \left(\frac{\partial f}{\partial \tau}(S(t),t)+\frac{1}{2}\cdot\frac{\partial^2 f}{\partial x^2}(S(t),t)\cdot\sigma^2\cdot S(t)^2\right) \,dt $$

From then, they argue that this process is locally riskless.

First Question: Since this integral can be defined as a pathwise Riemann integral, what stops me from definying it using right Riemann sums? In this case, I cannot see how it would be riskless at all.

Then, the authors declare this portfolio should earn the risk-free rate, so they get the following identity:

$$ \int_{t=0}^{T} \left(\frac{\partial f}{\partial \tau}(S(t),t)+\frac{1}{2}\cdot\frac{\partial^2 f}{\partial x^2}(S(t),t)\cdot\sigma^2\cdot S(t)^2\right) \,dt = \int_{t=0}^{T} \left(r\cdot\left(V(S(t),t)-\frac{\partial f}{\partial x}(S(t),t)\cdot S(t)\right)\right) \,dt $$

Fair enough. However, based on this, them they assume the expression within the integrals are equal at al times t.

Second Question: Why is this true?

Thank you.

EDIT: Attempt to answer Second Question: The identity actually is:

$$ \int_{t=0}^{\lambda} \left(\frac{\partial f}{\partial \tau}(S(t),t)+\frac{1}{2}\cdot\frac{\partial^2 f}{\partial x^2}(S(t),t)\cdot\sigma^2\cdot S(t)^2\right) \,dt = \int_{t=0}^{\lambda} \left(r\cdot\left(V(S(t),t)-\frac{\partial f}{\partial x}(S(t),t)\cdot S(t)\right)\right) \,dt $$

For all $0\leq \lambda \leq T$

So equality of functions within the parenthesis holds for all $0\leq t \leq T$.


2 Answers 2


Ito process representation uniqueness states that, if $$ \int_0^t Y_u du + \int_0^t Z_u dW_u = 0 $$ for all $T\geq t\geq 0$, then $$ Y = Z = 0 $$
almost surely (aka up to a set of measure $0$).

In your case, you can take $Z$ to be $0$ and $Y_u:=g(S_u,u)$, where $S$ is your Ito process and $g$ is a deterministic function defined by:

$$ g(x,t) := (\partial_2 V)(x,t) +0.5 (\partial_{11} V)(x,t)\sigma^2 x^2 - r(V(x,t)-(\partial_1 V)(x,t)x). $$

If $g$ is continuous, we then get that $g(x,t)=0$ for all $x$ and $t$ (null everywhere) by using the following fact: for a random variable $X$ with strictly positive density, if $l$ is continuous such that $l(X)=0$ a.s., then $l=0$ (everywhere).


Regarding your first question; nothing prevents you I think, and even something close exists. Without dealing with all the technicalities (I am neither an expert on that), the opposite way of doing stochastic integration, rather than using the Ito-integral, is known as the Stratonovich integral.

Similar to your approach of using the right Riemann sums, the Stratonovich integral is not applicable in finance as it needs future information. One reason for choosing the Ito-integral as it is, comes from the fact that the integral can be calculated with all the information about the process that is know untill a certain point in time. Given the same information set, the Ito-integral can and your approach cannot be computed as it needs at least one value of the process in future time.

Regarding your second question; it comes due to the no-arbitrage condition. I will give you the full derivition in a slightly different notation, which will be a bit easier to follow I think.

Consider the portfolio $P$ with the European style option $C(t, S(t))$ and $\Delta$ shares with share price $S(t)$, $$ P(t, S(t)) = C(t, S(t)) - \Delta S(t) $$ Following the Black-Scholes approach, assume a constant interest rate $r$ and let the risky asset $S(t)$ be following a geometric Brownian motion with drift $r$ and constant volatility $\sigma$, $$ dS(t) = r S(t) dt + \sigma S(t) dW(t), $$ where $W(t)$ is a Brownian motion under the risk-neutral measure $\mathbb{Q}$. Write $C\equiv C(t, S(t))$, $P \equiv P(t, S(t))$ and $S \equiv S(t)$. The change of this portfolio, over an infenitesimal small time interval $t + dt$ will be, $$ dP = dC - \Delta dS $$ From Ito's lemma, it is known that, \begin{align} dC = \frac{\partial C}{\partial t} dt + \frac{\partial C}{\partial S} dS + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} (dS)^2 \end{align} Substitution of this result, in the Equation for the evolution of the portfolio yields, $$ dP = \frac{\partial C}{\partial t} dt + \left(\frac{\partial C}{\partial S} - \Delta \right) dS + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} (dS)^2 $$ Here comes the argumentation with respect to the risk of the portfolio (part of your first question). Note that $(dS)^2 = \sigma^2 S^2 dt$ (I leave the derivation up to you). Hence, the only risk factor, resulting from the Brownian motion, comes from the term $S(t)$. Hence, setting, $$ \Delta = \frac{\partial C}{\partial S}, $$ i.e. selling $\frac{\partial C}{\partial S}$ shares, will remove the risk due to the Brownian motion within the portfolio. This phenomena is called delta hedging. The change of the portfolio is therefore given by $$ dP = \left(\frac{\partial C}{\partial t} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} S(t)^2 \sigma^2 \right) dt $$ Here comes the part regarding to your second question. Suppose that the growth of the portfolio exceeds the growth of the risk-free asset, i.e. $dP > r P dt$. At time $t$, we sell the risk-free and buy the portfolio. Note that we have a investment strategy that does not need any initial capital. As the growth of the portfolio exceeds the growth of the risk-free, we sell the portfolio at time $t + dt$ and buy back the risk-free asset. Furthermore, we keep the profit comming from the strategy. As a consequence, we have created an arbitrage and hence $dP > r P dt$ cannot be true. A similar argument holds for $dP < r P dt$, I leave that to you. As a consequence, we must have that $$ dP = r P dt $$ Combining both results yield, $$ \left(\frac{\partial C}{\partial t} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} S^2 \sigma^2 \right) dt = r P dt = r (C - \Delta S) dt $$ Dividing by $dt$ and using the known expression for $\Delta$ yields the Black-Scholes expression; $$ \frac{\partial C}{\partial t} + \frac{1}{2}\frac{\partial^2 C}{\partial S^2} S^2 \sigma^2 + rS\frac{\partial C}{\partial S} - rC = 0 $$

I hope you find it useful.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.