Commit cf03eb17 authored by Erik Strand's avatar Erik Strand

Editing for clarity

parent 755f048b
......@@ -35,9 +35,10 @@ continuity of entropy with respect to the topologies of $$\mathbb{R}_{\geq 0}^n$
$$\mathbb{R}$$.
First let's show that $$x \log x$$ is continuous. I take as given that $$\log(x)$$ is a continuous
function on its domain. Then $$x \log(x)$$ is also continuous, since finite products of continuous
functions are continuous. This suffices for $$x > 0$$. At zero, $$x \log x$$ is continuous because
we have defined it to be equal to the limit we found above.
function on its domain (after all it's the inverse of $$e^x$$, which is strictly monotonic and
$$C^\infty$$). Then $$x \log(x)$$ is also continuous, since finite products of continuous functions
are continuous. This suffices for $$x > 0$$. At zero, $$x \log x$$ is continuous because we have
defined it to be equal to the limit we found above.
Thus each term of the entropy function is a continuous function from $$\mathbb{R}_{\geq 0}$$ to
$$\mathbb{R}$$. But we can also view each term as a function from $$\mathbb{R}_{\geq 0}^n$$ to
......@@ -431,9 +432,9 @@ $$
\begin{align*}
\langle \hat{x_0} \rangle
&= \left \langle \frac{1}{n} \sum_{i = 1}^n x_i \right \rangle \\
&= \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty
\left( \prod_{i = 1}^n \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x_i - x_0)^2}{2 \sigma^2}}
\right) \left( \frac{1}{n} \sum_{i = 1}^n x_i \right) \mathrm{d} x_1 \ldots \mathrm{d} x_n \\
&= \int_{\mathbb{R}^n} \left( \prod_{i = 1}^n \frac{1}{\sqrt{2 \pi \sigma^2}}
e^{-\frac{(x_i - x_0)^2}{2 \sigma^2}} \right)
\left( \frac{1}{n} \sum_{i = 1}^n x_i \right) \mathrm{d} x_1 \ldots \mathrm{d} x_n \\
&= \frac{1}{n} \sum_{i = 1}^n \prod_{j = 1}^n
\int_{-\infty}^\infty
\frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x_j - x_0)^2}{2 \sigma^2}}
......@@ -488,15 +489,15 @@ $$
\begin{align*}
\left \langle \hat{x_0}^2 \right \rangle
&= \left \langle \left( \frac{1}{n} \sum_{i = 1}^n x_i \right)^2 \right \rangle \\
&= \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty
\left( \prod_{i = 1}^n \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x_i - x_0)^2}{2 \sigma^2}}
\right) \left( \frac{1}{n} \sum_{i = 1}^n x_i \right)^2 \mathrm{d} x_1 \ldots \mathrm{d} x_n \\
&= \frac{1}{n^2} \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty
\left( \prod_{i = 1}^n \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x_i - x_0)^2}{2 \sigma^2}}
\right) \left( \sum_{i, j} x_i x_j \right) \mathrm{d} x_1 \ldots \mathrm{d} x_n \\
&= \frac{1}{n^2} \sum_{i, j} \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty
\prod_{k = 1}^n \frac{1}{\sqrt{2 \pi \sigma^2}}
e^{-\frac{(x_k - x_0)^2}{2 \sigma^2}} x_i x_j \mathrm{d} x_1 \ldots \mathrm{d} x_n \\
&= \int_{\mathbb{R}^n} \left( \prod_{i = 1}^n \frac{1}{\sqrt{2 \pi \sigma^2}}
e^{-\frac{(x_i - x_0)^2}{2 \sigma^2}} \right)
\left( \frac{1}{n} \sum_{i = 1}^n x_i \right)^2 \mathrm{d} x_1 \ldots \mathrm{d} x_n \\
&= \frac{1}{n^2} \int_{\mathbb{R}^n} \left(
\prod_{i = 1}^n \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x_i - x_0)^2}{2 \sigma^2}} \right)
\left( \sum_{i, j} x_i x_j \right) \mathrm{d} x_1 \ldots \mathrm{d} x_n \\
&= \frac{1}{n^2} \sum_{i, j} \int_{\mathbb{R}^n} \left( \prod_{k = 1}^n
\frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{(x_k - x_0)^2}{2 \sigma^2}} \right) x_i x_j
\mathrm{d} x_1 \ldots \mathrm{d} x_n \\
&= \frac{1}{n^2} \left(
\sum_{i=j} \int_{-\infty}^\infty \frac{1}{\sqrt{2 \pi \sigma^2}}
e^{-\frac{(x_i - x_0)^2}{2 \sigma^2}} x_i^2 \mathrm{d} x_i \right. \\
......
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