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    "# Setting the beta hyper-parameters\n",
    "\n",
    "Suppose $\\theta \\sim \\beta(\\alpha_1,\\alpha_2)$ and we believe that $E[\\theta] = m$ and $P(l < \\theta < u) = 0.95$. Write a program that can solve for $\\alpha_1$ and $\\alpha_2$ in terms of $m$, $l$, and $u$.\n",
    "\n",
    "The mean for the beta distribution is well-known, so we have that\n",
    "\n",
    "\\begin{equation}\n",
    "E[\\theta] = m = \\frac{\\alpha_1}{\\alpha_1 + \\alpha_2}.\n",
    "\\end{equation}\n",
    "\n",
    "Thus, this implies that \n",
    "\n",
    "\\begin{equation}\n",
    "\\alpha_2 = \\alpha_1\\frac{1-m}{m}.\n",
    "\\end{equation}\n",
    "\n",
    "Now, as $\\alpha_1$ increases the strength of our prior increases, so $P(l < \\theta < u)$ is a monotonic function of $\\alpha_1$. And so, we can solve for $\\alpha_1$, and thereby, $\\alpha_2$ with a binary search."
   ]
  },
  {
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   "execution_count": 1,
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     "text": [
      "4.506062413888118\n",
      "25.534353681276208\n"
     ]
    }
   ],
   "source": [
    "from scipy import stats\n",
    "\n",
    "m = 0.15\n",
    "l = 0.05\n",
    "u = 0.3\n",
    "\n",
    "alpha_1_lower = 0.01\n",
    "alpha_1_upper = 1000\n",
    "alpha_1 = (alpha_1_lower + alpha_1_upper)/2\n",
    "tol = 1e-15\n",
    "while alpha_1_upper - alpha_1_lower >= 1e-9:\n",
    "    alpha_2 = alpha_1*(1-m)/m\n",
    "    density = stats.beta.cdf(u, a = alpha_1, b = alpha_2) - stats.beta.cdf(l, a = alpha_1, b = alpha_2)\n",
    "    if density > 0.95:\n",
    "        alpha_1_upper = alpha_1\n",
    "    else:\n",
    "        alpha_1_lower = alpha_1\n",
    "    alpha_1 = (alpha_1_lower + alpha_1_upper)/2\n",
    "\n",
    "print(alpha_1)\n",
    "print(alpha_2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The problem asks for $\\alpha_1$ and $\\alpha_2$ if $m = 0.15$, $l = 0.05$, and $u = 0.3$. We find that\n",
    "\n",
    "\\begin{align}\n",
    "\\alpha_1 &\\approx 4.506062413888118 \\\\\n",
    "\\alpha_2 &\\approx 25.534353681276208.\n",
    "\\end{align}\n"
   ]
  }
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