MAP

MAP_Estimation.ipynb

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    "### MAP Estimation and Conjugate prior distribution\n",
    "- https://qiita.com/jyori112/items/80b21422b15753a1c5a4\n",
    "- https://mathwords.net/kyoyakujizen\n",
    "- https://to-kei.net/bayes/conjugate-prior-distribution/\n",
    "- https://to-kei.net/bayes/beta-bernoulli-distribution/\n",
    "\n",
    "MAP(Maximum a Posteriori)推定とは何かを考える前に、まず最尤推定から考える。  \n",
    "最尤推定は、[ここ](https://gist.github.com/shotahorii/7355db2bc49aa746abe85dc24fa70838)でも書いたように、あるデータが観測された時、そのデータから鑑みて最も尤もらしいパラメータを推定する方法である。"
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      "text/latex": [
       "例えば、ガチャを繰り返して20回連続で当たりを引いたとする。このガチャの真の当たり率が0.1であるというのはどれほど尤もらしいか？\n",
       "これは以下のように式で表される。<br><br>\n",
       "$$P(20 win|\\theta = 0.1) = 0.1^{20} = 0.000000000000000000001$$<br>\n",
       "\n",
       "ほぼあり得そうもない。では、真の当たり率が0.9であるというのはどれほど尤もらしいか。<br><br>\n",
       "\n",
       "$$P(20 win|\\theta = 0.9) = 0.9^{20} = 0.12157665459056935$$<br>\n",
       "これはあり得るかもしれない。<br><br>\n",
       "\n",
       "と、このように、観測されたデータから、各パラメータの尤もらしさを計算することができる。<br>\n",
       "ここで、この尤もらしさが最も大きいパラメータを選ぼう、というのが最尤推定である。式にすると以下。<br><br>\n",
       "$$argmax_\\theta P(D|\\theta)$$<br><br>\n",
       "\n",
       "上で出たガチャのような二項分布を例に最尤法によるパラメータ推定を一般化して考える。例えば、N回ガチャを行ってそのうちn回当たりを引いたとする。\n",
       "このガチャの真の当たり率を最尤推定してみる。真の当たり率をθとするとN回試行してn回当たりが出る確率は以下である。<br><br>\n",
       "\n",
       "$$P(n|N,\\theta) = \\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right) \\theta^n (1-\\theta)^{N-n}$$<br><br>\n",
       "\n",
       "これを最大化するθを見つけ出したいので、この式を微分する。<br><br>\n",
       "$$\\frac{d}{d\\theta}P(n|N,\\theta) = \\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right)(n\\theta^{n-1}(1-\\theta)^{N-n}-(N-n)\\theta^n(1-\\theta)^{N-n-1})$$<br>\n",
       "$$=\\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right)\\theta^{n-1}(1-\\theta)^{N-n-1}(n(1-\\theta)-(N-n)\\theta)$$<br><br>\n",
       "\n",
       "これを0とするθを求めたいが、ここで、\n",
       "$$\\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right)$$\n",
       "や\n",
       "$$\\theta^{n-1}(1-\\theta)^{N-n-1}$$\n",
       "は0にならないので、以下の式を解けば良い。<br><br>\n",
       "$$n(1-\\theta)-(N-n)\\theta = 0$$<br><br>\n",
       "つまり、$$\\theta = \\frac{n}{N}$$"
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    "%%latex\n",
    "例えば、ガチャを繰り返して20回連続で当たりを引いたとする。このガチャの真の当たり率が0.1であるというのはどれほど尤もらしいか？\n",
    "これは以下のように式で表される。<br><br>\n",
    "$$P(20 win|\\theta = 0.1) = 0.1^{20} = 0.000000000000000000001$$<br>\n",
    "\n",
    "ほぼあり得そうもない。では、真の当たり率が0.9であるというのはどれほど尤もらしいか。<br><br>\n",
    "\n",
    "$$P(20 win|\\theta = 0.9) = 0.9^{20} = 0.12157665459056935$$<br>\n",
    "これはあり得るかもしれない。<br><br>\n",
    "\n",
    "と、このように、観測されたデータから、各パラメータの尤もらしさを計算することができる。<br>\n",
    "ここで、この尤もらしさが最も大きいパラメータを選ぼう、というのが最尤推定である。式にすると以下。<br><br>\n",
    "$$argmax_\\theta P(D|\\theta)$$<br><br>\n",
    "\n",
    "上で出たガチャのような二項分布を例に最尤法によるパラメータ推定を一般化して考える。例えば、N回ガチャを行ってそのうちn回当たりを引いたとする。\n",
    "このガチャの真の当たり率を最尤推定してみる。真の当たり率をθとするとN回試行してn回当たりが出る確率は以下である。<br><br>\n",
    "\n",
    "$$P(n|N,\\theta) = \\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right) \\theta^n (1-\\theta)^{N-n}$$<br><br>\n",
