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While you can construct mathematical models as simple as integrating white noise which evolve trajectories that have the appearance of Brownian motion, physical Brownian motion occurs due to a more complex, stochastic dynamic equilibrium between macroscopic particles and atoms/molecules. So if you are interested in real Brownian motion you need to consider ...


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To make proper sense of this equation, of course it is helpful to put it in some physical context. I will relate some of the explanation to Brownian motion, since it is a very typical example. I think you are basically asking two questions. 1) How can something with mean 0 have non-zero correlation? Quite generally, a random noise is considered a ...


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The claim follow from central limit theorem. If you consider your energy per particle $E/N$ to be defined as $$\frac{E}{N} = \frac{1}{N} \sum_i \frac{mv_i^2}{2} $$ then using the central limit theorem (https://en.wikipedia.org/wiki/Central_limit_theorem) you will find that the width of the peak of the distribution of $P(E/N )$ distribution is then $\approx 1/...


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So you have measurements of the time it takes to fall, not the acceleration of a falling mass. (Hence the units: counts per second vs seconds). If there are $N_i$ counts in a bin, your estimate of the uncertainly is $\sqrt{N_i}$. If you scale that by a parameter $\alpha$ to get a histogram value: $$ y_i = \alpha N_i$$ and $$ \Delta y_i = \alpha \sqrt{N_i}...


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Please refer to the page number 6-6 in the chapter where deviation of the NH from its expected value (N/2) is explained. Here he put the value of 'N' as 30 and finds the typical value of k as 2.8 units from 15(expected value). From the Fig 6-2 it is seen that width of the curve (most of the outcomes falls) is measured 3 units from the center. Similarly ...


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The jack knife or bootstrap methods are good for estimating means and errors for dependent data. The basic idea is that a sampling (with replacement) of your data is to the data as the data is to the population.


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Your choice of an example considering resistance complicates matters which I will mention later. If the error is the maximum error then using the first equation you are finding an error based on the assumption that both values have a maximum error simultaneously which is not a very likely event. This means that the error found this way is an over ...


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So take a look at the definition of the Fisher Information: $$ \mathcal{I}(\theta) = E\left[ \left. \left( \frac{\partial}{\partial \theta} \log f(X;\theta) \right)^2 \right| \theta \right] = \int \left( \frac{\partial}{\partial \theta} \log f(x;\theta) \right)^2 f(x;\theta) dx $$ Nowhere in the above will you find any $p(\theta)$. That is because the ...


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