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0

Try modelling samples every 20,000 ticks, instead of 2 hours (or any such number like that). Markets are often less fat tailed in terms of the trade- or volume-clock. See http://www.amazon.ca/Introduction-High-Frequency-Finance-Ramazan-Gen%C3%A7ay/dp/0122796713 and http://papers.ssrn.com/sol3/papers.cfm?abstract_id=2034858


5

The first process is a BM. The second does not exist in continuous time. The variance goes down too slowly with dt and the process blows up at the limit. You can break the (0,1) interval into 1, 100, 1000, 1000000 steps and see that happening. Variance of a martingale has to scale with dt: if it is too fast then the process dies, if it is too slow then ...


4

The first process $$ B_{t+dt} = B_t + Z $$ where $Z$ is independent of $(B_s)_{s \le t}$ and follows a Gaussian distribution with mean $0$ and varince $dt$ is a standard Brownian motion (thus the variance of $B_t$ is $t$). For the second process let us recall the definition from your link: $$ E[B^H_t B^H_s] = \frac12 ( t^{2H} + s^{2H} - |t-s|^{2H}), $$ thus ...


0

the answer is simple: look at key differences between these two models. GBM is diffusion, OU is mean-reversion


3

Given efficient markets, asset prices should be unpredictable in the sense that any upcoming returns are uncorrelated with current or past returns. Hence for traded assets the price should follow something more similar to a GBM than an O-U process. However, many financial metrics are not prices; for example interest rates or volatility. O-U processes may ...


1

The two processes are not pathwise equal. Here is a simulation (sample path) of the two processes $(t-\tau)dW_{\tau}$ and $W_\tau d\tau$: Note that both processes have the same value at the final time.


1

If $X_t$ is square integrable, then the integral \begin{align*} \int_0^t X_{\tau} dW_{\tau} \end{align*} is a martingale. Here, the integrand $X_{\tau}$ does not depend on the integral limit $t$. However, in your case, the integrand, $t-\tau$, depends on $t$, then the condition for the martingality of the integral fails.


0

For Q2, let $\lambda = \mu/\sigma$. Moreover, we define the measure $Q$ on $(\Omega, \mathcal{F})$ such that \begin{align*} \frac{dQ}{dP}\big|_{\mathcal{F}_t} = \exp\Big(-\frac{1}{2}\lambda^2 t - \lambda W_t\Big), \mbox{ for } t \ge 0. \end{align*} Then, by Girsanov theorem, $W^*$, where \begin{align*} W_t^* = \lambda + W_t, \end{align*} is a standard ...



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