12_hw_cp220_Q.ipynb

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    "<a href=\"https://colab.research.google.com/github/AshrithSagar/CP220-M4RAS-2024/blob/main/12_hw_cp220_Q.ipynb\" target=\"_parent\"><img src=\"https://colab.research.google.com/assets/colab-badge.svg\" alt=\"Open In Colab\"/></a>"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "cGc8gmov8c2h"
   },
   "source": [
    "---\n",
    "**HW12**\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "r57KCixm8hw0"
   },
   "source": [
    "---\n",
    "**PDF of a function of a rv**\n",
    "\n",
    "Let X be a rv, and $Y=g(X)$\n",
    "\n",
    "Then to find $f_Y(y)$, we can find its CDF and then get the pdf via differentiation (or differencing for discrete rv)\n",
    "\n",
    "$F_Y(y) = P(Y \\leq y) = P(g(X) \\leq y) = \\int_{x|g(x)\\leq y} f_X(x)dx$\n",
    "\n",
    "$f_Y(y) = \\frac{dF_Y(y)}{dy}$\n",
    "\n",
    "Example:\n",
    "\n",
    "$Y=X^2$\n",
    "\n",
    "$F_Y(y) = \\int_{-\\sqrt{y}}^{\\sqrt{y}} f_X(x)dx = \\int_{-∞}^{\\sqrt{y}} f_X(x)dx - \\int_{-∞}^{-\\sqrt{y}} f_X(x)dx = F_X(\\sqrt{y}) - F_X(-\\sqrt{y})$\n",
    "\n",
    "Therefore:\n",
    "\n",
    "$f_Y(y) = \\frac{1}{2\\sqrt{y}}(f_X(\\sqrt{y})+f_X(-\\sqrt{y}) )$ for $y \\geq 0$\n",
    "\n",
    "$\\ \\ \\ \\ =0$ elsewhere\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "JMx3YFgn-g3H"
   },
   "source": [
    "---\n",
    "*$\\underline{Linear\\ functions}$*\n",
    "\n",
    "If $Y=aX+b$\n",
    "\n",
    "then\n",
    "\n",
    "$F_Y(y) = P(Y \\leq y) = P(X \\leq \\frac{y-b}{a}) = \\int_{-∞}^{\\frac{y-b}{a}} f_X(x)dx$\n",
    "\n",
    "$F_Y(y+h)-F_Y(y) = \\int_{\\frac{y-b}{a}}^{\\frac{y-b}{a}+\\frac{h}{a}} f_X(x)dx$\n",
    "\n",
    "$\\approx f_X(\\frac{y-b}{a})*h/a$ if $a > 0$\n",
    "\n",
    "and\n",
    "\n",
    "$\\approx -f_X(\\frac{y-b}{a})*h/a$ if $a < 0$ (we need to reverse the limits of the integral)\n",
    "\n",
    "Hence\n",
    "\n",
    "$f_Y(y)=\\frac{DF_Y(y)}{dy} = lim_{h→0} \\frac{F_Y(y+h)-F_Y(y)}{h} = \\frac{1}{|a|} f_X(\\frac{y-b}{a})$\n",
    "\n",
    "*$\\underline{Strictly\\ monotonic\\ functions}$*\n",
    "\n",
    "If $Y=g(X)$ is strictly monotonic, i.e. for every $Y=y$, there is a single $X=x$ such that $y=g(x)$,\n",
    "then\n",
    "\n",
    "there exists a function $g^{-1}$ such that $x = g^{-1}(y)$\n",
    "\n",
    "Then we can use the above derivation to get\n",
    "\n",
    "$f_Y(y) = f_X(g^{-1}(y))|\\frac{dg^{-1}(y)}{dy}|$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "bf5thbWKCD0D"
   },
   "source": [
    "*Example*\n",
    "\n",
    "If $X\\ \\ \\sim\\ \\ N(\\mu, \\sigma)$\n",
    "\n",
    "and $Y = aX + b$\n",
    "\n",
    "then,\n",
    "\n",
    "$f_Y(y) = \\frac{1}{a} \\frac{1}{\\sqrt{2\\pi}\\sigma} e^{-\\frac{(\\frac{(y-b)}{a}-\\mu)^2}{2\\sigma^2}} = \\frac{1}{\\sqrt{2\\pi}a\\sigma} e^{-\\frac{(y -a\\mu-b)^2}{a^2\\sigma^2}}$\n",
    "\n",
    "Hence $Y$ is Normal with mean $a\\mu+b$ and standard deviation $a\\sigma$\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "0aGj84C7Gd9d"
   },
   "source": [
    "---\n",
    "**Q1**\n",
    "\n",
    "Problem 1 . If X is a random variable that is uniformly distributed between - 1 and 1. find the PDF of $\\sqrt{|X|}$ and the PDF of $-ln(|X|)$."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "nNv6YcMgGuvN"
   },
   "source": [
    "Ans:\n",
    "\n",
    "We're given that $X$ is a uniformly distributed random variable between -1 and 1, i.e., the PDF of $X$ is given by\n",
    "\n",
    "$\n",
    "\\displaystyle\n",
    "f_{X}(x)\n",
    "= \\begin{cases}\n",
    "\\frac{1}{2} & \\text{if } x\\in [-1,1] \\\\\n",
    "0 & \\text{otherwise}\n",
    "\\end{cases}\n",
    "$\n",
    "\n",
    "The CDF of $X$ is given by, for $x\\in [-1,1]$ as\n",
    "\n",
    "$\n",
    "\\displaystyle\n",
    "\\implies\n",
    "F_{X}(x)\n",
    "= P[X\\leq x]\n",
    "= \\int_{-1}^{x}f_{X}(t)dt\n",
    "= \\int_{-1}^{x}\\frac{1}{2}dt\n",
    "= \\frac{t}{2}\\bigg|_{-1}^{x}\n",
    "= \\frac{x+1}{2}\n",
    "$\n",
    "\n",
    "$\n",
    "\\displaystyle\n",
    "F_{X}(x)\n",
    "= \\begin{cases}\n",
    "0 & \\text{if } x<-1 \\\\\n",
    "\\frac{x+1}{2} & \\text{if } x\\in [-1,1] \\\\\n",