    "\n",
    "これを最大化するθを見つけ出したいので、この式を微分する。<br><br>\n",
    "$$\\frac{d}{d\\theta}P(n|N,\\theta) = \\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right)(n\\theta^{n-1}(1-\\theta)^{N-n}-(N-n)\\theta^n(1-\\theta)^{N-n-1})$$<br>\n",
    "$$=\\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right)\\theta^{n-1}(1-\\theta)^{N-n-1}(n(1-\\theta)-(N-n)\\theta)$$<br><br>\n",
    "\n",
    "これを0とするθを求めたいが、ここで、\n",
    "$$\\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right)$$\n",
    "や\n",
    "$$\\theta^{n-1}(1-\\theta)^{N-n-1}$$\n",
    "は0にならないので、以下の式を解けば良い。<br><br>\n",
    "$$n(1-\\theta)-(N-n)\\theta = 0$$<br><br>\n",
    "つまり、$$\\theta = \\frac{n}{N}$$"
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    "ここまでが最尤推定の説明。では、MAP推定とは何か。  \n",
    "最尤推定は尤度、つまりあるデータが観測された時のパラメータθの尤もらしさ(=あるθにおいてそのデータが観測される確率)を最大化するθを選ぶ、という推定方法であった。  \n",
    "  \n",
    "一方で、MAP推定で最大化するのは尤度ではなく、事後確率である。これはどういうことか。  \n",
    "最尤推定では、θそのものの取り得る値の分布を考慮しなかった。どういうことかというと、例えば上のガチャ当たり率の最尤推定では、真の当たり率が0.1と0.9とどっちが可能性として高いのか、ということは考えずに、純粋に観測されたデータから最も尤度の高い真の当たり率θを選んだ。  \n",
    "しかし、現実には、θの分布が(はっきりではないにしろ)少し予測できていて、恣意的な重み付けを行いたいことがある。  \n",
    "  \n",
    "データの観測数が少ない場合などが顕著。例えば、カジノに行ったとする。コイントスを5回行って4回表が出た。これは、実際にコインが真正な0.5のコインであっても15%程度の確率で起こり得る。が、最尤法でこのコインのパラメータを推定すると0.8となる。  \n",
    "カジノで毎日全てのコインが事前にチェックされている、という予備知識を鑑みると、（観測されたデータに関わらず）このコインの真のパラメータが本当に0.8である可能性は、0.5である可能性よりも低そうに思われる。  \n",
    "  \n",
    "ここで、この「このコインの真のパラメータが本当に0.8である可能性は、0.5である可能性よりも低そうだ」というものを事前分布と呼ぶ。つまりデータが観測される前の、パラメータの分布。MAP推定では、「データが観測されることによってこの事前分布が更新されて、事後確率になる」と捉え、この事後確率を最大化するパラメータを選ぶ。"
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      "text/latex": [
       "式で考える。データDが観測された条件下でのパラメータθの確率は$$P(\\theta|D)$$であるが、ベイズの定理により以下のように書き換えられる。<br><br>\n",
       "$$P(\\theta|D) = \\frac{P(D|\\theta)P(\\theta)}{P(D)}$$<br><br>\n",
       "ここで、目的はθを変数としたP(θ|D)の最大化である。右辺分母のP(D)はθによらない値なので、この最大化問題は以下のように考えることができる。<br><br>\n",
       "$$P(\\theta|D) \\propto P(D|\\theta)P(\\theta)$$ であり、$$argmax_\\theta P(\\theta|D) = argmax_\\theta P(D|\\theta)P(\\theta)$$\n",
       "<br><br>\n",
       "P(D|θ)は尤度であり、P(θ)はパラメータθの事前分布である。つまりMAP推定は、尤度と事前分布の掛け合わせを最大化するパラメータを選ぶ。"
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   "source": [
    "%%latex\n",
    "式で考える。データDが観測された条件下でのパラメータθの確率は$$P(\\theta|D)$$であるが、ベイズの定理により以下のように書き換えられる。<br><br>\n",
    "$$P(\\theta|D) = \\frac{P(D|\\theta)P(\\theta)}{P(D)}$$<br><br>\n",
    "ここで、目的はθを変数としたP(θ|D)の最大化である。右辺分母のP(D)はθによらない値なので、この最大化問題は以下のように考えることができる。<br><br>\n",
    "$$P(\\theta|D) \\propto P(D|\\theta)P(\\theta)$$ であり、$$argmax_\\theta P(\\theta|D) = argmax_\\theta P(D|\\theta)P(\\theta)$$\n",
    "<br><br>\n",
    "P(D|θ)は尤度であり、P(θ)はパラメータθの事前分布である。つまりMAP推定は、尤度と事前分布の掛け合わせを最大化するパラメータを選ぶ。"
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    "では、先ほどのカジノのコインのパラメータをMAP推定してみる。  \n",
    "ここで、このパラメータの事前分布はどのようにして決めればよいのか。なんとなく、0.5をピークにした釣鐘型っぽい気はするが。。。 \n",
    "  \n",
    "ここで登場するのが、共役事前分布(Conjugate prior distribution)という概念である。  \n",
    "これは、データの母集団の確率分布によって決まる分布で、「尤度をかけて事後分布を求めると、事前分布と事後分布の分布形が同じになる」\n",
    "ような事前分布のことを指す。\n",
    "  \n",
    "事前分布と事後分布が同じ分布形になって何が嬉しいかというと、計算が容易になること。特に、データを得ることで何度も確率分布を更新していく\n",
