Fixes some typos

HW6/P1.ipynb

@@ -13,42 +13,42 @@
    "metadata": {},
    "source": [
     "The two net-gain equations are just\n",
-    "\\begin{align}\n",
+    "\\begin{align*}\n",
     "G_A &= Y_A - b_A Y_A - b_B Y_B \\\\\n",
     "G_B &= Y_B - b_A Y_A - b_B Y_B\n",
-    "\\end{align}\n",
+    "\\end{align*}\n",
     "\n",
     "From these, we would like to prove that if Jill chooses $\\left\\{b_A, b_B\\right\\}$ such that either $G_A\\ge0$ or $G_B\\ge0$ (ie. the bet is not a sure loss), then $b_A \\in [0,1]$.\n",
     "\n",
     "Stated logically, this is:\n",
-    "\\begin{equation}\n",
+    "\\begin{equation*}\n",
     "(G_A\\ge0 \\vee G_B\\ge0) \\rightarrow b_A \\in [0,1] \\quad \\quad (1)\n",
-    "\\end{equation}\n",
+    "\\end{equation*}\n",
     "\n",
     "Clearly, only the signs of $G_A$, and $G_B$ are important so let's scale them by dividing through by $Y_A$. The new $G$s are \n",
-    "\\begin{align}\n",
+    "\\begin{align*}\n",
     "G_A &= 1 - b_A - b_B \\gamma \\\\\n",
     "G_B &= \\gamma - b_A - b_B \\gamma\n",
-    "\\end{align}\n",
+    "\\end{align*}\n",
     "\n",
     "where, \n",
-    "\\begin{equation}\n",
+    "\\begin{equation*}\n",
     "\\gamma \\equiv \\frac{Y_B}{Y_A}\n",
-    "\\end{equation}\n",
+    "\\end{equation*}\n",
     "\n",
     "At this point, it is necessary to note the ranges of the involved variables. We assume that all $Y$ and $b$ are non-negative, which implies that $\\gamma$ is also non-negative.\n",
     "\n",
     "According to (1), it is sufficient to show individually that $G_A\\ge0 \\rightarrow b_A \\in [0,1])$, and $G_B\\ge0 \\rightarrow b_A \\in [0,1])$. Since if both of those conditions hold, then (1) holds. Let's do the $G_A$ case first.\n",
     "\n",
     "If $G_A\\ge 0$, then\n",
-    "\\begin{equation}\n",
+    "\\begin{equation*}\n",
     "1 - b_A - b_B\\gamma \\ge 0\n",
-    "\\end{equation}\n",
+    "\\end{equation*}\n",
     "\n",
     "Or, written slightly differently,\n",
-    "\\begin{equation}\n",
+    "\\begin{equation*}\n",
     "1 - b_A \\ge b_B\\gamma\n",
-    "\\end{equation}\n",
+    "\\end{equation*}\n",
     "\n",
     "The right hand side is strictly non-negative, which implies that $(1-b_A) \\ge 0$. This coupled with the assumption that $b_A$ itself must be non-negative implies that $b_A \\in [0,1]$. \n",
     "\n",
@@ -63,58 +63,58 @@
     "\n",
     "The net-gain equations are\n",
     "\n",
-    "\\begin{align}\n",
+    "\\begin{align*}\n",
     "G_1 &= (1-b_a)Y_A - b_B Y_B + (1-b_C)Y_C\\\\\n",
     "G_2 &= -b_A Y_A + (1-b_B) Y_B + (1-b_C)Y_C\\\\\n",
     "G_3 &=  -b_A Y_A - b_B Y_B - b_B Y_C\n",
-    "\\end{align}\n",
+    "\\end{align*}\n",
     "\n",
     "Let's first rewrite the above equations with the assumption that $b_C=b_A+b_B$ and gathering common terms in the $b$'s.\n",
     "\n",
-    "\\begin{align}\n",
+    "\\begin{align*}\n",
     "G_1 &= (Y_A + Y_C) - (Y_A + Y_C)b_A - (Y_B + Y_C)b_B \\\\\n",
