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1) Does the following algorithm (my question is math, not programming-related):

n = 1000000
dt = 0.01
B = zeros(n)        # [0, 0, 0, ..., 0]
for i in range(1,n):
    B[i] = B[i-1] + random.normal(0, sqrt(dt)) 

simulate a standard brownian motion, since we have $B(t+dt) - B(t) \sim \mathcal{N}(0, \sqrt{dt}^2)$?

According to my tests, I would say yes, but I wanted to be sure.


2) What's the name of the process obtained by doing:

B[i] = B[i-1] + random.normal(0, dt ** 0.3)

i.e.

$$B(t+h) - B(t) \sim \mathcal{N}(0, \sigma^2)$$ with, for example, $\sigma = h^{0.3}$ instead of $\sigma = h^{0.5}$ for the standard brownian.

It seems it would be not far from a fractional brownian motion with Hurst exponent $H = 0.3$ (but the covariance formula - first formula of this page - would apply only for $t = s$ or am I wrong?)

What's the name of such a stochastic process, for which increments are gaussian with standard deviation $(dt)^{0.3}$ instead of $(dt)^{0.5}$?

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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 the process blows up.

Edit: Suppose that you partition the $(0,1)$ interval into $N$ steps of length $1/N$, and define the process $X_n$ over these points. Now say that each step has variance that scales as $$var(X_{n+1}-X_n) \sim (1/N)^\alpha$$

If the process is a random walk, then the variance of the process at $t=1$ (that is to say $X_N$) would be $$var(X_n) \sim \sum_{n=1}^N (1/N)^\alpha = N^{1-\alpha}$$ You can see that as $N\rightarrow \infty$ this goes to zero ($\alpha>0$), blows up ($\alpha<1$) or just balances at one for $\alpha=1$ that corresponds to the standard BM.

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  • $\begingroup$ Thanks! I understand the global idea, but can you precise mathematically what you mean by "the process blows up at the limit" ? Why does the fact the variance grows too slowly imply that? $\endgroup$ – Basj Jan 25 '16 at 15:17
  • $\begingroup$ When you go from a discrete process to a continuous time, you got to be careful or you lose continuity of the process, it "breaks up". $\endgroup$ – noob2 Jan 25 '16 at 16:47
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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 for $t=s$ this is $\frac12 ( t^{2H} + t^{2H} - |0|^{2H}) = t^{2H}$. Note that for $H=\frac12$ this equals $t$ as in the Brownian motion case above.

For such a process the increment after $t$ is not independent of the process at time $t$. Look at the case $s>t$: $$ E[B^H_t (B^H_s-B^H_t)] = E[B^H_t B^H_s] - E[B^H_t B^H_t], $$ and this is, using the formulae above $$ \frac12 ( t^{2H} + s^{2H} - (s-t)^{2H}) - t^{2H}. $$ The above expression equals 0 for $H = \frac12$: $$ \frac12 ( t + s - (s-t)) - t = \frac12 ( t + s - s + t) - t = 0. $$

Finally to answer you question: You got the variance term right, but sampling the increment you have to take the correlation/dependence of the increment to the process into account. So as you say the process is not fractional Brownian motion. Finally I wonder whether this process exists in continuous time.

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