    "1 & \\text{if } x>1\n",
    "\\end{cases}\n",
    "$\n",
    "\n",
    "To find the PDF of $Y=\\sqrt{|X|}$, we first find the CDF of $Y$.\n",
    "\n",
    "If $y < 0$, then $F_{Y}(y) = P[\\sqrt{|X|}\\leq y] = 0$\n",
    "\n",
    "If $y\\in [0,1]$, then\n",
    "\n",
    "$\n",
    "\\displaystyle\n",
    "F_{Y}(y)\n",
    "= P[Y\\leq y]\n",
    "= P[\\sqrt{|X|}\\leq y]\n",
    "= P[|X|\\leq y^{2}]\n",
    "= P[-y^{2}\\leq X\\leq y^{2}]\n",
    "= F_{X}(y^{2})-F_{X}(-y^{2})\n",
    "= \\int_{-y^{2}}^{y^{2}}f_{X}(x)dx\n",
    "$\n",
    "\n",
    "$\n",
    "\\displaystyle\n",
    "\\implies\n",
    "F_{Y}(y)\n",
    "= \\int_{-y^{2}}^{y^{2}}\\frac{1}{2}dx\n",
    "= \\frac{x}{2}\\bigg|_{-y^{2}}^{y^{2}}\n",
    "= \\frac{y^{2}+y^{2}}{2}\n",
    "= y^{2}\n",
    "$\n",
    "\n",
    "$\n",
    "\\displaystyle\n",
    "\\implies\n",
    "F_{Y}(y)\n",
    "= \\begin{cases}\n",
    "0 & \\text{if } y<0 \\\\\n",
    "y^{2} & \\text{if } y\\in [0,1] \\\\\n",
    "1 & \\text{if } y>1\n",
    "\\end{cases}\n",
    "$\n",
    "\n",
    "Hence, the PDF of $Y$ is given by\n",
    "\n",
    "$\n",
    "\\displaystyle\n",
    "f_{Y}(y)\n",
    "= \\frac{dF_{Y}(y)}{dy}\n",
    "= \\begin{cases}\n",
    "0 & \\text{if } y<0 \\\\\n",
    "\\frac{d}{dy}(y^{2}) & \\text{if } y\\in [0,1] \\\\\n",
    "0 & \\text{if } y>1\n",
    "\\end{cases}\n",
    "\\implies\n",
    "\\boxed{\n",
    "f_{\\sqrt{|X|}}(x)\n",
    "= \\begin{cases}\n",
    "2x & \\text{if } x\\in [0,1] \\\\\n",
    "0 & \\text{otherwise}\n",
    "\\end{cases}\n",
    "}\n",
    "$\n",
    "\n",
    "For $Z = -\\ln{(X)}$, we first find the CDF of $Z$,\n",
    "\n",
    "$\n",
    "\\displaystyle\n",
    "F_{Z}(z)\n",
    "= P[Z\\leq z]\n",
    "= P[-\\ln{(X)}\\leq z]\n",
    "= P[\\ln{(X)}\\geq -z]\n",
    "= P[X\\geq e^{-z}]\n",
    "= 1-P[X<e^{-z}]\n",
    "= 1-F_{X}(e^{-z})\n",
    "$\n",
    "\n",
    "where we've used the property that $\\ln()$ is a strictly increasing function.\n",
    "\n",
    "$\n",
    "\\displaystyle\n",
    "\\implies\n",
    "F_{Z}(z)\n",
    "= \\begin{cases}\n",
    "0 & \\text{if } z<0 \\\\\n",
    "\\frac{1-e^{-z}}{2} & \\text{if } z\\geq 0\n",
    "\\end{cases}\n",
    "$\n",
    "\n",
    "Hence, the PDF of $Z$ is given by\n",
    "\n",
    "$\n",
    "\\displaystyle\n",
    "f_{Z}(z)\n",
    "= \\frac{dF_{Z}(z)}{dz}\n",
    "= \\begin{cases}\n",
    "0 & \\text{if } z<0 \\\\\n",
    "\\frac{d}{dz}\\left(\\frac{1-e^{-z}}{2}\\right) & \\text{if } z\\geq 0\n",
    "\\end{cases}\n",
    "= \\begin{cases}\n",
    "0 & \\text{if } z<0 \\\\\n",
    "\\frac{1}{2}e^{-z} & \\text{if } z\\geq 0\n",
    "\\end{cases}\n",
    "\\implies\n",
    "\\boxed{\n",
    "f_{-\\ln{(X)}}(z)\n",
    "= \\begin{cases}\n",
    "0 & \\text{if } z<0 \\\\\n",
    "\\frac{1}{2}e^{-z} & \\text{if } z\\geq 0\n",
    "\\end{cases}\n",
    "}\n",
    "$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "yVow7c7Hj-RA"
   },
   "source": [
    "---\n",
    "Complex distributions can be calculated from functions of simpler distributions, and hence can provide a way for generating samples for these.\n",
    "\n",
    "The exponential can be generated as follows:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "id": "_6EPo27Xh--V"
   },
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "import matplotlib.pyplot as plt"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "id": "owWITTVxiE1t"
   },
   "outputs": [],
   "source": [
    "rng = np.random.default_rng(100)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
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     "height": 430
    },
    "id": "AVLmbhc-a2Ge",
    "outputId": "02f37771-cb99-4c6a-83ab-825804221d52"
   },
   "outputs": [
    {
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",
      "text/plain": [
       "<Figure size 640x480 with 1 Axes>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "nSamples = 1000\n",
    "#generate uniform random between -1 to 1\n",
    "X_D = 2*rng.random(nSamples)-1.0\n",
    "\n",
    "#generate exponential by taking natural log\n",
    "Y_D = -np.log(np.abs(X_D))\n",
    "\n",
    "#get histogram\n",
    "Y_D_Hist,_ = np.histogram(Y_D,20,(0,20))\n",
    "\n",
