Commit cf03eb17 by 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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