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Correct sign in product of Gaussian derivation

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"&\\propto \\exp \\Big[-\\frac{1}{2\\sigma_z^2\\bar\\sigma^2}[\\bar\\sigma^2(z-x)^2-\\sigma_z^2(x-\\bar\\mu)^2]\\Big]\n",
changed to
"&\\propto \\exp \\Big[-\\frac{1}{2\\sigma_z^2\\bar\\sigma^2}[\\bar\\sigma^2(z-x)^2+\\sigma_z^2(x-\\bar\\mu)^2]\\Big]\n",
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Forrest Edwards 2019-09-14 18:34:42 -05:00 committed by GitHub
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03-Gaussians.ipynb
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@ -1976,7 +1976,7 @@
"$$\\begin{aligned}\n",
"p(x \\mid z) &\\propto \\exp \\Big[-\\frac{(z-x)^2}{2\\sigma_z^2}\\Big]\\exp \\Big[-\\frac{(x-\\bar\\mu)^2}{2\\bar\\sigma^2}\\Big]\\\\\n",
"&\\propto \\exp \\Big[-\\frac{(z-x)^2}{2\\sigma_z^2}-\\frac{(x-\\bar\\mu)^2}{2\\bar\\sigma^2}\\Big] \\\\\n",
"&\\propto \\exp \\Big[-\\frac{1}{2\\sigma_z^2\\bar\\sigma^2}[\\bar\\sigma^2(z-x)^2-\\sigma_z^2(x-\\bar\\mu)^2]\\Big]\n",
"&\\propto \\exp \\Big[-\\frac{1}{2\\sigma_z^2\\bar\\sigma^2}[\\bar\\sigma^2(z-x)^2+\\sigma_z^2(x-\\bar\\mu)^2]\\Big]\n",
"\\end{aligned}$$\n",
"\n",
"Now we multiply out the squared terms and group in terms of the posterior $x$.\n",