Update files at 2023-11-30 23:14:31

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@@ -7493,7 +7493,7 @@ <h1 id="Part-2:-Simulations">Part 2: Simulations<a class="anchor-link" href="#Pa
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-<p>From Chebyshev [1] we know that for a random variable with variance $\sigma^2$ $\Pr(|X-\mu |\geq k )\leq {\frac {\sigma^2}{k^{2}}}$. For an ensemble average over $N$ reservoirs, if the ensemble has variance $\sigma_0^2$ then the random variable $X_N = \frac{1}{N} X_i$ has variance $\sigma^2 = \sigma_0^2/N$ so that $\Pr(|X_N-\mu|\geq k\sigma_0 ) \leq {\frac {1}{Nk^{2}}}$. Whereas before we might expect $\Pr(|X-\mu|\geq \sigma_0) \leq 68\%$ if $X$ were Gaussian distributed, we now have, for $k \sim 1$ and $N\sim 500$, a bound of $.02\%$. That's much tighter! This depends on the actual shape of the distribution and its actual moment, but we expect such an improvement to be sufficient... and we will empirically observe that it is!</p>
+<p>From Chebyshev [1] we know that for a random variable with variance $\sigma^2$ $\Pr(|X-\mu |\geq k )\leq {\frac {\sigma^2}{k^{2}}}$. For an ensemble average over $N$ reservoirs, if the ensemble has variance $\sigma_0^2$ then the random variable $X_N = \frac{1}{N} \sum_{i=1}^N X_i$ has variance $\sigma^2 = \sigma_0^2/N$ so that $\Pr(|X_N-\mu|\geq k\sigma_0 ) \leq {\frac {1}{Nk^{2}}}$. Whereas before we might expect $\Pr(|X-\mu|\geq \sigma_0) \leq 68\%$ if $X$ were Gaussian distributed, we now have, for $k \sim 1$ and $N\sim 500$, a bound of $.02\%$. That's much tighter! This depends on the actual shape of the distribution and its actual moment, but we expect such an improvement to be sufficient... and we will empirically observe that it is!</p>
 <p>[1] <a href="https://en.wikipedia.org/wiki/Chebyshev%27s_inequality">https://en.wikipedia.org/wiki/Chebyshev%27s_inequality</a></p>
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