# Integration and Sampling, Discrete and Continuous Distributions, Errors evaluation for a Markov chain_Questions

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You can insert math formulæ using LaTeX code: e.g. typing
[[math]]
\pi_C(t+1)=\sum_{C'} \pi_{C'}(t) p_{C' \to C}
[[math]]
yields:
$\pi_C(t+1)=\sum_{C'} \pi_{C'}(t) p_{C' \to C}$

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## Questions and Remarks:

• Q: What exactly mean Var(\xi_i)? Is it mean
$Var(\xi_i)=\frac{1}{N}\sum_{i=0}^N \xi_i^2-(\frac{1}{N}\sum_{i=0}^N \xi_i)^2$ ?
A:
$Var(\xi_i)$
is the variance, essentially the square of the standard deviation (l' écart-type) of my data distribution

• Q: I am not sure to understand question I-2: are we suppose to do as if each step of the markovian walk gives a independent random variable (which is of course wrong: the position between two steps are correlated). ( Ok seems to be "yes" according to part II)

• Q: I have a problem with the formula (1), namely the formula which gives the statistical error in our Markov chain data from the autocorrelation function. I would expect that we recover the usual formula ( sqrt(var/N) ) if the data are uncorellated (ie if C(n) = 0 for n > 0), but here I get sqrt(2var/N).
• A: I guess one has to count for each n 2 possible distances between couples of sites, one in the positive, the other in the negative direction, hence a factor 2, except for the distance 0 point....I guess. Ask Alberto.

• A: Yes you are right, there is a small mistake in Eq.(1). The correct formula is
$\langle\xi\rangle= \frac{1}{N}\sum_{n=0}^N \xi_n \pm \frac{\sqrt{ C(0) +2 \sum_{n=1}^N C(n)}}{\sqrt{N}}\qquad\qquad (1)$

- Shouldn't there be a minus sign in front of C(0) ?
$\langle\xi\rangle= \frac{1}{N}\sum_{n=0}^N \xi_n \pm \frac{\sqrt{ - C(0) +2 \sum_{n=0}^N C(n)}}{\sqrt{N}}\qquad\qquad (1)$
Oups sorry I just realised you change the beginning of your sum, so this is completely equivalent...

- I don't know where to put this put the lecture class for this week is not online (even though the link exist, the corresponding page does not)
A: Link should be operational now