    "#compare with exponential of numpy\n",
    "E_D = rng.exponential(1.0,nSamples)\n",
    "E_D_Hist,_ = np.histogram(E_D,20,(0,20))\n",
    "\n",
    "plt.plot(Y_D_Hist)\n",
    "plt.plot(E_D_Hist)\n",
    "plt.legend(['Derived_From_Uniform','Numpy Exponential'])\n",
    "plt.show()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "kLO7b2tH0U_A"
   },
   "source": [
    "---\n",
    "**Q2**\n",
    "Let $U,V$ be zero mean, unit variance, Normal random variables.\n",
    "\n",
    "Show that their ratio: $X=U/V$ has pdf $f_X(x) = \\frac{1}{\\pi(1+x^2)}$. This distribution is called the Cauchy distribution."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "bzfjgruk02HK"
   },
   "source": [
    "Ans:\n",
    "\n",
    "We have that $U, V \\sim \\mathcal{N}(0, 1)$ and $X = U/V$ and we need to show that $f_X(x) = \\frac{1}{\\pi(1+x^2)}$.\n",
    "\n",
    "$\n",
    "\\displaystyle\n",
    "f_U(u) = \\frac{1}{\\sqrt{2\\pi}} e^{-\\frac{1}{2} u^2}\n",
    ", \\quad\n",
    "f_V(v) = \\frac{1}{\\sqrt{2\\pi}} e^{-\\frac{1}{2} v^2}\n",
    "$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "8JoqiIYlGZG7"
   },
   "source": [
    "---\n",
    "**Variance of sum of RV**\n",
    "\n",
    "\n",
    "Let $X=X_1+X_2$\n",
    "\n",
    "$EX = EX_1 + EX_2$\n",
    "\n",
    "$EX^2 = EX_1^2 + EX_2^2+2EX_1X_2$\n",
    "\n",
    "$Var(X) = EX^2 - (EX)^2 = Var(X_1)+Var(X_2) + 2(EX_1X_2 = EX_1EX_2) = Var(X_1) + Var(X_2) + 2Cov(X_1,X_2)$\n",
    "\n",
    "*$\\underline{Law\\ of\\ Iterated\\ Expectation}$*\n",
    "\n",
    "\n",
    "$EX = \\int_x f_X(x)dx = \\int_x \\int_y f_{X|Y}(x)f_Y(y)dydx$\n",
    "\n",
    "By changing order of integration,\n",
    "\n",
    "$\\ \\ \\ =\\int_y f_Y(y) \\int_x f_{X|y}(x)dx dy = \\int_y f_Y(y) E_{X|Y} (X) = E_Y E_{X|Y} (X)$ also written as $E E[X|Y]$\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "3VXmlwhqD5EX"
   },
   "source": [
    "---\n",
    "**Transforms of RV**\n",
    "\n",
    "$M_X(s) = E(e^{sX}) = \\int_{-∞}^∞ e^{sX}f_X(x)dx = \\int_{-∞}^∞ (1+sX + \\frac{s^2X^2}{2!}+...)f_X(x)dx= 1+sEX + \\frac{s^2}{2}EX^2+...$\n",
    "\n",
    "Hence to find the $k^{th}$ raw moment,\n",
    "we just need to find\n",
    "\n",
    "$EX^k = \\frac{d^kM_X(s)}{ds^k}|_{s=0}$\n",
    "\n",
    "*$\\underline{Examples}$*\n",
    "\n",
    "*1) Uniform*\n",
    "\n",
    "$X\\ \\ \\tilde \\ \\ Unif(a,b)$\n",
    "\n",
    "$f_X(x)=\\frac{1}{b-a}$ for $a \\le X \\le b$ and 0 otherwise\n",
    "\n",
    "$M_x(s) = \\int_a^be^{sx}\\frac{1}{b-a}dx = \\frac{e^{sb}-e^{sa}}{s(b-a)} = 1 + \\frac{s(b+a)}{2}+ \\frac{s^2(b^2+ab+a^2)}{6} + O(s^3)$\n",
    "\n",
    "\n",
    "We can then see that\n",
    "\n",
    "$EX = \\frac{dM_x(s)}{ds}|_{s=0} = \\frac{b+a}{2}$\n",
    "\n",
    "$EX^2 = \\frac{d^2M_x(s)}{ds^2}|_{s=0}= \\frac{b^2+ab+a^2}{3}$\n",
    "\n",
    "$Var(X) = EX^2-(EX)^2=\\frac{(b-a)^2}{12}$\n",
    "\n",
    "*2) Exponential*\n",
    "\n",
    "$f_X(x) = \\lambda e^{-\\lambda x}$ for $X \\ge 0$ and 0 otherwise.\n",
    "\n",
    "$M_X(s) = \\int_0^∞ e^{(s-\\lambda)x}dx = \\frac{-\\lambda}{s-\\lambda}$\n",
    "\n",
    "$EX = \\frac{dM}{ds}|_{s=0} = \\frac{\\lambda}{(s-\\lambda)^2}|_{s=0}=\\lambda$\n",
    "\n",
    "$EX^2 = \\frac{d^2M}{ds^2}|_{s=0}= \\frac{-2\\lambda}{(s-\\lambda)^3}|_{s=0} = \\frac{2}{\\lambda^2} $\n",
    "\n",
    "$Var(X) = EX^2 - (EX)^2 = \\frac{1}{\\lambda^2}$\n",
    "\n",
    "*3) Gaussian*\n",
    "\n",
    "$f_X(x) = \\frac{1}{\\sqrt{2\\pi}\\sigma}e^{-\\frac{(x-\\mu)^2}{2\\sigma^2}}$\n",
    "\n",
    "$M_X(s) = \\frac{1}{\\sqrt{2\\pi}\\sigma}\\int_{-∞}^∞ e^{sx}{-\\frac{(x-\\mu)^2}{2\\sigma^2}}dx$\n",
    "\n",
    "One trick to handle this is to complete the square of the exponent.\n",
    "\n",
    "i.e.\n",
    "\n",
    "$sx  -\\frac{(x-\\mu)^2}{2\\sigma^2} = \\mu s + \\frac{s^2\\sigma^2}{2} -\\frac{(x-\\mu - s\\sigma^2)^2}{2\\sigma^2}$\n",
    "\n",
    "Therefore\n",
    "\n",
    "$M_X(s) = e^{\\mu s + \\frac{s^2\\sigma^2}{2}} \\int_{-∞}^{∞} f_Z(x)dx$\n",
    "\n",
    "Where $Z$ is Gaussian with pdf $N(\\mu + s\\sigma^2,\\sigma)$ and hence integrates to 1.\n",
    "\n",
    "Therefore\n",
    "\n",
    "$M_X(s) = e^{\\mu s + \\frac{s^2\\sigma^2}{2}} $\n",
    "\n",
