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I'm trying to estimate the hedging costs relating to a financial derivative which moves like a Wiener process, and I'm struggling to find the correct setup to solve the problem.

Suppose I have a process $X_t \sim N(0,t) $ starting at $0$, and we "hedge" every time $X_t$ reaches our position limit $\pm L$. When I say we "hedge", what I mean is we pay some cost, say $\$ (1-\lambda)$, and $X_t$ is sent back to $+ \lambda L$ if it hit the barrier $+L$, or sent back to $-\lambda L$ if it hit the barrier $-L$, with $0 < \lambda < 1$ fixed; $X_t$ then starts moving again from this reset position with the same distribution, until it hits either $\pm L$ again. The barriers are not reset relative to $\lambda L$ - that is, you will reset back to a distance $(1-\lambda)L$ from the barrier. (I know strictly speaking a Wiener process should start from 0 but ignore this for now.) I essentially want to know what the expected long-term hedging cost is, as a function of the time elapsed $t$ and the re-hedge level $\lambda$.

I know that your expected time to hit the barrier is $L^2$, and I gather that if $\lambda = 0$ (i.e. if you reset $X_t$ back to $0$ every time) then in time $T$ you would expect to reach the barrier approximately $T(1/L^2 + o(T))$ times. However, is $T(1-\lambda)/L^2$ really the true hedging cost? A friend suggested you might need a factor of $1/2$ - if in any infinitesimal time period you expect to reach the barrier $k$ times, then you may only expect to cross it half of those times in any small time period, so you might expect $\$ (T/(2\cdot L^2)) (1-\lambda)$ to be the correct estimate of the hedging cost.

I guess the underlying picture I have in my head here is the usual binomial tree model, where you can move either above or below the barrier in the next step. Is there a flaw in this reasoning? I'm very wary of making invalid assumptions with these infinitesimal movements - obviously there's a need to be a bit cautious if you're talking about hedging whenever you go an infinitesimally small amount over a barrier with a Wiener process exhibiting infinitesimal movements. Does it fundamentally alter your model if you start looking at a binomial tree with small finite moves $dX$ in a small finite time period $dt$, and then let these $\to 0$, or can you safely take such limits given the right constraints? Hopefully my concerns do make some sense - I want to use these results to model some hedging costs in real life, so realistically I'm only going to be concerned with discrete time periods/movements anyway.

More generally, if rather than rehedging back to $0$ we rehedge back to $\pm \lambda L$, is there a closed-form solution in terms of $\lambda$ for the expected hedging costs? I would be particularly interested to see what happens as $\lambda \to 1$ and we hedge back to the barrier very frequently for a very small cost. More generally it would be quite nice to know, given some cost function $c(\lambda)$ to hedge, is there an optimal choice $\lambda$ which minimizes the amount you expect to pay? Obviously the first step to that is understanding exactly how often I expect to hedge, and then it's more of an optimisation problem. Will be interesting to see anyway.

I couldn't find any papers or articles answering my question but if anyone knows of one then I'd be very happy to read up on a solution myself, just point me in the right direction. Thanks for your help!

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2 Answers

The factor $\frac12$ is not relevant here, due to the following fact: assume that $|X_0|\lt L$ and consider the random times $T=\inf\{t\geqslant0\,;\,|X_t|=L\}$ and $S=\inf\{t\geqslant0\,;\,|X_t|\gt L\}$. Then $T=S$ with full probability.

In words, when it reaches the barriers $\pm L$, the Brownian particle instantaneously goes out of the interval $[-L,+L]$.

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More generally it would be quite nice to know, given some cost function c(λ) to hedge, is there an optimal choice λ which minimizes the amount you expect to pay? Obviously the first step to that is understanding exactly how often I expect to hedge, and then it's more of an optimisation problem. Will be interesting to see anyway.

The answer is yes - see this paper http://ssrn.com/abstract=1865998

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