Let $(B_t)$ be a brownian motion on [0,1]. For the following, let $\omega$ be fixed.

Let's compute the total absolute variation when **sampling period = $\delta$** is fixed:

$$V(\delta) = \sum_{i=0}^{N-1} |B_{t_{i+1}}(\omega) - B_{t_i}(\omega)|. $$

(i.e. $0 = t_0 < t_1 < t_2 < ... < t_N = 1$ with a constant step $\delta = t_{i+1} - t_i$ for all $i$)

I noticed experimentally that:

$$ V(\delta) \sim c\ \delta^{-1/2}$$

*It confirms the common-sense feeling that the smaller the sampling period (=the higher the sampling rate), the higher the total absolute variation.*

**Is it a well-known result?**
If so, where could I find a proof?

____

Some Python code to show this:

    # -*- coding: utf-8 -*-
    
    import numpy as np
    import matplotlib.pyplot as plt
    
    # GENERATION OF BROWNIAN MOTION
    X = 2 * np.random.binomial(1, 0.5, 2*1000*1000) - 1
    cumsumX = np.cumsum(X)
    n = 1000*1000
    x = np.linspace(0, 1, num=1000*1000)
    Y = 1/np.sqrt(n) * np.array([cumsumX[int(n*t)] for t in x])
    
    plt.plot(x,Y)
    plt.show()
    
    # ABSOLUTE VARIATION FOR EACH DIFFERENT SAMPLING PERIOD
    print('Sampling period, absolute variation')
    SP = []
    ABSVAR = []
    for k in range(1,15):
        sp  = 2 ** k
        Z = Y[::sp]
        absvar=sum(abs(Z[1:]-Z[:-1]))
        SP.append(sp)
        ABSVAR.append(absvar)
        print sp, absvar
    
    print('Coefficient:')
    print((np.log(ABSVAR)[-1]-np.log(ABSVAR)[0])/(np.log(SP)[-1]-np.log(SP)[0]))
    
    # LOGARITHMIC PLOT
    plt.plot(SP, ABSVAR, marker='o')
    plt.xscale('log')
    plt.yscale('log')
    plt.show()


[![enter image description here][1]][1]

Logarithmic plot of total absolute variation, in function of sampling period (both axis are log):

[![enter image description here][2]][2]


  [1]: https://i.sstatic.net/IRhEp.png
  [2]: https://i.sstatic.net/ze5Rv.png