    "We can then check that mean and variance come out to be as expected.\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "C6_qZdZrNyQg"
   },
   "source": [
    "*$\\underline{Connection\\ with\\ Laplace\\ Transform}$*\n",
    "\n",
    "Note that the Laplace transform of a function $f(x)$ is\n",
    "\n",
    "$F(s) = ℒ(f(x)) = \\int_{-∞}^∞ e^{-sx}f(x)dx  $\n",
    "\n",
    "Hence our moment generating function is:\n",
    "\n",
    "$M_X(-s) = ℒ(f_X(x))$\n",
    "\n",
    "So we can obtain the pdf as inverse laplace transform\n",
    "\n",
    "$f_X(x) = ℒ^{-1}(M_X(-s))$\n",
    "\n",
    "From laplace theory,\n",
    "\n",
    "If $F(s) = ℒ(f(x))$ and $G(s) = ℒ(g(x))$ then $H(s) = F(s)G(s) = ℒ(f(x)*g(x))$ where $*$ is the convolution operator defined as:\n",
    "\n",
    "$h(x) = f(x)*g(x) = \\int f(u)g(x-u)du$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "2iH6vY6zM2FT"
   },
   "source": [
    "*$\\underline{Sums\\ of\\ Indpendent\\ rvs}$*\n",
    "\n",
    "\n",
    "Let $Z = X + Y$ with X,Y independent (and not necessarily identical)\n",
    "\n",
    "Then $M_Z(s) = Ee^{sZ} = Ee^{s(X+Y)} = Ee^{sX}e^{sY} = Ee^{sX}Ee^{sY} = M_X(s)M_Y(s)$ the product of the two moment generating functions.\n",
    "\n",
    "Hence $f_Z(z) = f_X(z)*f_Y(z) = \\int_{-\\infty}^∞ f_X(x)f_Y(z-x)dx$\n",
    "\n",
    "\n",
    "\n",
    "This can be extended to find sum of any number of independent random variabls as:\n",
    "\n",
    "Let $Z=X_1+X_2+...X_n$ be a sum of independent RVs (not necessarily identical)\n",
    "\n",
    "$M_Z(s) = ∏_1^n M_{X_i}(s)$\n",
    "\n",
    "\n",
    "*Example*\n",
    "\n",
    "$X_1, X_2, ...X_n$ be independent and indentically distributed as Gaussian $N(\\mu,\\sigma)$.\n",
    "\n",
    "Then let $Z = \\sum_{i=1}^n X_i$\n",
    "\n",
    "$M_Z(s) = e^{n\\mu s + \\frac{ns^2\\sigma^2}{2}}$\n",
    "\n",
    "We can see from above that $Z$ is Gaussian $N(n\\mu,\\sqrt{n}\\sigma)$\n",
    "\n",
    "*$\\underline{Sum\\ of\\ random\\ number\\ of\\ independent\\ rvs}$*\n",
    "\n",
    "Let $N$ be a discrete rv $\\in \\{1,2,....\\}$\n",
    "\n",
    "Let $Z = \\sum_{i=1}^N X_i$ where $X_i$ are independent identical rv.\n",
    "\n",
    "Then $M_Z(s) = E_Z e^{sX} = E_N E_{Z|N} e^{s\\sum_{i=1}^N X_i}$ by law of iterated expectation."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "7_8yaOEy2XB9"
   },
   "source": [
    "---\n",
    "**Q3**\n",
    "\n",
    "Problem 12. The random variables X, Y. and Z are independent and uniformly distributed between zero and one. Find the PDF of X + Y + Z.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "Kng5Y2v42nQH"
   },
   "source": [
    "Ans:\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 56
    },
    "id": "kQSWMvGpmP1G",
    "outputId": "1360e2b8-1c0e-42db-cb3d-f3e7818b40e1"
   },
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$\\displaystyle \\frac{x^{2} \\theta\\left(x\\right)}{2} - \\frac{\\left(x - 3\\right)^{2} \\theta\\left(x - 3\\right)}{2} + \\frac{3 \\left(x - 2\\right)^{2} \\theta\\left(x - 2\\right)}{2} - \\frac{3 \\left(x - 1\\right)^{2} \\theta\\left(x - 1\\right)}{2}$"
      ],
      "text/plain": [
       "x**2*Heaviside(x)/2 - (x - 3)**2*Heaviside(x - 3)/2 + 3*(x - 2)**2*Heaviside(x - 2)/2 - 3*(x - 1)**2*Heaviside(x - 1)/2"
      ]
     },
     "execution_count": 15,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "#Trying Sympy\n",
    "import sympy\n",
    "from sympy import inverse_laplace_transform, exp, Symbol\n",
    "from sympy.abc import s, x,z\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "ESNUYV5VrvNw"
   },
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "ZV2XGbC2uC4W"
   },
   "source": [
    "---\n",
    "**Q4**\n",
    "\n",
    "Find the pdf of Z=X+Y when X, Y are independent exponential rv with parameter $\\lambda$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "dIAvI1kGuRjA"
   },
   "source": [
    "Ans:\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 39
    },
    "id": "PaiEDrhKtUV2",
    "outputId": "d098ac90-d8b1-41a9-acb1-27a146c27ffd"
   },
   "outputs": [
    {
     "data": {
      "text/latex": [
       "$\\displaystyle a^{2} z e^{- a z} \\theta\\left(z\\right)$"
      ],
      "text/plain": [
