Error in Derivation of BS PDE

[ruban]

This proof does not make sense.

The notes and Hull say that:

pi = - V + delta * S => d(pi) = -dV + delta*dS but I disagree as this implies that delta is a constant.

What do you think?

 

[ronankeane]

That's a good point but I can follow the proof if I make a mental adjustment to the equations:

(Hull, page 292)

Since the portfolio is not self-financing, d(Pi) has an extra term
+d(df/ds) * S

As you read the proof, add this term to the RHS of equation (13.13). It carries through to the RHS of (13.14).

This next step is a little hazy, but it makes sense to me that you can also add the same term [+d(df/ds) * S] to the RHS of equation (13.15). The share and the derivative holding cancel out the direct dependence on small changes in S, leaving you with growth at the risk-free rate plus the change in portfolio value resulting from changing the volume of shares held.

If you accept that, you can still equate the RHS of (13.14) and (13.15) to get (13.16).

Hope that makes sense.

 

[TheOke]

I think the argument is that the delta is constant over a short period of time. I have an extremely similar question, which I can't reconcile:

Page 26, chapter 06 of the full reading, or page 590 of hull, where they try show that the market price of risk is equal for 2 derivatives that have the same source of randomness:

If you hold f1 of f2, and f2 of f1, then you essentially have 2*f1*f2 and
d(f1*f2) = f1*d(f2) + f2*d(f1) + d <f1,f2 > /// The last term essentially means d(f1)*d(f2).
Point is that they actually leave out 2 terms, and moreover, if you take these into account, the dZ (the random terms) do NOT cancel. (and the proof does not work)

It is like saying you hold S1 of S1, and hence d(S1*S1) = S1*d(S1). Makes no sense to me... you would either have to have 2*S1*d(S1)+d<S1,S1> or use Ito's formula on F[S1] = (S1)^2

Any help from anyone?

 

[TheOke]

I had a thought if this helps anyone answer the question.

You can propose that you follow any strategy you like. So if you say you hold S(t) of S(t) over [t, t+dt], then at time t+dt, you would have S(t)*S(t+dt). Let Pi(t) denote this portfolio. Then according to this strategy, d[Pi(t)]=S(t)*d[S(t)]. You would ignore the other terms (another S(t)*d(S(t)) and the quadratic variation term d<S(t),S(t)> )
However, this strategy would not be self-financing, since the true change in value would have these extra terms which are non-zero.

In the case of the derivation of the original B-S partial differentially equation, the strategy followed can be proved to be self-financing. Hence, the logic would be correct.
In the case I previously provided, I can't see how the strategy is self-financing. The problem is that if the strategy is not self-financing, then the argument that the portfolio grows at , r , the risk-free rate cannot hold - since you are then adding and removing money every time you re-balance the portfolio.
Hence, even along these lines of reasoning, this proof is invalid.

 

[AngelFace]

I think you may have missed a very important assumption of the BS PDE - namely that it is possible to continuously delta hedge. This implies that over the very small interval [t, t+dt), delta of the portfolio is constant and therefore can be hedged.

In the real world of course rebalancing happens less frequently, due to the costs of rebalancing. This is where gamma hedging comes in.

 

[TheOke]

Ignoring the issues with real-life trading, the reason you hold Delta constant is the self-financing condition, which states exactly that - that the change in the value of the portfolio only comes about through the change in the underlying assets. So if the portfolio is self-financing, then mathematically, D[Pi]= -d[V] + Delta * d.
The proof should actually state that the portfolio must be self-financing by construction, and then solve for the Delta which allows this. Turns out to be partial[V]/partial as expected.

 

[mkone]

This proof does not make sense.

The notes and Hull say that:

pi = - V + delta * S => d(pi) = -dV + delta*dS but I disagree as this implies that delta is a constant.

What do you think?

Don't know if you got a satisfactory answer, but the short answer is that the derivation is correct. Delta is constant because it is the amount of shares you buy, which remain constant over the short time interval dt, rather than a rate of change of the value of the shares. So after a very short interval, over which you have not changed the composition of your portfolio, you have delta shares, whose value have changed by dS per share, hence the total change in the value of the shares in the portfolio is delta * dS. So yes, delta, as in the number of shares, remains constant.