still trying to figure out BDA 5.9.8

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)"
    ]
   },
   {