    "ような場合、事後分布が次の更新の際の事前分布になるため、同じ分布形であるとこの更新が容易になる。"
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      "text/latex": [
       "コインの例、つまりデータの母集団の分布が二項分布である場合、共役事前分布はベータ分布である。\n",
       "つまり、θの事前分布はベータ分布で表される。<br>\n",
       "ベータ分布は、α,βをハイパーパラメータとした分布であり、以下のように表される。<br><br>\n",
       "$$P(\\theta)=\\frac{\\Gamma(\\alpha+\\beta)}{\\Gamma(\\alpha)\\Gamma(\\beta)}\\theta^{\\alpha-1}(1-\\theta)^{\\beta-1} $$<br><br>\n",
       "\n",
       "ここで、θの取り得る範囲は0<=θ<=1であり、α,βは共に正の実数である。<br>\n",
       "このベータ分布が二項分布の共役事前分布であることを数式で見てみよう。つまり、事前分布であるベータ分布と、尤度である二項分布を掛け合わせた\n",
       "事後分布がベータ分布となることを確かめる。<br><br>\n",
       "$$P(\\theta|D) \\propto P(D|\\theta)P(\\theta)$$<br>\n",
       "$$P(\\theta|D) \\propto \\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right) \\theta^n (1-\\theta)^{N-n} \n",
       "\\frac{\\Gamma(\\alpha+\\beta)}{\\Gamma(\\alpha)\\Gamma(\\beta)}\\theta^{\\alpha-1}(1-\\theta)^{\\beta-1}$$<br>\n",
       "ここで、$$\\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right)$$と\n",
       "$$\\frac{\\Gamma(\\alpha+\\beta)}{\\Gamma(\\alpha)\\Gamma(\\beta)}$$は$$\\theta$$に対して定数とみなせる。<br>\n",
       "$$P(\\theta|D) \\propto \\theta^n (1-\\theta)^{N-n}\\theta^{\\alpha-1}(1-\\theta)^{\\beta-1}$$<br>\n",
       "$$P(\\theta|D) \\propto \\theta^{\\alpha+n-1}(1-\\theta)^{\\beta+N-n-1}$$<br>\n",
       "$$\\alpha'=\\alpha+n$$とおき、$$\\beta'=\\beta+N-n$$とおくと、$$P(\\theta|D) \\propto \\theta^{\\alpha'-1}(1-\\theta)^{\\beta'-1}$$<br>\n",
       "となり、事後分布がベータ分布に従うことがわかる。"
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   "source": [
    "%%latex\n",
    "コインの例、つまりデータの母集団の分布が二項分布である場合、共役事前分布はベータ分布である。\n",
    "つまり、θの事前分布はベータ分布で表される。<br>\n",
    "ベータ分布は、α,βをハイパーパラメータとした分布であり、以下のように表される。<br><br>\n",
    "$$P(\\theta)=\\frac{\\Gamma(\\alpha+\\beta)}{\\Gamma(\\alpha)\\Gamma(\\beta)}\\theta^{\\alpha-1}(1-\\theta)^{\\beta-1} $$<br><br>\n",
    "\n",
    "ここで、θの取り得る範囲は0<=θ<=1であり、α,βは共に正の実数である。<br>\n",
    "このベータ分布が二項分布の共役事前分布であることを数式で見てみよう。つまり、事前分布であるベータ分布と、尤度である二項分布を掛け合わせた\n",
    "事後分布がベータ分布となることを確かめる。<br><br>\n",
    "$$P(\\theta|D) \\propto P(D|\\theta)P(\\theta)$$<br>\n",
    "$$P(\\theta|D) \\propto \\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right) \\theta^n (1-\\theta)^{N-n} \n",
    "\\frac{\\Gamma(\\alpha+\\beta)}{\\Gamma(\\alpha)\\Gamma(\\beta)}\\theta^{\\alpha-1}(1-\\theta)^{\\beta-1}$$<br>\n",
    "ここで、$$\\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right)$$と\n",
    "$$\\frac{\\Gamma(\\alpha+\\beta)}{\\Gamma(\\alpha)\\Gamma(\\beta)}$$は$$\\theta$$に対して定数とみなせる。<br>\n",
    "$$P(\\theta|D) \\propto \\theta^n (1-\\theta)^{N-n}\\theta^{\\alpha-1}(1-\\theta)^{\\beta-1}$$<br>\n",
    "$$P(\\theta|D) \\propto \\theta^{\\alpha+n-1}(1-\\theta)^{\\beta+N-n-1}$$<br>\n",
    "$$\\alpha'=\\alpha+n$$とおき、$$\\beta'=\\beta+N-n$$とおくと、$$P(\\theta|D) \\propto \\theta^{\\alpha'-1}(1-\\theta)^{\\beta'-1}$$<br>\n",
    "となり、事後分布がベータ分布に従うことがわかる。"
   ]
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   "source": [
    "ということで、ベータ分布をθの事前分布としてMAP推定を行っていくが、その前にベータ分布がどんな形なのか見てみる。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 98,
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   "source": [
    "import numpy as np\n",