       "a**2*z*exp(-a*z)*Heaviside(z)"
      ]
     },
     "execution_count": 14,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "#your code"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "gpDTchG1ux-J"
   },
   "source": [
    "\n",
    "Note that $\\theta(z)$ is the heaviside\n",
    "step function $\\theta(z) =1 $ for $z > 0$; $=1/2$ for $z=0$; and 0 elsewhere.\n",
    "\n",
    "\n",
    "hence $f_Z(z) = \\lambda^2 z e^{-\\lambda z}$ for $z \\geq 0$ and 0 elsewhere."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "oF2NVYmWGVER"
   },
   "source": [
    "---\n",
    "**Q5**\n",
    "\n",
    "Problem 14. Competing exponentials. The lifetimes of two light bulbs are modeled as independent and exponential random variables X and Y, with parameters $\\lambda$ and $\\mu$, respectively. The time at which a light bulb first burns out is\n",
    "$Z= min\\ \\{X,Y\\}$.\n",
    "Show that Z is an exponential random variable with parameter $\\lambda + \\mu$ ."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "6xhfBQisGhUr"
   },
   "source": [
    "Ans:\n",
    "\n",
    "The exponential random variable $X \\sim \\text{Exp}(\\lambda)$ has the distribution function\n",
    "\n",
    "\\begin{equation*}\n",
    "F_X(x)\n",
    "= P[X \\leq x]\n",
    "= \\begin{cases}\n",
    "1 - e^{-\\lambda x} & \\text{if } x \\geq 0 \\\\\n",
    "0 & \\text{otherwise}\n",
    "\\end{cases}\n",
    "\\end{equation*}\n",
    "\n",
    "With $Z = min\\{X,Y\\}$, we have\n",
    "\n",
    "\\begin{align*}\n",
    "F_Z(z)\n",
    "&\n",
    "= P[Z \\leq z] = P[min\\{X,Y\\} \\leq z]\n",
    "\\\\ &\n",
    "= 1 - P[X > z, Y > z]\n",
    "= 1 - P[X > z]P[Y > z]\n",
    "\\\\ &\n",
    "= 1 - [1 - F_X(z)][1 - F_Y(z)]\n",
    "= 1 - [1 - (1 - e^{-\\lambda z})][1 - (1 - e^{-\\mu z})]\n",
    "\\\\ &\n",
    "= 1 - [e^{-\\lambda z}] [e^{-\\mu z}]\n",
    "= 1 - e^{-(\\lambda + \\mu)z}\n",
    "\\end{align*}\n",
    "\n",
    "Thereby, since the distribution function uniquely determines the random variable, we have that $\\boxed{Z \\sim \\text{Exp}(\\lambda + \\mu)}$.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "btxDT5S2H5XG"
   },
   "source": [
    "---\n",
    "**Q6**\n",
    "\n",
    "Problem 17. Suppose that X and Y are random variables with the same variance. Show that X - Y and X + Y are uncorrelated."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "FgBvY5r6H-M7"
   },
   "source": [
    "Ans:\n",
    "\n",
    "Given that $X$ and $Y$ are random variables with $\\operatorname{Var}(X) = \\operatorname{Var}(Y)$, we have\n",
    "\n",
    "\\begin{align*}\n",
    "\\operatorname{Cov}(X+Y, X-Y)\n",
    "&\n",
    "= E[(X+Y)(X-Y)] - E[X+Y] E[X-Y]\n",
    "\\\\ &\n",
    "= E[X^2-Y^2] - (E[X]+E[Y])(E[X]-E[Y])\n",
    "\\\\ &\n",
    "= E[X^2]-E[Y^2] - ((E[X])^2 - (E[Y])^2)\n",
    "\\\\ &\n",
    "= (E[X^2] - (E[X])^2) - (E[Y^2] - (E[Y])^2)\n",
    "\\\\ &\n",
    "= \\operatorname{Var}(X) - \\operatorname{Var}(Y)\n",
    "= 0\n",
    "\\end{align*}\n",
    "\n",
    "As a result, the correlation coefficient between $X+Y$ and $X-Y$ is zero.\n",
    "Hence, $X+Y$ and $X-Y$ are uncorrelated.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "LD7-mkmWIZ0X"
   },
   "source": [
    "---\n",
    "**Q7**\n",
    "Problem 22. Consider a gambler who at each gamble either wins or loses his bet with probabilities $p$ and $1-p$ independent of earlier gambles.\n",
    "\n",
    "When $p > 1/2$, a popular\n",
    "gambling system, known as the Kelly strategy, is to always bet the fraction $2p - 1$ of the current fortune. Compute the expected fortune after $n$ gambles, starting with $x$ units and employing the Kelly strategy."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "C8gsx9wtPk4u"
   },
   "source": [
    "Ans:\n",
    "\n",
    "Starting with an initial fortune of $x$ units, we can model the fortune after $n$ gambles as a random variable $F_n$ given by\n",
    "\n",
    "\\begin{align*}\n",
    "F_0\n",
    "&\n",
    "= x\n",
    "\\\\\n",
    "F_1\n",
    "&\n",
    "= \\begin{cases}\n",
    "x + (2p-1)F_0\\text{ with probability }p\n",
    "\\\\\n",
    "x - (2p-1)F_0\\text{ with probability }1-p\n",
    "\\end{cases}\n",
    "\\\\\n",
    "F_2\n",
    "&\n",
