Wednesday, July 16, 2008

What to expect: Trend and Volatility

I assume here that the price evolution is modelised by a Fractional Brownian Motion (FBM) of index-$\alpha$ (0<$\alpha$<1): href="http://www.codecogs.com/">$X:[0,\infty)\rightarrow&space;\mathbb{R}$

where X(t) represents the price at time t, so that we have the following equality (E1) about the expectation of dependent price increments (demonstration in [FALC03]pp267-268):

$E[(X(t)-X(0))(X(t+h)-X(t))]=\frac{1}{2}[(t+h)^{2\alpha&space;}-t^{2\alpha&space;}-h^{2\alpha&space;}]$

Clearly the value $\alpha$=1/2 seems to play a very specific role in that equation, since it cancels out its right-side term.
$\alpha$=1/2 indeed consists in the classical Brownian Motion (Wiener Brownian Motion:WBM) where the increments over time of the variable X are independent.
This index $\alpha$ is directly linked to the Fractal Dimension Df by the relation:
$\alpha=2-D_{f}$

Therefore, when $\alpha$=1/2, which is happening when Df=1.5, we have a genuine Random Walk.
When such is not the case, however, we can say:

1) Df<1.5
This case is equivalent to $\alpha$>1/2, and we can then expect from the equality (E1) that X(t+h)-X(t) tend to be of the same sign as X(t)-X(0), therefore, if X(t) has an history of increasing, the next move X(t+h) will be more likely to be up, similarly if X(t) has an history of decreasing, the next move will tend to be down. In this case, we are in a trend.

2) Df>1.5
This case is equivalent to $\alpha$<1/2.> In this case, X(t+h)-X(t) tend to be of the opposite sign of X(t)-X(0), therefore, following the same logic as above, we are in a trend reversal period.

1 comment:

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