    "from scipy import stats\n",
    "import matplotlib.pyplot as plt\n",
    "%matplotlib inline\n",
    "\n",
    "def drawBeta(a,b):\n",
    "    x = np.linspace(stats.beta.ppf(0.001, a, b), stats.beta.ppf(0.999, a, b), 100)\n",
    "    fig, ax = plt.subplots(1,1)\n",
    "    chart = ax.plot(x, stats.beta.pdf(x, a, b), lw=3, alpha=0.6)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 99,
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9B/Bt4PLFrnsRnxfL6S+trui2f2ux617EsfgL4POvjwPwU2DZYtd+hMbjQ8CFwK7DtM85\nO8c58x/mYrH1wO0AVfUQsDzJyWOscVwGjkVVPVhVP+s2H+TXr61YKoZ5XgB8GrgL+O9xFjdmw4zF\nlcA3q2ofQFW9OOYax2WYsSj6f0FI9+9Pq+q1MdY4NlX1APDSLF3mnJ3jDP9hLhab3mffDH2WgmHG\nYqo/Br5zRCtaPAPHIskpwO9X1d+xtK8VGeZ58V7gnUm+m+QHST4xturGa5ixuBU4N8kL9C80vWFM\ntR2N5pydA//aR4srye8C19B/29eqLwNT13yX8i+AQZYBFwEfBd4KfD/J96vq6cUta1Gsof+dYh9N\ncibwb0nOr6pfLHZhbwbjDP999L/24XUru33T+5w6oM9SMMxYkOR8YDOwtqpme8v3ZjbMWLwfuDNJ\n6K/tXprkQFUttW+JHWYsngderKpfAr9M8j3gAvrr40vJMGNxDfB5gKr6UZIfA2cDj4ylwqPLnLNz\nnMs+PwDek+T0JMcDHwemv3jvAa6CN64sfrmq9o+xxnEZOBZJTgO+CXyiqn60CDWOy8CxqKrf7m7v\npr/u/ydLMPhhuNfIPwEfSnJskrfQ/3BvN0vPMGPxLPB7AN369nuBZ8Za5XjN9hU5c87Osc38a4iL\nxarq3iTrkjwNvEL/N/uSM8xYAH8JvBP4227Ge6CqVi1e1UfGkGPx/w4Ze5FjMuRr5Ikk/wrsAg4C\nm6tqyX1d+pDPi78G/mHKnz/+eVX9zyKVfEQluQPoAScleQ7YCBzPArLTi7wkqUH+N46S1CDDX5Ia\nZPhLUoMMf0lqkOEvSQ0y/CWpQYa/JDXI8JekBv0fzHijJLG1JOMAAAAASUVORK5CYII=\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x10ec0eeb8>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# alpha = beta = 1\n",
    "drawBeta(1,1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 100,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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nkFpgc2viuU+ZWQ/gaufcbwHLXniZ2b7db9HlY4EJE/IdgYiUmosu8utwAPbs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\npr1GOyCkcggAAAAASUVORK5CYII=\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x10ebce4a8>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# alpha = beta = 2\n",
    "drawBeta(2,2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 101,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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yZVT/5jfhmGPii0ckWyor4ZproEOHUN+6Fe67L96YykEmw3hbgAEp9X6J5z7j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\nAAAAAElFTkSuQmCC\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x10f32e978>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# alpha = beta = 10\n",
    "drawBeta(10,10)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 104,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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JzZvDTTfBsGHBQ9nKSpgwwa+UTe11JXsXWpLfurXqJiEdO4YViYhEkRmUlfm7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\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x10f60f4a8>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# alpha = 2 , beta = 4\n",