    "= \\begin{cases}\n",
    "F_1 + (2p-1)F_1\\text{ with probability }p\n",
    "\\\\\n",
    "F_1 - (2p-1)F_1\\text{ with probability }1-p\n",
    "\\end{cases}\n",
    "\\\\\n",
    "& \\vdots\n",
    "\\\\\n",
    "F_n\n",
    "&\n",
    "= \\begin{cases}\n",
    "F_{n-1} + (2p-1)F_{n-1}\\text{ with probability }p\n",
    "\\\\\n",
    "F_{n-1} - (2p-1)F_{n-1}\\text{ with probability }1-p\n",
    "\\end{cases}\n",
    "\\end{align*}\n",
    "\n",
    "The expected fortune after the $i$-th gamble can be found out by, where $i = 1, 2, \\ldots, n$,\n",
    "\n",
    "\\begin{align*}\n",
    "\\implies\n",
    "\\operatorname{E}[F_i]\n",
    "&\n",
    "= p\\operatorname{E}[F_{i-1} + (2p-1)F_{i-1}] + (1-p)\\operatorname{E}[F_{i-1} - (2p-1)F_{i-1}]\n",
    "\\\\ &\n",
    "= p(2p)\\operatorname{E}[F_{i-1}] + (1-p)(2-2p)\\operatorname{E}[F_{i-1}]\n",
    "\\\\ &\n",
    "= (2p^2 + 2 - 2p - 2p + 2p^2)\\operatorname{E}[F_{i-1}]\n",
    "\\\\ &\n",
    "= (4p^2 - 4p + 2)\\operatorname{E}[F_{i-1}]\n",
    "\\end{align*}\n",
    "\n",
    "which is a recurrence relation.\n",
    "The expected fortune after $n$ gambles is then given by\n",
    "\n",
    "\\begin{equation*}\n",
    "\\operatorname{E}[F_n]\n",
    "= {(4p^2 - 4p + 2)}^n\\operatorname{E}[F_0]\n",
    "= {(4p^2 - 4p + 2)}^n x\n",
    "\\end{equation*}\n",
    "\n",
    "Hence, the expected fortune after $n$ gambles, starting with $x$ units and employing the Kelly strategy, is $\\boxed{{(4p^2 - 4p + 2)}^n x}$.\n",
    "\n",
    "We can see that $(4p^2 - 4p + 2) = 2(p^2 + (p-1)^2)$ and hence the expected fortune grows exponentially with $n$.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "UTBVzEHyQfCt"
   },
   "source": [
    "---\n",
    "**Q8**\n",
    "\n",
    "Problem 24. A retired professor comes to the office at a time which is uniformly distributed between 9 a.m. and 1 p.m., performs a single task, and leaves when the task is completed. The duration of the task is exponentially distributed with parameter $\\lambda(y) = 1/(5 - y)$, where y is the length of the time interval between 9 a.m. and the time of his arrival.\n",
    "\n",
    "a) What is the expected amount of time that the professor devotes to the task?\n",
    "\n",
    "b) What is the expected time at which the task is completed?\n",
    "\n",
    "If a student comes to meet the professor at a time that is uniformly distributed between 9am to 5pm, find\n",
    "\n",
    "c) Probability that the student comes before the professor\n",
    "\n",
    "d) (Harder) Probability student comes after professor leaves. (Will need numerical integration - for which we can use scipy)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "d0vROuMuWwi3"
   },
   "source": [
    "Ans 8:\n",
    "\n",
    "\n",
    "(you can use scipy for integration)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "id": "tEgAEeyzsZ24"
   },
   "outputs": [],
   "source": [
    "import scipy"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "yll9o8bwsuhu",
    "outputId": "5211ae05-f467-442d-c363-bf531be8cb06"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "0.4299500980755558\n"
     ]
    }
   ],
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "c3L7eo7LtlsE"
   },
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "hBM1jLVedURc"
   },
   "source": [
    "---\n",
    "**Q9**\n",
    "\n",
    "Let $p_i$ be the chance of scoring penalty by $i^{th}$ player where $i=1,2,3$ and it is independent of other players\n",
    "\n",
    "Let $X$ be the rv for number of penalties scored. So $X \\in {0,1,2,3}$\n",
    "\n",
    "Find pmf of X by finding is transform first and then inverting it."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "jIKf-w6hds7y"
   },
   "source": [
    "Ans:\n",
    "\n",
    "We're given independent probabilities $p_i$ of scoring a penalty by the $i^{th}$ player, with $i=1,2,3$.\n",
    "\n",
    "With $X \\in \\{0,1,2,3\\}$ as the random variable for the number of penalties scored, we can construct the probability generating function (pgf) of $X$ as\n",
    "\n",
    "\\begin{align*}\n",
    "G_X(t)\n",
    "&\n",
    "= \\operatorname{E}[t^X]\n",
    "= \\sum_{x=0}^3 t^x P[X=x]\n",
    "\\\\ &\n",
