## Wednesday, April 30, 2008

We now assume that the bead displacement d_n during time interval n is the sum of k_n unresolved microsteps, each of (average) width w_1:

$d_n&space;=&space;w_1&space;\;k_n$
Let the k_n fluctuate according to a Poisson-distribution with average event number lambda (given by some event rate times the discretization time interval):

$P_{\lambda}(k)=\frac{\lambda^k&space;e^{-\lambda}}{k!}$
which has a variance of

$\sigma_k^2=\lambda$
Then one obtains for the mean bead displacement per discretization time interval

$\overline{d}=\lambda\;w_1$
and for the variance

$\sigma_d^2=\sigma_k^2\;w_1^2=\lambda\;w_1^2$