    "drawBeta(2,4)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 105,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "image/png": 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04p44pGQuUkYyZ3uqVZ5fapmLSEGsXu0PUKs8Cu1b5oVed1DJXKQMOAd//WtQv+gitcrz\nrbraHwAHD/qlcAtJyVykDKxcmb0GyzXXxBtPqYqzq0XJXKTEOQezZgX1sWM12zMqcY41VzIXKXHL\nl8P77/tyZSVMnhxrOCUtzrHmSuYiJax9q/zyy7VeeZSUzEUkEn/7W/benlqDJVqZ3Vc7dxb2vZXM\nRUpUa2t2q3z8+CNvNCy569s3KCuZi0hevPYabN3qyz17am/PQshcObG+vrBjzZXMRUpQc3P2bM8J\nE6BXr/jiKRdVVcFY85aWwo41D53MzazCzBaZ2ayjny0icXr5Zdi1y5dranwXixRGZldL+ntQCJ1p\nmd8NrIwqEBHJjwMH4JlngvqkSb7FKIURV795qGRuZkOAycBvog1HRHI1e3bw7/1xx/nhiFI4mf3m\nSWyZ/1/gu0CBl44Rkc5oaPDJPO3aa/1EISmcuLpZjvptNrNrgK3OuSVmlgLscOfOmDHj43IqlSKV\nSuUeoYiE9tRT0NTky4MHwwUXxBtPOcpsmXfUzVJXV0ddXV3e39fcUcbOmNm/AbcCLUBPoAZ43Dn3\npXbnuaO9lohEZ9s2mD7djy8HuOsuOPfceGMqR6tXw8yZvnzqqfC97x35fDPDOXfYRnJYR+1mcc79\nb+fcSc65U4CbgTntE7mIxO+vfw0S+emnwznnxBtPuUr0DVARSbb33oOFC4P6Zz8LlnNbT7oiM5nv\n3h38gY1ap5K5c+4l59y0qIIRkc5zDh59NKiPGgWnnBJfPOWue3c/th98It+9uzDvq5a5SJFbsiR7\n44nrros3HolnRIuSuUgRa2mBxx8P6qmUtoNLgjj6zZXMRYrYyy/7USzg1wTRdnDJEMfEISVzkSK1\nbx88+WRQnzxZi2klhbpZRCS0J5/0CR383pPjxsUbjwSONnEoCkrmIkVoyxbInER4/fWatp8kapmL\nSCiPPRaMXz7jDDj//HjjkWy6ASoiR7VihT/ATwy68UZNEEqa2trge7Jnjx91FDUlc5Ei0tKSPUHo\nkktg6ND44pGOdesGffr4snN+C7moKZmLFJG5c+HDD325Rw+/xK0kU6H7zZXMRYrE7t3Z+3pOmQK9\ne8cXjxxZofvNlcxFisTjjwdrlZ94IlxxRbzxyJEVeuKQkrlIEXjnHXj99aB+002+X1aSS90sIpKl\ntRUeeSSojxoFZ50VXzwSTqEnDimZiyTc3LmwYYMvd+8On/tcvPFIOLW1QbkQy+AqmYskWH2930Eo\n7Zprslt8klzpoYmgZC5S9v77v4ObngMHwlVXxRuPhJc50qihwY83j5KSuUhCvfUW/O1vQf0LX9D6\nK8Wke3e/LDH4+x5790b7fkrmIgnU3AwPPxzUx4zxa7BIcSlkV0uoZG5mVWa2wMwWm9lyM5sebVgi\n5e2JJ+Cjj3y5uhpuuCHeeKRrCpnMQ/3T5pxrMrNxzrlGM+sGzDezZ5xzb0Qbnkj52bABZs8O6p/9\nbLBBsBSXxLXMAZxzjW3FKvwfgYi780XKT2sr/OEP2cvbjh0bb0zSdYlM5mZWYWaLgQ+B2c65N6ML\nS6Q8zZkDH3zgy5WVcOutWt62mCWumwXAOdcKnG9mvYH/Z2YjnHMrM8+ZMWPGx+VUKkUqlcpTmCKl\nb/v27DHlU6bAgAHxxSO56yiZ19XVUZe5TVSemOvC4Ecz+yGwzzk3M+Mx15XXEhE/BnnmTFizxtcH\nD4Z/+ietv1Ls1qyBn/zEl089Fb73vU+eY2Y453L+/yvsaJb+ZtanrdwTuApYleubi4j30ktBIq+o\ngNtuUyIvBZkTh5LSzXIi8Hszq8D/AfiTc+7p6MISKR/bt/vlbdOuvhpOPjm+eCR/MrtZ0rNAo7oH\nEnZo4nJgVDQhiJQv5+CBB7LXKZ8yJd6YJH969PAzQZub4eBB/33u0SOa99IMUJEYzZ0Lq1f7shl8\n+cuasl9KzAo3okXJXCQmW7Z8sntl2LDYwpGIKJmLlLBDh+D++/2/3wBDhsDUqfHGJNFQMhcpYU89\nlT056CtfUfdKqVIyFylR69bBM88E9b//exg0KL54JFpK5iIlaP9++M1vstdeufLKeGOSaCmZi5QY\n5+DBB2HHDl+vrobbb9faK6VOyVykxLz2GryZsTzdF7+o/TzLQfuJQ1FRMhcpgK1b4ZFHgvrYsTBK\n0/DKQqGm9CuZi0Ts4EG4997sjZlvvDHemKRwamr8ejsA+/ZBS0s076NkLhKxRx6BTZt8ubISvvY1\nqKqKNyYpHLPCtM6VzEUi9NprMH9+UL/5Zj9BSMpLIW6CKpmLRGTTJj96Je3CC7UFXLkqxE1QJXOR\nCDQ2wn/+ZzBdf+BA+MIXNAyxXKllLlKEnIP77oOPPvL1qiq48071k5cz9ZmLFKEnn4QVK4L6bbdp\nun65U8tcpMgsXeqTedrVV8Po0fHFI8mgZC5SRDZu9N0racOH+0W0RHQDVKRI7NkDv/pVMDGof38/\nnrxCv2GCnziUtmdPNO8R6kfNzIaY2Rwze8vMlpvZt6IJR6T4tLT4kSs7d/p6jx7wjW/AscfGG5ck\nR/tk7lz+3yNsu6EF+J/OubOBi4BvmNnw/IcjUlzSGzK/+66vm8FXv6obnpLtmGP8xs4QbO6cb6GS\nuXPuQ+fckrbyXuBtYHD+wxEpLrNmwYIFQf366+Hcc+OLR5LJLPqulk736JnZMGAksODIZ4qUtnnz\n4Omng/pll2mjCTm8zGS+d2/+X79Tuw6a2bHAY8DdbS30LDNmzPi4nEqlSKVSOYYnkkwrVmRP1T/n\nHPj85zXDUw4vfQ9l8+Y67rmnjgED8vv65kL2xJtZJfAk8Ixz7mcdfN6FfS2RYrZuHcycGUzVHzoU\nvvtdzfCUI7v/fnj9dV++7Ta4+GJfNjOcczk3AzrTzfJbYGVHiVykXGzeDL/4RZDI+/WDu+5SIpej\nS0SfuZldAnwBuMLMFpvZIjObmP9wRJJrxw742c/8Ilrgfznvvhtqa+ONS4pD1Mk8VJ+5c24+0C3/\nby9SHOrrfddKfb2vV1XBN79J3vs9pXQlomUuUs4aGnwi377d1ysr4R//EU4+Od64pLhkTiJTMhcp\nsH374Kc/9Rsyg5+ef+edft0Vkc5I1NBEkXKyZ49P5On9O838eivnnRdvXFKcEtFnLlJu9uzxXSub\nN/u6Gdx+O4waFW9cUrw6Wp8ln/MS1M0i0k5DA/zkJ9mJ/Mtf9nt4inRV1OuzKJmLZNixA/7932HL\nFl9Pt8jHjIk3Lil+Ua/PomQu0mbLFvjxj2HbNl+vqICvfEUtcsmfKJO5+sxFgA8+gJ//PBhlUFnp\nb3aOHBlvXFJalMxFIrRiBdx7b9CHWVXlx5Fr+KHkW5RjzZXMpazNnw9//CO0tvp6dTV861vwqU/F\nG5eUpijHmiuZS1lyzm8skbkeeb9+PpEPHBhfXFLa1M0ikkdNTfC738GiRcFjQ4f6RN67d2xhSRlQ\nMhfJk/p6+NWvYP364LGzz4Y77vAbMYtESX3mInmwdq2/0Zn5SzR+PNxwgx+GKBK1zP/8lMxFOsk5\nmDsXHn00uNFZUeG3ebvssnhjk/KilrlIFx044EervPlm8FhNjR9DfuaZ8cUl5an9aJZ87rSpZC4l\na8MG+PWvgxmdAMOGwde/Dn37xhaWlLGqKr8+S3OzP5qa8vfaSuZScpyDl17y3SotLcHjl14KN90U\nLHYkEoeaGti505fzOdY8VDI3s/uAKcBW55xWc5bEamiA3//ez+pMq6qCW2+FCy6ILy6RtMxkns9+\n87At8/uBXwAP5O+tRfJryRL4wx+yWztDhvhhh9qrU5IiqrHmYTd0nmdm2vFQEmnvXnjkkeybnABX\nXgnXXecXzRJJiqhGtOjHXIqWc7B4MTz0UPYvRW2tX4NcC2VJEkU11lzJXIrSjh3w8MOwfHn242PG\nwI03Qq9e8cQlcjRF0TKfMWPGx+VUKkUqlcrny4vQ0gJz5sATT2Rvu1Vb629ynntufLGJhLF2bR0L\nF9YB8OGH+XtdcyFHrZvZMOAJ51yHvy5m5sK+lkhXvPUW/OlPsHVr9uOXXeb7xqur44lLpDOWL4df\n/tKXzz4b7r7bcM7lvLVz2KGJDwEpoJ+ZrQemO+fuz/XNRcLYsgUefxyWLct+fNAg3xo/9dR44hLp\nirhHs9ySv7cUCaehwXenzJsXrKkCfnXDqVNh3Djo1i2++ES6IqpkHrqb5agvpG4WyZPGRnj+eXjx\nxex+cTO46CLfpaJ1x6VYNTf7TVFqavy9ntGj89PNomQuidHY6G9uvvAC7N+f/bnhw/1StUOHxhOb\nSFTMCthnLhKlvXt9K3zOHL/KYaYhQ3xL/OyzfctcRDqmZC6x2b7dt8Lnz8/uTgE//X7qVPi7v1MS\nFwlDyVwKyjlYt863xBct+uR6zgMGwOTJflEs7f4jEp6SuRTEwYOwcKHf8Sdz/820IUNg0iQYNUpJ\nXKQrlMwlUhs3+qGFr7/+yZuaAGedBRMm+I/qThHpOiVzybu9e+GNN+DVV/1uP+117w4XXujHiQ8Z\nUvj4REqRkrnkRVOTX0/8jTdg5crsST5pJ5zgd/u55BIthCWSb0rm0mUHDvgp9osW+Z19mps/eU73\n7jBypE/iZ5yhrhSRqCiZS6fs2gVLl/okvnp19h6bmU45BS6+GEaP1gJYIoWgZC5H1NwM777rVyxc\nsQI2bz78uYMG+SGFn/kM9O9fuBhFRMlc2mlt9UMHV6+GVatg7dqOu0/Shg71re9Ro7TPpkiclMzL\nXFMTvPeen8jzzjv+aGo6/PmVlb7v+9OfhvPOg+OOK1ysInJ4SuZl5NAhvzb4Bx/4BP7ee77bpKOR\nJ5lOOAFGjPDro5x5JlRVFSZeEQlPybxENTb6RL1xox/rvXGjPw53wzJT376+9X3mmX61wn79oo9X\nRHKjZF7EnPOL22/d6vcS/PBD3/LevNmPOgnDDE48EU47ze/Yc9ppPnlrCKFIcVEyT7hDh3xi3rHD\nrzL40UfBsXXrJ5eMPZp+/eCkk2DYsODo0SOCwEWkoJTMY9TcDLt3B8euXVBfDzt3+vLOnb7elT0/\nKith4EA/XX7wYD/qZOhQOPbY/H8dIhK/0MnczCYCPwUqgPucc/dEFlURcs4n5337/LF3b/axZ48/\nGhqCo6OFpzqrRw9/g/LEE33yHjjQJ+/jj9fqgyLlJNS2cWZWAawBxgObgTeBm51zqzLOKYpt4+rq\n6kilUh/XW1v98qxNTf5Ilw8cCD6mj/37/cfGRl/ev9+X00eYm4thbd5cx6BBPs7aWt890q+fT9LH\nH+8n5QwY4PcRjLN/u/31TCrFmT/FECMUT5yF3jbuAmCtc+6Dtjd/BLgWWHXEZ0Vs71546SWfRFta\nfMs4s5yuHzwY1F94oY6LLkpx8KB/PJ8JuLMqKvzGxH36+IRdW+tHktTWwoMP1jF9eoq+fX2XSVIV\nyy+M4syfYogRiifOfAmbJgYDmYuZbsQn+Fg1NsKsWZ17zr59vh86CpWVfjXA9FFT4/uo0+X00adP\n8LnDtaqfe863wEVEwkhwm+/o8tFiNYNjjvFHVVX20aNH8LFnT39UVfmFo9L16urg6N5dQ/pEJB5h\n+8zHADOccxPb6j8AXOZNUDNLfoe5iEgC5aPPPGwy7wasxt8A3QK8AXzeOfd2rgGIiEjuQnVUOOcO\nmdldwPMEQxOVyEVEEiJUy1xERJIt1LQSM5toZqvMbI2Zff8w56TMbLGZrTCzuZ15br7kGOf7Zra0\n7XNvxBmnmX2nLY5FZrbczFrMrDbMcxMSY5KuZW8zm2VmS9ri/HLY5yYoziRdz1oze7wtntfNbETY\n5yYozoJcTzO7z8y2mtmyI5zzczNb2/Z9H5nxeOevpXPuiAc+4b8DnAx0B5YAw9ud0wd4CxjcVu8f\n9rn5OnKJs628DugbRWydjbPd+VOAFwp5PXOJMWnXEvhfwI/S329gB757MWk/mx3GmcDr+WPgh23l\nMwv9s5lrnAW+nmOBkcCyw3x+EvBUW/lC4PVcrmWYlvnHE4acc81AesJQpluAPzvnNgE457Z34rn5\nkkucAEbI/1QKEGemzwMPd/G5ccQIybqWDqhpK9cAO5xzLSGfm4Q4IVnXcwQwB8A5txoYZmbHh3xu\nEuKEAl29jQ1rAAACiUlEQVRP59w84Ejrl14LPNB27gKgj5kNoIvXMswX1NGEocHtzjkDOM7M5prZ\nm2b2xU48N19yiRP8L9Pstse/FlGMYeMEwMx6AhOBP3f2uTHGCMm6lr8ERpjZZmApcHcnnpuEOCFZ\n13Mp8FkAM7sAOAkYEvK5SYgTCnc9j+ZwX0eXrmW+Jg1VAqOAK4BewGtm9