    "= P[X=0] + t P[X=1] + t^2 P[X=2] + t^3 P[X=3]\n",
    "\\end{align*}\n",
    "\n",
    "Given that $X$ is the sum of the penalties scored by the three players, and that the probabilities of scoring a penalty are independent, we have\n",
    "\n",
    "\\begin{align*}\n",
    "G_X(t)\n",
    "&\n",
    "= (1-p_1)(1-p_2)(1-p_3) + t(p_1(1-p_2)(1-p_3) + (1-p_1)p_2(1-p_3) + (1-p_1)(1-p_2)p_3)\n",
    "\\\\ &\n",
    "\\quad + t^2(p_1p_2(1-p_3) + p_1(1-p_2)p_3 + (1-p_1)p_2p_3) + t^3 p_1p_2p_3\n",
    "\\end{align*}\n",
    "\n",
    "With this, we have\n",
    "\n",
    "\\begin{equation*}\n",
    "P[X=k]\n",
    "= \\frac{1}{k!} \\frac{d^k G_X(t)}{dt^k} \\bigg|_{t=0}\n",
    "\\end{equation*}\n",
    "\n",
    "for $k \\in \\{0,1,2,3\\}$.\n",
    "\n",
    "\\begin{align*}\n",
    "\\implies\n",
    "P[X=0]\n",
    "&\n",
    "= (1-p_1)(1-p_2)(1-p_3)\n",
    "\\\\\n",
    "\\implies\n",
    "P[X=1]\n",
    "&\n",
    "= p_1(1-p_2)(1-p_3) + (1-p_1)p_2(1-p_3) + (1-p_1)(1-p_2)p_3\n",
    "\\\\\n",
    "\\implies\n",
    "P[X=2]\n",
    "&\n",
    "= p_1p_2(1-p_3) + p_1(1-p_2)p_3 + (1-p_1)p_2p_3\n",
    "\\\\\n",
    "\\implies\n",
    "P[X=3]\n",
    "&\n",
    "= p_1p_2p_3\n",
    "\\end{align*}\n",
    "\n",
    "Hence, the pmf of $X$ is given by\n",
    "\n",
    "\\begin{equation*}\n",
    "\\boxed{\n",
    "f_X(x)\n",
    "= \\begin{cases}\n",
    "(1-p_1)(1-p_2)(1-p_3) & \\text{if } x=0\n",
    "\\\\\n",
    "p_1(1-p_2)(1-p_3) + (1-p_1)p_2(1-p_3) + (1-p_1)(1-p_2)p_3 & \\text{if } x=1\n",
    "\\\\\n",
    "p_1p_2(1-p_3) + p_1(1-p_2)p_3 + (1-p_1)p_2p_3 & \\text{if } x=2\n",
    "\\\\\n",
    "p_1p_2p_3 & \\text{if } x=3\n",
    "\\\\\n",
    "0 & \\text{otherwise}\n",
    "\\end{cases}\n",
    "}\n",
    "\\end{equation*}\n",
    "\n",
    "We can indeed verify that this is a valid probability mass function by checking that the probabilities sum to 1,\n",
    "\n",
    "\\begin{align*}\n",
    "\\sum_{x=0}^3 f_X(x)\n",
    "&\n",
    "= (1-p_1)(1-p_2)(1-p_3) + p_1(1-p_2)(1-p_3) + (1-p_1)p_2(1-p_3) + (1-p_1)(1-p_2)p_3\n",
    "\\\\ &\n",
    "\\qquad + p_1p_2(1-p_3) + p_1(1-p_2)p_3 + (1-p_1)p_2p_3 + p_1p_2p_3\n",
    "\\\\ &\n",
    "= (1-p_2)(1-p_3) + (1-p_1)p_2(1-p_3) + (1-p_1)(1-p_2)p_3\n",
    "\\\\ &\n",
    "\\qquad + p_1p_2(1-p_3) + p_1(1-p_2)p_3 + (1-p_1)p_2p_3 + p_1p_2p_3\n",
    "\\\\ &\n",
    "= 1 - p_2 - p_3 + p_2p_3 + p_2 - p_1p_2 - p_2p_3 + p_1p_2p_3 + p_3 - p_1p_3 - p_2p_3 + p_1p_2p_3\n",
    "\\\\ &\n",
    "\\qquad + p_1p_2 - p_1p_2p_3 + p_1p_3 - p_1p_2p_3 + p_2p_3 - p_1p_2p_3 + p_1p_2p_3\n",
    "\\\\ &\n",
    "= 1\n",
    "\\end{align*}\n",
    "\n",
    "as expected.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "rg9HeMi3gToI"
   },
   "source": [
    "---\n",
    "**Q10**\n",
    "Problem 41. At a certain time, the number of people that enter an elevator is a Poisson random variable with parameter $\\lambda$. The weight of each person is independent of every other person's weight, and is uniformly distributed between 100 and 200 lbs. Let $Y_i$ be the fraction of 100 by which the $i^{th}$ person exceeds 100 lbs, e.g., if the 7th person weighs 175 lbs., then Y_7=0.75. Let Y be the sum of the $Y_i$.\n",
    "\n",
    "(a) Find the transform associated with Y.\n",
    "\n",
    "(b) Use the transform to compute the expected value of Y.\n",
    "\n",
    "(c) Verify your answer to part (b) by using the law of iterated expectations.\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "kEmdA6aRgv_7"
   },
   "source": [
    "Ans:\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "w_Z8im35KTUY"
   },
   "source": [
    "*$\\underline{Law\\ of\\ Total\\ Variance}$*\n",
    "\n",
    "Just like iterated expectation law allows us to use conditional densities to more easily calculate means, we can also do something similar for Variance\n",
    "\n",
    "Let $X,Y$ be two rvs which might not be independant.\n",
    "\n",
    "Let $Z = E_{X|Y}X = $$\\int_x f_{X|Y=y}(x)dx$. Note that $Z$ is a function of $y$ and is hence a r.v. with density determined by density of $Y$\n",
    "\n",
    "$E_YZ = \\int_y f_Y(y) Z dy = \\int_y f_Y(y)\\int_x f_{X|Y=y}(x)X dx dy = \\int_x X\\int_y f_Y(y)f_{X|Y=y}(x)dy dx = \\int_x Xf_X(x)dx = E_XX$\n",
    "\n",
