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Suppose we have a Brownian motion $B_t$ with $B_0 = 0$ and $B_t - B_s \sim N(0,t-s)$. Every time $B_t$ hits $\pm h$, where $h$ is some "barrier" $>0$, I pay someone £1 and the brownian motion resets back to 0, and we start again. Is there a good estimate for how much I would expect to pay over time? (A finite time that is, clearly over arbitrarily large spans of time you'd expect to hit one of $\pm h$ infinitely often.)

I guess equivalently you could say suppose $B_t \sim N(0,\sigma^2 t)$ with $B_t$ smooth in $t$: roughly how frequently would you expect to hit $\pm 1$ given a reset to $0$ each time you hit the barrier? I don't need a fully rigorous answer or even fully accurate, just an estimate is good.

My first thought was that after time $t$, you "expect" to be roughly $\sigma \sqrt t$ away from the origin, so after time $1/\sigma^2$ you expect to be roughly 1 away from the origin: so maybe you expect to hit the barrier approximately every $1/\sigma^2$ units of time. However, I'm very wary of my use of the word "expect" here - the expectation we want for time per unit movement, and obviously expectation doesn't always commute with taking inverses to go from the "movement per unit time" we start with to "time per unit movement". Maybe this inversion is within reason of the correct calculation or maybe it's totally inaccurate. (I'm also cautious about the possibility I may have missed a factor of 2 in there or something like that...)

I'm afraid it's a while since I've done this sort of maths so any help would be greatly appreciated!

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up vote 1 down vote accepted

Paying events occur after i.i.d. times distributed as the first hitting time $T$ of $\pm h$ starting from $0$. One knows that $\mathbb E(T)=h^2$ (the easiest approach to this result being possibly that $(B_t^2-t)_{t\geqslant0}$ is a martingale). Hence, by the standard law of large numbers, the number of events occuring during a time interval of length $t$ is $(t/h^2)+t\varepsilon_t$, where $\varepsilon_t\to0$ when $t$ is large.

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Thanks very much! That makes perfect sense. –  Tom Jan 9 '13 at 18:38
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