lqeXjufOozTOfcOcIlz\nbkvbX+/ZZvZ221/WOE0F5jnnIpqzmhcdxZika3k1sNg5d4WZndoWz3kxxXIkHcbpnNtLsq7n/wF+\nZmaLgOXAYuBQTLEcyZHiTNL1zJTTWPMwyXwT/q9a2pC2xzJtBLY75w4AB8zsZeDTIZ+bL7nE+Y5z\nbguAc+4jM/sL/l+dKL7BnbkmN5PdfVGo65lLjCTsWt4O/KgtnnfN7D1geMjnJiHOhUm6ns65PcD/\nSNfb4lwHVB/tuQmJs5A/n0ezCRiaUU9/HcfQlWsZohO/G0Fn/DH4zviz2p0zHJjddm41/i/hiDDP\nzePNhlzirAaObTunFzAfmBBXnG3n9cHfBOvZ2efGHGOiriXwK2B6W3kA/t/X4xL4s3m4OJN2PfsA\n3dvKXwN+V8ifzTzEWbDr2fYew4Dlh/ncZIIboGMIboB26VqGDWgifgboWuAHbY/dCdyRcc538CNF\nlgHfPNJzI7xwXYoT+FTbBVuMT/BJiPM24KEwz01SjEm7lsCJwHNt3+9l+JnLifvZPFycCbyeY9o+\n/zbwGNAnodezwzgLeT2Bh/BLhjcB6/H/fbX/HfolPnEvBUblci01aUhEpARoLxoRkRKgZC4iUgKU\nzEVESoCSuYhICVAyFxEpAUrmIiIlQMlcRKQEKJmLiJSA/w9zwy/tp8RqtAAAAABJRU5ErkJggg==\n",
      "text/plain": [
       "<matplotlib.figure.Figure at 0x10f736cf8>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# alpha = 20 , beta = 2\n",
    "drawBeta(20,2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "α,βが1の時完全に一様分布、α,βが大きいほど尖度が大きくなり、α,βの値が離れているほど歪度が大きくなる。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 144,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/latex": [
       "さて、コインのパラメータのMAP推定に戻る。事後分布を最大化したいので、式にすると以下になる。<br><br>\n",
       "$$argmax_\\theta P(\\theta|D) = argmax_\\theta P(D|\\theta)P(\\theta)$$<br>\n",
       "$$= argmax_\\theta \\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right) \\theta^n (1-\\theta)^{N-n} \n",
       "\\frac{\\Gamma(\\alpha+\\beta)}{\\Gamma(\\alpha)\\Gamma(\\beta)}\\theta^{\\alpha-1}(1-\\theta)^{\\beta-1}$$<br>\n",
       "ここで、$$\\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right)$$と\n",
       "$$\\frac{\\Gamma(\\alpha+\\beta)}{\\Gamma(\\alpha)\\Gamma(\\beta)}$$は$$\\theta$$に対して定数とみなせるため、最大化の式から除外できる。<br>\n",
       "$$=argmax_\\theta \\theta^n (1-\\theta)^{N-n}\\theta^{\\alpha-1}(1-\\theta)^{\\beta-1}$$<br>\n",
       "$$=argmax_\\theta \\theta^{\\alpha+n-1}(1-\\theta)^{\\beta+N-n-1}$$<br><br>\n",
       "\n",
       "この最大化式を解く。<br>\n",
       "θで微分して0となる時のθの値が最大化する点であるので、式は以下のようになる。<br><br>\n",
       "$$(\\theta^{\\alpha+n-1}(1-\\theta)^{\\beta+N-n-1})'=0$$<br>\n",
       "ここで、可読性のため$$A=\\alpha+n-1$$とおき、$$B=\\beta+N-n-1$$とおく。<br>\n",
       "$$(\\theta^A(1-\\theta)^B)'=A\\theta ^{A-1} (1-\\theta)^B - \\theta ^A B (1-\\theta)^{B-1}$$<br>\n",
       "$$=\\theta ^{A-1}(1-\\theta)^{B-1}(A(1-\\theta)-B\\theta)$$<br>\n",
       "ここで、$$\\theta ^{A-1}(1-\\theta)^{B-1}$$は0にならないので、以下を解けば良い。<br>\n",
       "$$A(1-\\theta)-B\\theta = 0$$<br>\n",
       "つまり、$$\\theta = \\frac{A}{A+B} = \\frac{n + \\alpha-1}{N + \\alpha+\\beta-2}$$"
      ],
      "text/plain": [
       "<IPython.core.display.Latex object>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "%%latex\n",
    "さて、コインのパラメータのMAP推定に戻る。事後分布を最大化したいので、式にすると以下になる。<br><br>\n",
    "$$argmax_\\theta P(\\theta|D) = argmax_\\theta P(D|\\theta)P(\\theta)$$<br>\n",
    "$$= argmax_\\theta \\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right) \\theta^n (1-\\theta)^{N-n} \n",
    "\\frac{\\Gamma(\\alpha+\\beta)}{\\Gamma(\\alpha)\\Gamma(\\beta)}\\theta^{\\alpha-1}(1-\\theta)^{\\beta-1}$$<br>\n",
    "ここで、$$\\left(\\begin{array}{cc} N \\\\ n \\end{array} \\right)$$と\n",
