Wednesday, April 30, 2008

[AD 15] Poisson-distributed SWD

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 = w_1 \;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 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

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