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I am trying to derive an optimal hedging interval for a delta-replicated European call option from the Black-Scholes model. To do that, I would like to compare the resulting hedging error with my trading costs.

I am currently hedging (i.e. adjusting the delta of) the position every day and am trying to find out whether more frequent intraday hedging makes sense.

Unfortunately, I am making a mistake in my assumptions somewhere which I am hoping you can point out.

What I tried so far:
Lets assume volatility and time to maturity stay the same and risk-free-rate = dividends = 0.
I'll also use $\Delta \delta \approx \gamma \cdot \Delta S$ to describe the change in Delta for a small change in price.

I can roughly calculate the hedging error for a small price change using a Taylor expansion of the option price. $C(S) \approx C(S_0) + \delta(S_0) \cdot (S - S_0) + \frac{\gamma(S_0)}{2} \cdot (S - S_0)^2+... $
Since I hold $\delta(S_0)$ in my portfolio, the error for a small change in price is roughly $\frac{1}{2} \gamma (\Delta S)^²$

Additionally, I can calculate the trading costs as: $c \cdot \left| (\delta_t - \delta_0) \cdot S_{t} \right| \approx c \cdot \left| \gamma \cdot \Delta S \cdot (S_0 + \Delta S) \right|$
c is a constant representing my trading costs per trading volume.

Now, using geometric brownian motion $dS_t = \mu S_t \, dt + \sigma S_t \, dW_t$, I find the expected value for my hedging error and trading costs.

$\mathbb{E}[\frac{1}{2} \gamma (\Delta S)^²] \approx \frac{1}{2} \gamma S^2 \sigma^2 \Delta t$

$\mathbb{E}[c \left| \gamma \Delta S (S_0 + \Delta S) \right|] \approx c \gamma S^2 \sigma \sqrt{\Delta t} \sqrt{\frac{2}{\pi}}$

This seems to be correct in my simulations.
However, when I try to calculate the total costs over a greater timespan T (one day for example), the hedging error seems to be the same no matter the hedging frequency. Let $n = \frac{T}{\Delta t}$ the number of intervals/hedges, then $\sum_{i=1}^{n} \frac{1}{2} \gamma S^2 \sigma^2 \Delta t = n \cdot \frac{1}{2} \gamma S^2 \sigma^2 \Delta t = \frac{1}{2} \gamma S^2 \sigma^2 T$

Adding this error to my costs (which decrease by increasing $\Delta t$) doesn't make much sense. Are my calculations wrong or my assumptions too simple? Do I have to calculate it completely differently? What am I doing wrong?

Please only use Black-Scholes and GBM in your answer and please also try to simplify it as much as possible. Thank you very much!

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    $\begingroup$ It is totally counter-intuitive that (neglecting costs) the hedging error over $T$ should not depend on the trading interval $\Delta t\,.$ The authors of this write up produce a number of graphs that show what we expect: hedging errors go down when we increase the number of trading dates. $\endgroup$ Commented Oct 18, 2024 at 17:17
  • $\begingroup$ Hi, thanks for your comment and the paper. Yes, Figure 3 shows the decrease of the expected hedging error when n is increased. However, this is just the expected error for one interval. I am trying to calculate the total error over T. $\endgroup$ Commented Oct 18, 2024 at 20:30
  • $\begingroup$ I do not think so. They present four Figures all showing the same error decrease. On p. 14 they mention that Figure 1 is for an option maturity of $T=1$ and for number of trading days in that interval (denoted by $N$) ranging from one to 65. $\endgroup$ Commented Oct 19, 2024 at 8:43
  • $\begingroup$ Think the intuitive logic is that hedging error will decrease with 1/sqrt(N) but costs increase with N where N is number of hedges over option life. So sharpe ratio will likely be some negative parabola. $\endgroup$ Commented Oct 28, 2024 at 8:33
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    $\begingroup$ I think the best solution would be to run monte carlo sims of real numbers. It sorts the academic theoretical chaff from the real word wheat. $\endgroup$ Commented Oct 28, 2024 at 19:23

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Assuming proportional transaction costs (to the size of the trade):

The number of shares, N, to be hedged at each interval is given by $$ N = \Delta(S_t + dS, t + dt) - \Delta(S_t, t) \approx \Gamma_t * dS$$ which expands to: $$ N \approx \Gamma * \sigma * S * dW $$

The incremental cost is given by the number shares, N, multiplied by the share price, multiplied by the transaction cost fraction, $\alpha$: $$ Cost = |N*S| * \alpha $$ $$ Cost = |\Gamma * \sigma * S^2 * dW| * \alpha $$

$|dW| > 0$, so costs are, obviously, non-zero. Rewriting $dW$ as $z\sqrt{dt}$, we can see that for an option with a $\frac{T}{dt}$ number of hedges over the life, T, total costs will scale with $\frac{T}{dt} * \sqrt{dt} = \frac{T}{\sqrt{dt}}$. So as the hedging interval moves to 0, costs move to infinity.

The expected return of a hedged short option position is given by: $$\frac{\Gamma S^2}{2} * (\sigma^2_i - \sigma^2_r) * dt$$ and the volatility is given by: $$ \frac{\Gamma S^2 \sigma^2_r}{2} * \sqrt{\frac{\kappa -1}{N}} $$ where $\kappa$ is the raw kurtosis, under GBM $\kappa = 2$.

The expected return drag from increasing the hedging frequency is approximately: $$\Gamma * S^2 * \sigma_r \sqrt{\frac{2}{\pi * dt}}$$

Creating a Sharpe ratio gives: $$ \sqrt{\frac{N}{\kappa -1}} * (((\frac{\sigma^i}{\sigma_r})^2 -1) - \frac{\alpha\sqrt{\frac{8}{\pi}}}{\sigma_r * dt^\frac{3}{2}} )$$

You can then differentiate and maximise the sharpe ratio w.r.t hedging frequency.

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  • $\begingroup$ A joke is cited in Sigmund Freud's Der Witz und seine Beziehung zum Unbewußten: "I take a bath every <frequency>, whether I need it or not". Whilst this math is very pretty, I think there's something fundamentally wrong with the notion of re-hedging at any fixed cadence whether the book needs it or not. Re-hedge when some factor sensitivity or VaR or P&L under stress scenarios is too much. Set small limits if you don't want P&L from gammas. BAU re-hedge during business hours, emergency re-hedge overnight or on weekends/holidays if needed. $\endgroup$ Commented Dec 2, 2024 at 19:19

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