In the derivation of the Black Scholes equation, the value of the portfolio at time $t$ is given by
$$P_t = -D_t + \frac{{\partial D_t}}{{\partial S_t}}S_t $$ where $P_t$ is the value of the portfolio at time $t$, $D_t$ the price of the derivative and $S_t$ the price of the stock.
I guess $D_t$ means the price of the premium, i.e., the right to by a call/put option.

1) Why is that the value of the portfolio is equal to the value of the premium?
2) Does this equation assume that the underlying derivative is a call option?


3 Answers 3


1) Instead of asking why the portfolio is equal to the premium, ask why create it at all? I say that because it actually isn't equal to the premium, since that would just be D, and also because the actual value isn't as important as what the portfolio represents.

We form this particular portfolio because the laws of no-arbitrage guarantee it has a certain rate of return (within a set of necessary assumptions). That gives us a constant point which we can exploit to solve the equation; without this observation, the equations aren't grounded.

Specifically, the portfolio consists of a short derivative paired with an amount of stock such that the change in the derivative price due to the stock moving is exactly offset by these shares. That amount, represented here by the $\frac{{\partial D_t}}{{\partial S_t}}$ term, is called "delta". Because of this pairing, the resulting portfolio is not exposed to risk (i.e. it has been hedged) and therefore we know it must return the risk-free rate. In a later step of the derivation, we will differentiate the portfolio value with respect to time and use this information to figure out what that is (unsurprisingly, it is simply a proportionate piece of the risk-free rate).

So in sum: the portfolio is set up in such a way that we know how it behaves, a fact which we later use to anchor the differential equation.

2) This equation is valid for any derivative that is differentiable twice with respect to S (the stock price) and once with respect to t (time). We have made no assumptions yet about put or call. As you will see, delta for puts is between -1 and 0, so if the derivative were a put the stock component of this portfolio would be negative.

Hull's book really does a great job of explaining this, I suggest you pick up a copy if you haven't already.


You are refering to a portfolio consisting of a call option and $\frac{\partial D_t}{\partial S_t}$ (the so-called Delta) "shares" of the "stock" $S_t$.
This strategy is used to "replicate a bond" in the derivation.

Since $P_t$ should replicate a bond you will have to use this in your further derivation (an thus get an equation). Just continue to read the derivation and it will all come out.


Say you have a portfolio with $\alpha$ dollars in cash and $\beta$ stocks at time $t=0$. The value of your portfolio at time $t$ is

$$P_t = \alpha e^{rt} + \beta S_t \tag{1} $$

Black-Scholes approach solve the following problem

Can we find $(\alpha,\beta)$ such that the portfolio is self-financing and has final value equal to the payoff of the option ? If yes then -no arbitrage argument- the fair price of the option is the initial value of the portfolio.

The self-financing equation is (in continuous time) : $$ dP_t = \alpha r e^{rt} dt + \beta dS_t \tag{2}$$

The value of $\alpha$ and $\beta$ are found using a purely mathematical derivation. We assume that $P_t=P(t,S_t)$ and use Itô formula : $$ dP_t = \frac{\partial{P_t}}{{\partial t}} dt + \frac{\partial P_t}{\partial S} dS_t + \frac{1}{2} \frac{\partial^2 P_t}{{\partial S^2}} d\langle S,S \rangle_t \tag{3}$$

Taking $\beta=\frac{\partial P_t}{\partial S}$ in $(2)$ and equalizing $(2)$ and $(3)$ yields the Black-Scholes PDE. The value of $P_t$ is the solution of this PDE, and the delta is the derivative of this solution w.r.t. its second argument, $S$.

This approach does not use the fact that the option is a call option. You actually only need a payoff of form $f(S_T)$. If the payoff involves two underlying e.g. you can use similar approaches with $\alpha, \beta$ and $\gamma$ for the second underlying.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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