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Why a Binomial Hedge Uses the Previous Step’s Delta

Article Quant Q&A · Author: s5s

Summary

The document clarifies why a binomial model’s portfolio value at the next time step uses the delta chosen at the previous step. At a given node, the trader knows the current stock price and selects a stock holding to hedge the option against the next possible up or down move. That chosen position remains in place while the market moves, so the portfolio at the next node is valued using the earlier delta.

Once the move is observed, the trader can rebalance by selecting a new delta for the following interval. Thus, the delta indexed by a time step represents a decision made at that time, rather than a holding that updates automatically with the stock price. The answer uses a simple two-period example to explain the timing. It addresses the interpretation of the hedge notation, not the derivation of option values or the conditions under which a binomial hedge is exact.

Key ideas

  • The hedge for an interval is chosen before the next stock-price move is known.
  • The portfolio after a move uses the delta selected at the preceding node.
  • After observing the move, the trader may rebalance to a new delta for the next interval.
  • Delta indexes a chosen stock holding at a time step rather than updating automatically with price.

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Full text
# Binomial model and delta hedging


# Binomial model and delta hedging












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:

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

## Answer by Magic is in the chain (score 3)

https://quant.stackexchange.com/a/54539

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.

## Answer by stackoverblown (score 1)

https://quant.stackexchange.com/a/54545

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.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.