Timeline for Geometric Brownian Motion as the limit of a Binomial Tree?
Current License: CC BY-SA 4.0
9 events
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| Jun 23 at 21:46 | history | edited | Knabe | CC BY-SA 4.0 |
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| Jun 23 at 20:46 | comment | added | Knabe | Now I see what the questions is. I extended the answer. If I have time, I will derive those u, d, and p, for the relative returns. | |
| Jun 23 at 20:43 | history | edited | Knabe | CC BY-SA 4.0 |
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| Jun 23 at 20:38 | history | edited | Knabe | CC BY-SA 4.0 |
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| Jun 21 at 15:26 | comment | added | Bumblebee | In my mind, these formulas for up/down factors take the time value of money into account, hence more accurate than the ones given in the CRR model. Secondly, this natural probability is not a simplification of risk-neutral probability. It is a different probability that does not come from "arbitrage-free" assumption. | |
| Jun 21 at 15:17 | comment | added | Knabe | @Bumblebee Why do you call this expression for p a natural probability? This is just a simplification of a more correct formula. Maybe you can provide a reference where the equations for p, u, and d are coming from, as they are quite non-standard. They lead to clear errors, like 2mu in the exponent. With the given formulas, it is not possible to match either the first or the second moment of spot at time t+1. If you look in Luenberger "Investment science", you will see p like you gave, but then it is corrected to a more proper value. | |
| Jun 21 at 14:49 | comment | added | Bumblebee | I have seen this derivation with risk-neutral probability before. What I'm asking is how/why we cannot get the same geometric Brownian motion with the natural probability $p=\dfrac12\left(1+\dfrac{\mu}{\sigma}\sqrt{\Delta t}\right). $ | |
| S Jun 21 at 12:08 | review | First answers | |||
| Jun 22 at 10:08 | |||||
| S Jun 21 at 12:08 | history | answered | Knabe | CC BY-SA 4.0 |