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Consider a Brownian motion B_t with constant instantaneous volatility σ and zero drift

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where t is larger than zero and the brownian motion is equal to zero in the beginning.

We know that the increments are equal to

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and satisfy

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This leads to the moment condition:

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For which we can estimate σ based on a sample of returns. This is basically the same as close-to-close / traditional volatility. However, rather than measuring the size of Brownian increments over a fixed time interval, we can also measure the time it takes the Brownian motion to travel a given distance up or down. This can be written as:

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According to Borodin and Salminen (2002) we have the following moment condition

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Again we can calculate σ, when we re-arrange the formula:

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However, when volatility is not constant we can not assume that the observed passage times are identically distributed. According to this paper we can apply the following method to calculate the local volatility:

Consider a fixed time grid

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that consists of N, not necessarily equispaced, intervals. Each interval is equal to

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At a given point, t_i, we need to look for the next passage time enter image description here. Since the Brownian motion is time-reversible, we can also look backwards and find the previous passage time, enter image description here. Visually this looks as follows:

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Next, we have two independent estimators of local volatility at t_i, namely

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and

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We take the average of the two and have an estimation of the local volatility at the point t_i.

What I don't understand is that the N intervals are not necessarily equispaced intervals. Why is this the case? How to determine then the size of each individual interval?

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    $\begingroup$ Just a note: If $B(t)$ is defined as Brownian Motion with volatility $\sigma$, then the very first equation in your question should be $\int_{0}^{t}\sigma dW_t$ $\endgroup$ – Jan Stuller Oct 31 '20 at 16:52
  • $\begingroup$ yes, you're correct. Thanks for pointing out. Any chance you can also help me with my final question? $\endgroup$ – HJA24 Nov 2 '20 at 11:00
  • $\begingroup$ Let me see if I understood the question. Are you asking why the time intervals can be non-identical? And how to compute them? $\endgroup$ – rvignolo Nov 6 '20 at 22:29
  • $\begingroup$ Not sure what you mean with non-identical, I am focusing on non-equally sized $\endgroup$ – HJA24 Nov 10 '20 at 6:47
  • $\begingroup$ time intervals can be different because day count conventions can yield to different interval sizes. When you move from dates given in a term sheet to real numbers used for computations, you will probably get non-equally sized time intervals. $\endgroup$ – rvignolo Nov 10 '20 at 18:48

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