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I'm trying to understand the standard hedging argument to derive the Black--Scholes PDE. There's one aspect of the derivation which I can't get passed and I'd be very grateful for some clarification here.

We make the standard assumptions: underlying follows a geometric BM, i.e. $\text{d} S_t = \mu S_t \text{d}t + \sigma S_t \text{d}W_t$ and we let $V(S_t,t)$ be an option written on this security so that by Ito's Lemma, we have:

$\text{d} V_t = \left(\frac{\partial V}{\partial t} + \mu S_t \frac{\partial V}{\partial S} + \frac{1}{2} \sigma^2 S^2_t \frac{\partial^2 V}{\partial S^2}\right) \text{d}t + \left(\sigma S_t \frac{\partial V}{\partial S}\right) \text{d}W_t$

All good so far. We then set-up a portfolio consisting of a single option and some amount $\Delta(S_t,t)$ of the underlying whose value is given by $\Pi_t = V_t + \Delta(S_t,t) S_t$. It is then assumed that this portfolio be self-financing, i.e. that there are no additional inflows or outflows of cash, formalised as $\text{d} \Pi_t = \text{d} V_t + \Delta(S_t,t) \text{d} S_t$. We then choose $\Delta(S_t,t) = - \frac{\partial V}{\partial S}$ and then by substituting this into the above formula and using self-financing, we quickly obtain a riskless portfolio and the Black--Scholes PDE follows from there.

The point that confuses me is that it is not clear to me that we can definitely construct a portfolio satisfying $\Pi_t = V_t - \frac{\partial V}{\partial S} S_t$ which also must satisfy $\text{d} \Pi_t = \text{d} V_t - \frac{\partial V}{\partial S} \text{d} S_t$. Every text I can find simply assumes that you can and proceeds as above. However, the existence of such a portfolio implies that $\frac{\partial V}{\partial S} S_t = \int_0^t \frac{\partial V}{\partial S}(S_u,u) \text{d}S_u$ and I don't see that this is necessarily true, that is to say, I don't see how such a portfolio can exist? What am I missing?

The interesting thing is that I have followed a different derivation of Black--Scholes where the portfolio instead consists of some amount of the underlying and some amount of a risk-free instrument $B_t$, i.e. $\Pi_t = \alpha(S_t,t) S_t + \beta(S_t,t) B_t$. Here we choose $\alpha(S_t,t) = \frac{\partial V}{\partial S}$ as before to remove the risk but this time the freedom in $\beta(S_t,t)$ seems to mean that we can ensure the portfolio is self-financing and therefore I am happy with this version of the argument.

I would really like to understand both arguments and where it is that I am going wrong in the first so any help is gratefully appreciated!

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in addition to my answer: $\alpha$ and $\beta$ is exactly this trading strategy! specifying $\alpha$ and initial costs, determines $\beta$ uniquely to set up an self-financing strategy. –  user8 Jul 24 '13 at 12:36

3 Answers 3

According to wikipedia one chooses $\pi_t=-V_t+\frac{\partial V}{\partial S}S_t$. This means, you are shorten $V$ and long $\frac{\partial V}{\partial S}$ shares of $S$. The general theory of self-financing strategies assumes that your market consists of a $\mathbb{R}^{d+1}$ process $S$, with $d$ risky assets and one risk free (bank account). A trading strategy is then a function $\phi=(\vartheta,\eta)$, where $\vartheta$ is $\mathbb{R}^d$-valued and $\eta$ is real-valued. Of course you have to put some additional assumptions on both of them. The general value process $V$ is defined to be $\vartheta S+\eta$ (note we discount everything, so the bank account is constant $1$). Self-financing means that the costs are constant over time, where the costs $C_t(\phi)=V_t(\phi)-\int\vartheta_udS_u$. So to be self-financing means $C_t(\phi)=C_0(\phi)=V_0(\phi)$ for all $t$. The following result is easily verified:

There is a bijection between self-financing strategies and pairs $(V_0,\vartheta)$

So it is enough to set up the strategy for your risky asset and your initial cash amount to obtain a self-financing strategy. In your example you exactly specify the risky part of your asset, leading to a self-financing strategy.

However, in my opinion, there is more elegant way to derive the PDE. Assume very generally that your payoff is of the form $H=h(S_T)$, where $h$ is some measurable function. From risk neutral pricing you now that the value process of $H$ is given by


under an equivalent local martingale measure. Therefore $V^H$ is a $Q$-martingale. Using independence of Brownian Motion you will easily find that $$V_t^H=E_Q[h(S_T)|\mathcal{F}_t]=v(t,S_t)$$ One can prove that the function $v$ is sufficiently smooth. Applying Itô:

$$dV^H=dv=v_x(t,x)\sigma S_tdW^Q_t+(v_t(t,x)+\frac{1}{2}v_{xx}(t,x)\sigma^2S^2_t)dt$$

For a $Q$ BM $W^Q$ and using the dynamics under $Q$ of $S$. Since $V^H$ is a $Q$ martingale, the finite variation part has to vanish, so you get:

$$v_t(t,x)+\frac{1}{2}v_{xx}(t,x)\sigma^2x^2=0$$ with $v(T,\cdot)=h(\cdot)$. Using undiscoutend terms, $v(t,x)=\exp{(-rt)}\tilde{v}(t,x\exp{(rt)})$, plugging this into the PDE above, gives exactly the BS-PDE:

$$0=\frac{\partial\tilde{v}}{\partial t}+r\tilde{x}\frac{\partial\tilde{v}}{\partial \tilde{x}}+\frac{1}{2}\sigma^2\tilde{x}^2\frac{\partial^2\tilde{v}}{\partial \tilde{x}^2}-r\tilde{v}$$ with $\tilde{v}(T,\cdot)=\tilde{h}(\cdot)$.