     "G_2 &= (Y_B + Y_C) - (Y_A + Y_C)b_A - (Y_B + Y_C)b_B \\\\\n",
     "G_3 &= - (Y_A + Y_C)b_A - (Y_B + Y_C)b_B\n",
-    "\\end{align}\n",
+    "\\end{align*}\n",
     "\n",
     "Noticing that $(Y_A + Y_C)$ and $(Y_B + Y_C)$ are common in the above equations, lets replace them with new variables, $\\alpha$ and $\\beta$, respectively.\n",
     "\n",
-    "\\begin{align}\n",
+    "\\begin{align*}\n",
     "G_1 &= \\alpha - \\alpha b_A - \\beta b_B \\\\\n",
     "G_2 &= \\beta - \\alpha b_A - \\beta b_B \\\\\n",
     "G_3 &= -\\alpha b_A - \\beta b_B\n",
-    "\\end{align}\n",
+    "\\end{align*}\n",
     "\n",
     "For one last simplification, we note that we only care about the sign of the G's, so we are free to multiply/divide by any positive factor. So divide $G_1$ and $G_3$ by $\\alpha$ and $G_2$ by $\\beta$.\n",
     "\n",
-    "\\begin{align}\n",
+    "\\begin{align*}\n",
     "G_1 &= 1 - b_A - \\gamma^{-1} b_B \\\\\n",
     "G_2 &= 1 - \\gamma b_A - b_B \\\\\n",
     "G_3 &= -b_A - \\gamma^{-1} b_B\n",
-    "\\end{align}\n",
+    "\\end{align*}\n",
     "\n",
     "Where $\\gamma \\equiv \\frac{\\alpha}{\\beta}$. What remains is to show that no value of $\\gamma$ can make all three $G$'s simultaneously negative. $G_3$ is always negative so at least one of $G_1$ and $G_2$ must be non-negative. Let's first consider what values of $\\gamma$ make $G_1$ negative.\n",
     "\n",
-    "\\begin{align}\n",
+    "\\begin{align*}\n",
     "1 - b_A - \\gamma^{-1} b_B &< 0\\\\\n",
     "(1 - b_A)\\gamma - b_B &< 0 \\\\\n",
     "\\gamma &< \\frac{b_B}{1 - b_A} \\quad\\quad (2) \\\\\n",
-    "\\end{align}\n",
+    "\\end{align*}\n",
     "\n",
-    "Now, we do the same thing for G_2,\n",
-    "\\begin{align}\n",
+    "Now, we do the same thing for $G_2$,\n",
+    "\\begin{align*}\n",
     "1 - \\gamma b_A - b_B &< 0 \\\\\n",
     "1 - b_B &< \\gamma b_A \\\\\n",
     "\\gamma &> \\frac{1-b_B}{b_A} \\quad\\quad (3) \\\\\n",
-    "\\end{align}\n",
+    "\\end{align*}\n",
     "\n",
     "So, if (2) and (3) describe disjoint regions in $\\gamma$, then $G_1$ and $G_2$ cannot be simultaneously negative. Notice that (2) forms an upper bound on $\\gamma$ and (3) a lower bound. If the upper bound is smaller than the lower bound, then clearly no value can satisfy both. So let's see if this is the case.\n",
     "\n",
-    "\\begin{align}\n",
+    "\\begin{align*}\n",
     "&\\frac{b_B}{1 - b_A} - \\frac{1-b_B}{b_A} \\\\\n",
     "&= \\frac{b_B b_A}{b_A(1 - b_A)} - \\frac{(1-b_A)(1-b_B)}{b_A(1-b_A)} \\\\\n",
     "&= \\frac{(b_A + b_B) - 1}{b_A(1 - b_A)} < 0, \\quad \\mathrm{if }\\quad b_A + b_B < 1\\\\\n",
-    "\\end{align}\n",
+    "\\end{align*}\n",
     "\n",
     "So, finally, if we assume that $b_A + b_B < 1$, which seems reasonable, the set of $\\gamma$ that make $G_1$ and $G_2$ simultaneously negative is empty. Therefore, it is impossible for Jack to choose the Y's such that Jill has a sure loss."
    ]