Option Hedging Costs, Local Time, and the Tanaka Formula

# Cost of Hedging and Ito Calculus


# Cost of Hedging and Ito Calculus












In Dynamic Hedging by N. Taleb, at pag. 198, is presented a stop loss strategy that potentially could replicate an option. In particular, suppose one sells a call on an underlying and hedge it with a stop loss in the market to buy the underlying at the strike K and, conversely, if the price goes below the strike K the operator sells the underlying.

Why does the author states that the costs of hedging are $\sqrt{2/\pi}\sqrt{(\Delta S)^2}$ times the amount of time spent swinging between $S$ and $S+\Delta S$?

I made some research and I found this example very similar to the stop loss / start gain strategy paradox solved in Carr (1990) or to the Exercise 4.21 in Stochastic Calculus for Finance II by Shreve. Unfortunately, I am not so proficient with the math involved to determine if my question is related to this paradox.

Could anyone tell me if these are related? And if so, where does the term $\sqrt{2/\pi}\sqrt{(\Delta S)^2}$ come from?

Thanks for the help. Let me know if more details are needed.

## Answer by Frido (score 1, accepted)

https://quant.stackexchange.com/a/79967

I'll try to explain what Taleb means. I do think though Taleb's explanation is not entirely correct.

I assume you know what the Heaviside and Dirac-delta functions are/do, as they occur in the following decomposition of an option payout which is called the Tanaka-Ito-Meyer formula: $$ (S_T - K)_+ = (S_0 - K)_+ + \int_0^T \theta(S_t - K) dS_t + \frac12 \int_0^T \delta(S_t - K) (dS_t)^2 $$ The left-hand side is the terminal payout. The first term to the right of the equality is an option's intrinsic value, the second term is the P/L over $[0,T]$ of buying one unit of the asset when $S_t >K$ and selling that unit / not buying the asset if $S_t < K$. The third term is the "local time".

As $\delta(S_t - K)$ is 1 if $S_t = K$ and 0 otherwise, you can already see that the third term has something to do with the amount of time $S_t = K$, or more accurately $S_t \in [K,K+dK]$.

Let's assume we live in a Black-Scholes world and for simplicity take $r=q=0$: $$ dS_t = \sigma S_t dW_t, \enspace (dS_t)^2 = \sigma^2 dt, \enspace \sigma = \text{const.} $$ To find the cost of something as you know you need to take risk-neutral expectation. Then $$ C(S_0,K,T) = (S_0 - K)_+ + \frac12 \sigma^2 \int_0^T E_0 [S_t^2 \delta (S_t - K) ] dt $$

As in the Black-Scholes world the volatility is constant, we can write $$ \frac12 \sigma^2 \int_0^T E_0 [S_t^2 \delta (S_t - K) ] dt = \frac12 \sigma^2 K^2 \int_0^T \frac{ \partial^2 C(S_0, K, t) }{ \partial K^2} dt $$ But the integrand $\frac{ \partial^2 C(S_0, K, t) }{ \partial K^2}$ is nothing else than the probability that $S_t \in [K, K+dK]$ at time $t$ given $S_0$ today. The probability multiplied by $dt$ and integrated over the interval $[0,T]$ is nothing else than the total time spent by the spot in an infinitesimal interval about the strike $K$. Furthermore $\sigma^2 K^2 = (dS_t)^2_{S_t = K}/dt$.

So in my opinion, what Taleb should have written is that the cost of hedging is $$ \frac12 \times ((dS_t)^2_{S_t = K}/dt)\times \text{the time spent swinging between $K$ and $K\pm dK$}. $$