diff --git a/BDA 5.9.8.ipynb b/BDA 5.9.8.ipynb index 9a60fc0..d3ea631 100644 --- a/BDA 5.9.8.ipynb +++ b/BDA 5.9.8.ipynb @@ -37,7 +37,79 @@ "where \n", "$$\n", "c_{m}=\\frac{\\lambda_m p_{m}(\\{y_{i}\\})}{\\sum_{m} \\lambda_m p_{m}(\\{y_{i}\\}}\n", - "$$" + "$$\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "In the special case under consideration, $p_1$ is normal with mean $-1$ and $\\sigma=.5$, $p_2$ is normal with mean $1$ and $\\sigma=.5$ and we can set $\\lambda_1=.1$ and $\\lambda_2=.9$. The $p_m(\\{y_{i}\\})$ can be calculated from the $t$ distribution. Drawing a sample of size $10$ from $p_1$ and getting a sample mean of $-.25$ and a sample variance of $1$ gives a $t$-statistics of $\\sqrt{10}(-.25+1)$ in the first case and $\\sqrt{10}(-.25-1)$ in the second. " + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "0.041664931082753924\n", + "0.0035119750957915393\n" + ] + } + ], + "source": [ + "import numpy as np\n", + "from scipy.stats import norm, t\n", + "t_1=np.sqrt(9)*.75\n", + "t_2=np.sqrt(9)*1.25\n", + "print(t.pdf(t_1,df=9))\n", + "print(t.pdf(t_2,df=9))" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0.5655172413793104" + ] + }, + "execution_count": 18, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + ".1*.041/(.1*.041+.9*.0035)" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": {}, + "outputs": [ + { + "data": { + "text/plain": [ + "0.43448275862068964" + ] + }, + "execution_count": 17, + "metadata": {}, + "output_type": "execute_result" + } + ], + "source": [ + ".9*.0035/(.1*.041+.9*.0035)" ] }, {