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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:

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

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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.

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  • $\begingroup$ 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
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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.

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