Sampling error when measuring the average diffusion coefficient

<p>Sampling errors is: </p> <blockquote> <p>the error created when only a subset of a population is sampled - [<a class="litnote tooltip" href="/literature/@catipovic2013">@catipovic2013<span class="tooltiptext">Improving the quantification of Brownian motion</span></a>]</p> </blockquote> <p>If we want to know the accuracy when determining the average diffusion coefficient, the error will be dominated by the standard deviation of the measured variances (see: <a class="wikilink" href="/calculating_diffusion_coefficient_from_jump_statistics/">calculating diffusion coefficient from jump statistics</a>):</p> <p>$$\sigma_\textrm{sampling} = \sqrt{\frac{2}{\textrm{N} -1}}$$</p> <p>And, as expected, the more particles we measure the better sampling accuracy we'll get (although see: <a class="wikilink" href="/sources_of_error_in_measuring_diffusion_coefficient_through_nanoparticle_tracking_analysis/">Sources of error in measuring diffusion coefficient through nanoparticle tracking analysis</a>)</p> <p>There's a caveat here, though. N is the number of measurements, so for 15 particles acquired along 100 frames these are 1500 data points. It is like building a distribution of <em>all</em> the jumps. </p> <p>However, if we use the mean squared displacement (see: <a class="wikilink" href="/calculating_diffusion_coefficient_from_mean_squared_displacement_data/">Calculating diffusion coefficient from mean squared displacement data</a>) we have:</p> <p>$$\sigma_\textrm{sampling}=\frac{1}{\sqrt{\textrm{N}}}$$</p> <p>But in this case $N$ is the number of particles.</p>