Of course this uses a lot us (basic) stochastic calculus, but if you understand once, what is going on here, you get a better feeling of the theory.

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@MattWolf Thanks for adding another solution. I know that this approach is from the mathematical theory. However, I think it is just an advantage if one knows why exactly the whole theory works as it does. If one has all (mathematical) tools together, one really gets a better understanding of all these things (pricing, hedging etc) –  user8 Jul 24 '13 at 12:50
fair point, thanks for clarifying. Your answer probably is closer to what the asker is looking for, though I explained things more from a risk neutral pricing rather than PDE approach because I believe it is easier this way to appreciate the importance and functioning of the hedge portfolio. –  Matt Wolf Jul 24 '13 at 12:53
@MattWolf I totally agree with the simplicity and more intuitive manner of your solution. To be honest, as a mathematician I'm biased and clearly like more the mathematical derivation :) –  user8 Jul 24 '13 at 12:56
good to share our knowledge on such board because I appreciate to read about the more rigorous approaches as I am not a mathematician by heart. Cheers ;-) –  Matt Wolf Jul 24 '13 at 13:02
Thank you for your answers. In the second approach with the $\alpha$ and $\beta$, is it "obvious" that once we've fixed $\alpha(S_t,t) = \frac{\partial V}{\partial S}$, we can find a $\beta(S_t,t)$ such that the portfolio is self-financing? I mean one can use Ito's Lemma to find $\text{d} P_t$ from the portfolio definition and then choose $\beta$ to clear all of the extra terms. Doing this, I obtain that we must have $\frac{\partial \beta}{\partial S} B_t = -\frac{\partial \alpha}{\partial S} S_t$ and $\frac{\partial \beta}{\partial t} B_t = -\frac{\partial \alpha}{\partial t} S_t$? –  Jon Jul 24 '13 at 15:58

I believe the setup of the first part you presented is inaccurate.

The whole point of the hedge argument is that you can setup a self-financing portfolio that only holds a certain amount of stock and invests/borrows at a specific financing rate. It can be shown that such portfolio almost surely has the same payoff as the option at maturity. The option payoff is F(T) measurable, meaning that the payoff is path-dependent. So, only through investing/borrowing in an interest rate bearing instrument that finances a specific amount of risky asset can such portfolio be constructed and equals the option price at any t. There is more to it (risk neutral pricing and change of measure as part of the complete hedge argument to derive the share of risky asset to be traded to hedge the option payoff so I just presented the gist of it).

Your first portfolio equation assumes you trade the option which you actually try to hedge which kind of misses the point of the hedge argument. Where the first part goes wrong is that, as you stated, "there is no cash in or outflow", which is incorrect: The cash in and out flows that follow from changes in the amount of holding in the risky asset are invested/borrowed in an interest bearing instrument. So, it is true that the whole portfolio is self-contained, however, within the hedge portfolio there is clearly cash flows that have to be borrowed/invested.

I highly recommend to follow pages 217-220 of Stochastic Calculus for Finance II by Steven Shreve to understand the hedge argument and how it sets the stage to derive the Black Scholes pricing model (even though it takes the approach of risk neutral pricing rather than the PDE approach I find it much better to understand the risk neutral pricing approach in order to fully appreciate how the hedge portfolio works and why it is so important.)

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I agree that usual presentation of the replicating portfolio argument is deficient. I think this may actually go back to Merton's 1973 paper on Black-Scholes, because his presentation is unclear in the same way.

In fact the choice to buy $\partial V\over \partial S$ shares of the stock is forced by the use of a self-financing portfolio. I think the argument should go like the following.

Define a portfolio $$\Pi = V- {\partial V\over \partial S} S + \psi P$$ where $P$ is price of a \$1 bond with the same maturity date as the option, and $\psi$ is some function of $t$ and $S$ such that $\Pi=0$ initially. We assume a Mertonian demon continuously shuffles funds back and forth between the stock and bond investments so as to always have exactly $\partial V\over \partial S$ shares of the stock.

Since this portfolio is self-financing by construction, the change in its value is due only to capital gains, i.e. $$ {\mathrm d}\Pi = dV - {\partial V\over \partial S} {\mathrm d}S + \psi {\mathrm d}P. $$ The only stochastic terms are in ${\mathrm d}V$ and ${\partial V\over \partial S} {\mathrm d}S $ and they cancel exactly, so ${\mathrm d}\Pi$ is deterministic. Then, appealing to lack of arbitrage opportunities, $\Pi$ must grow at the safe prevailing interest rate.

But therefore, since initially $\Pi=0$, it must in fact always be zero. Hence $$ V={\partial V\over \partial S} S - \psi P $$ is a self-financing replicating portfolio of $V$.

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