I've got a question about theory which is probably a one line answer. I use to understand it but I'm stuck right now.

In the Binomial model, we define the progression of the price as:

$$ S_k = S_{k-1} e^{\alpha X_k} $$

where $P(X_k = 1) = p$ and $P(X_k = -1) = 1-p = q$

Now, to determine $\alpha$ and $p$ using the delta hedging argument (rather than risk-neutrality) we define a small time increment $h$ and at time $t_{n-1} = (n-1)h$ a portfolio $\Pi$

$$ \Pi_{n-1} = f(S_{n-1}, t_{n-1}) - \Delta_{n-1} S_{n-1} $$

where $S_t$ is the asset price and $f(S_t, t)$ is the payout. To determine $\alpha$ and $p$ we choose $\Delta_{n-1}$ so that the evolution of $\Pi_{n-1}$ is deterministic. My lecture notes define $\Pi_n$ at $t=nh$ like so

$$ \Pi_n = f(S_{n}, t_{n}) - \Delta_{n-1} S_{n} $$

Substitute $S_k = S_{k-1} e^{\alpha X_k}$

$$ \Pi_n = f(S_{n-1} e^{\alpha X_n}, t_{n}) - \Delta_{n-1} S_{n-1} e^{\alpha X_n} $$

Then we calculate $\Delta_{n-1}$ so that $\Pi_n$ is deterministic etc.

What I do not understand is why do we have $\Delta_{n-1}$ in the formula for $\Pi_n$? Shouldn't the formula for $\Pi_n$ be:

$$ \Pi_n = f(S_{n}, t_{n}) - \Delta_{n} S_{n} $$


Say there are just two periods: Payoff at n, and premium/price at $n-2$.

We know the current stock price, say $S_{n-2}$, and we know in the next epoch, it will either be: $S_{n-1}^u=uS_{n-2}$ (up state), or $S_{n-1}^d=dS_{n-2}$ (down state). We need to make the decision at epoch $n-2$ as to how many units of the stock to buy or sell to hedge the option, and assume we decided to buy $\Delta_{n-2}$ units of stock.

We then wait a bit to find out the true state of nature. Price has gone up: Our stock is worth $\Delta_{n-2} S_{n-1}^u$. Price has gone down: Our stock is worth $\Delta_{n-2} S_{n-1}^d$. So hoping the hedging has worked, we need to get ready for the next move, so we rebalance the portfolio, which means decide at $n-1$ how many units of the stock to hold, $\Delta_{n-1}$, to hedge the option position against the next move.

So the one liner could be: We decide at $n-1$ how many units of the stock to buy/sell to hedge against the next up/down move of the stock price.

| improve this answer | |
  • $\begingroup$ By convention the initial time is Time 0. The time interval starting at Time 0 and lasting one period is Interval 1 or Period 1, which ends at Time 1. So period n (during which the stock price moves) starts at time n-1 and ends at time n. $\endgroup$ – noob2 May 30 at 17:03

Unlike $S$ and $f$ which are driven by the market that are out of your control, $\Delta$ is the amount of stocks $S$ that you have decided to short in the previous time step for this portfolio $\Pi$. It is not an intrinsic time dependent quantity. So of course it is staying fixed until you decide to change or not to change it in the next time step.

| improve this answer | |

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.