    "$$\\frac{\\Gamma(\\alpha+\\beta)}{\\Gamma(\\alpha)\\Gamma(\\beta)}$$は$$\\theta$$に対して定数とみなせるため、最大化の式から除外できる。<br>\n",
    "$$=argmax_\\theta \\theta^n (1-\\theta)^{N-n}\\theta^{\\alpha-1}(1-\\theta)^{\\beta-1}$$<br>\n",
    "$$=argmax_\\theta \\theta^{\\alpha+n-1}(1-\\theta)^{\\beta+N-n-1}$$<br><br>\n",
    "\n",
    "この最大化式を解く。<br>\n",
    "θで微分して0となる時のθの値が最大化する点であるので、式は以下のようになる。<br><br>\n",
    "$$(\\theta^{\\alpha+n-1}(1-\\theta)^{\\beta+N-n-1})'=0$$<br>\n",
    "ここで、可読性のため$$A=\\alpha+n-1$$とおき、$$B=\\beta+N-n-1$$とおく。<br>\n",
    "$$(\\theta^A(1-\\theta)^B)'=A\\theta ^{A-1} (1-\\theta)^B - \\theta ^A B (1-\\theta)^{B-1}$$<br>\n",
    "$$=\\theta ^{A-1}(1-\\theta)^{B-1}(A(1-\\theta)-B\\theta)$$<br>\n",
    "ここで、$$\\theta ^{A-1}(1-\\theta)^{B-1}$$は0にならないので、以下を解けば良い。<br>\n",
    "$$A(1-\\theta)-B\\theta = 0$$<br>\n",
    "つまり、$$\\theta = \\frac{A}{A+B} = \\frac{n + \\alpha-1}{N + \\alpha+\\beta-2}$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 145,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/latex": [
       "ここで、最尤推定のθの算出式が$$\\frac{n}{N}$$だったことを考えると、このMAP推定における算出式の$$\\frac{\\alpha-1}{\\alpha+\\beta-2}$$\n",
       "の部分が事前分布による重み付けであることがわかる。<br>"
      ],
      "text/plain": [
       "<IPython.core.display.Latex object>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "%%latex\n",
    "ここで、最尤推定のθの算出式が$$\\frac{n}{N}$$だったことを考えると、このMAP推定における算出式の$$\\frac{\\alpha-1}{\\alpha+\\beta-2}$$\n",
    "の部分が事前分布による重み付けであることがわかる。<br>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 146,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/latex": [
       "では、実際に先ほどのカジノコインの例でパラメータをMAP推定してみる。<br>\n",
       "先ほどの例と同様、5回のコイントスのうち4回表が出たとする。<br><br>\n",
       "<b>β=1,α=1の場合</b><br>\n",
       "$$\\theta = \\frac{n + \\alpha-1}{N + \\alpha+\\beta-2} = \\frac{4 + 1-1}{5 + 1+1-2} = \\frac{4}{5}$$<br>\n",
       "これは最尤推定と同様になる。<br><br>\n",
       "\n",
       "<b>β=2,α=2の場合</b><br>\n",
       "$$\\theta = \\frac{n + \\alpha-1}{N + \\alpha+\\beta-2} = \\frac{4 + 2-1}{5 + 2+2-2} = \\frac{5}{7}$$<br>\n",
       "この、重み部分が1/2となるケースを特別に<b>ラプラススムージング</b>という。<br><br>\n",
       "\n",
       "<b>β=10,α=10の場合</b><br>\n",
       "$$\\theta = \\frac{n + \\alpha-1}{N + \\alpha+\\beta-2} = \\frac{4 + 10-1}{5 + 10+10-2} = \\frac{13}{23}$$<br>\n",
       "ハイパーパラメータが大きいほど修正が大きくなっていることがわかる。<br><br>"
      ],
      "text/plain": [
       "<IPython.core.display.Latex object>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "%%latex\n",
    "では、実際に先ほどのカジノコインの例でパラメータをMAP推定してみる。<br>\n",
    "先ほどの例と同様、5回のコイントスのうち4回表が出たとする。<br><br>\n",
    "<b>β=1,α=1の場合</b><br>\n",
    "$$\\theta = \\frac{n + \\alpha-1}{N + \\alpha+\\beta-2} = \\frac{4 + 1-1}{5 + 1+1-2} = \\frac{4}{5}$$<br>\n",
    "これは最尤推定と同様になる。<br><br>\n",
    "\n",
    "<b>β=2,α=2の場合</b><br>\n",
    "$$\\theta = \\frac{n + \\alpha-1}{N + \\alpha+\\beta-2} = \\frac{4 + 2-1}{5 + 2+2-2} = \\frac{5}{7}$$<br>\n",
    "この、重み部分が1/2となるケースを特別に<b>ラプラススムージング</b>という。<br><br>\n",
    "\n",
    "<b>β=10,α=10の場合</b><br>\n",
    "$$\\theta = \\frac{n + \\alpha-1}{N + \\alpha+\\beta-2} = \\frac{4 + 10-1}{5 + 10+10-2} = \\frac{13}{23}$$<br>\n",
    "ハイパーパラメータが大きいほど修正が大きくなっていることがわかる。<br><br>"
   ]
  }
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