# 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.

• Clue: For a zero mean normal variate the expected absolute value is equal to $\sqrt{\frac{2}{\pi }}\sigma$ where $\sigma$ is the standard deviation. Taleb is using this well known approximation to derive this equation. See blog.arkieva.com/relationship-between-mad-standard-deviation Commented Jul 7 at 11:26
• Yes they are related, search for "local time" questions in this forum. Unless somebody else posts a proper answer I'll answer your question in the coming day or two (no time at the moment). Commented Jul 7 at 11:30
• Thanks, I suspected it was related to it. Is thus the author only multiplying it with the number of times the underlying price exceeds the strike K? Like expected number of "buy actions" times the average "buy" price over K? Or is he considering both buy and sell price in this way? Commented Jul 7 at 11:36
• @Frido, I will appreciate so much if you find the time. Thank you. Commented Jul 7 at 11:36
• The treatment by Carr is more mathematically rigorous; he made an attempt at an intuitive explanation here quant.stackexchange.com/questions/38582/… Commented Jul 7 at 11:38

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 $$\frac12 \times ((dS_t)^2_{S_t = K}/dt)\times \text{the time spent swinging between K and K\pm dK}.$$

• Thank you @Frido. Let me digest it and I will come back with some questions in the next few hours. Commented Jul 8 at 8:19
• Here we are. 1) How is the risk-neutral expectation linked to the second derivative wrt $K$? 2) Why could this second derivative be interpreted as the probability of $S_t \in [K, K+dK]$? 3) I do not understand properly this notation $(dS_t)^2_{S_t=K}/dt$ 4) I suppose that "time" for Taleb means the expcted number of days/hours/... obtained by multiplying the last intergral (a probabilty as you said) for the option duration in days/hours/..., correct? Thank you for the help. Commented Jul 8 at 9:20
• @Enrico Re the second derivative, it's called the Breeden-Litzenberger equation. You should read about it and try to understand why it's the risk-neutral prob. density as you'll encounter it more often in quant finance. Now $(dS_t)^2 = \sigma^2 S_t^2 dt$. When $S_t = K$ this becomes $(dS_t)^2|_{S_t = K} = \sigma^2 K^2 dt$, and then divide by $dt$. Commented Jul 8 at 9:38
• To be precise: $$\frac12 \sigma^2 \int_0^T E_0 [S_t^2 \delta (S_t - K) ] dt = \frac12 \sigma^2 \int_0^T\int_0^{+\infty} x^2 \delta (x - K)\frac{ \partial^2 C(S_0, x, t) }{ \partial K^2} dtdx= \frac12 \sigma^2 K^2 \int_0^T \frac{ \partial^2 C(S_0, K, t) }{ \partial K^2} dt$$. Regarding $[K, K+dK]$, as strictly speaking we are talking about probability density instead of discrete probability, a probability density is by definition that something is within an infinitesimal interval as a single point has measure zero. Hope this helps. Commented Jul 8 at 13:57
• OK, how you defined $C(S_0,K,T)$ confounded me :) Thank you for the help! Commented Jul 8 at 14:02