    "Note the above is nothing but law of iterated expectation: $E_XX = E_YE_{X|Y}X$\n",
    "\n",
    "We can now calculate variance also using conditional statistics.\n",
    "\n",
    "$Var(X) = E_XX^2 - (E_XX)^2 = E_XX^2 - E_YZ^2 + E_YZ^2 - (E_XX)^2$\n",
    "\n",
    "Using the fact from above that $E_YZ = E_XX$, and we can replace the last term to get (and writing the first term as iterated expectation, we get:\n",
    "\n",
    "$Var_X(X) = E_YE_{X|Y}X^2 - E_YZ^2 + E_YZ^2 - (E_YZ)^2 = E_Y(E_{X|Y}X^2 - (E_{X|Y}X)^2) + Var_Y(E_{X|Y}X)$\n",
    "\n",
    "\n",
    "Hence we can write the above as\n",
    "\n",
    "$Var_X(X) = E_Y(Var_{X|Y}(X)) + Var_Y(E_{X|Y}X)$\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "hZXAwxxdDJZZ"
   },
   "source": [
    "**Concentration Inequalities**\n",
    "\n",
    "These essentially allow one to obtain bounds on tail probabilities\n",
    "\n",
    "---\n",
    "*$\\underline{Markov\\ Inequality}$*\n",
    "\n",
    "If $X$ is a non-negative r.v., then\n",
    "\n",
    "$P(X \\geq a) \\leq \\frac{EX}{a}$ for $a > 0$\n",
    "\n",
    "We can see this by:\n",
    "\n",
    "$aP(X \\geq a) = \\int_a^∞ af_x(x)dx$\n",
    "\n",
    "$EX = \\int_0^∞ xf_X(x)dx = \\int_0^a xf_X(x)dx + \\int_a^∞ xf_X(x)dx$\n",
    "\n",
    "Hence $EX - aP(X\\geq a) = \\int_0^a xf_X(x)dx + \\int_a^∞ (x-a)f_X(x)dx \\geq 0$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "mCPEp8-h_HuS"
   },
   "source": [
    "---\n",
    "Example:\n",
    "\n",
    "Let $X_i$ be 1 if the $i^{th}$ person says they will vote for BJP in the national elections 2024 and 0 otherwise. Let the sequence { $X_i$} be independent and identical bernoulli rvs $\\sim\\ Ber(p)$\n",
    "\n",
    "Let $B_n = \\frac{\\sum_{i=1}^n X_i}{n}$ estimates the chance of BJP winning based on a sampling of $n$ people.\n",
    "\n",
    "We would like to know how many people to ask for their opinion inorder to get some estimate with some accuracy and confidence.\n",
    "\n",
    "\n",
    "$EB_n = EX_i = p$\n",
    "\n",
    "\n",
    "$P((B_n - p) \\geq ϵ) \\leq \\frac{p}{ϵ}$\n",
    "\n",
    "Unfortunately - we dont get any insight as to the number of samples needed from this."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "6HAC9fPyEhAi"
   },
   "source": [
    "---\n",
    "*$\\underline{Chebychev\\ Inequality}$*\n",
    "\n",
    "Let $X$ be r.v. with mean $\\mu$ and variance $\\sigma^2$.\n",
    "\n",
    "Then, the chance that it is more than $k$ std. deviations from the mean is upper bounded as:\n",
    "\n",
    "\n",
    "$P(|X-\\mu| \\geq k\\sigma) \\leq \\frac{1}{k^2}$\n",
    "\n",
    "This can be seen by considering $Z = (X-\\mu)^2$ and by Markov inequality:\n",
    "\n",
    "$P(Z \\geq (k\\sigma)^2) \\leq \\frac{EZ}{(k\\sigma)^2} = \\frac{E(X-\\mu)^2}{(k\\sigma)^2} = \\frac{1}{k^2}$\n",
    "\n",
    "and $P(Z \\geq (k\\sigma)^2) = P((X-\\mu)^2 \\geq (k\\sigma)^2) = P(|X-\\mu| \\geq k\\sigma)$\n",
    "\n",
    "We can also write this (equivalently) as:\n",
    "\n",
    "$P(|X-\\mu| \\geq c) \\leq \\frac{\\sigma^2}{c^2}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "sJWuIr0FBFLP"
   },
   "source": [
    "---\n",
    "Example continued:\n",
    "\n",
    "We know that $\\sigma_{X_i}^2 = p(1-p)$\n",
    "\n",
    "and\n",
    "\n",
    "$\\sigma_{B_n}^2 = \\frac{p(1-p)}{n}$\n",
    "\n",
    "\n",
    "From Chebychev's inequality\n",
    "\n",
    "$P(|B_n - p| \\geq ϵ) \\leq \\frac{\\sigma_{B_n}^2}{ϵ^2} = \\frac{p(1-p)}{nϵ^2} \\leq \\frac{1}{4n\\epsilon^2}$\n",
    "\n",
    "Since $p(1-p)$ achieves peak value of $1/4$ for $p=1/2$\n",
    "\n",
    "So if we want to be within $0.01$ of $p$, with $95\\%$ confidence, then\n",
    "\n",
    "$\\epsilon = 0.01$ and $\\frac{1}{4n\\epsilon^2} < 0.05$ which means:\n",
    "\n",
    " $n > 500$\n",
    "\n",
    " In other words, if you ask 500 independent people for their opinion of if they will vote for BJP, then you can say that\n",
    "\n",
    " $B_{500}-0.01 < p < B_{500}+0.01$ with 95% confidence.\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "iWUCrBA85m3z"
   },
   "source": [
    "*$\\underline{Chernoff\\ Bound}$*\n",
    "\n",
    "Let X be a rv and\n",
    "\n",
    "$M_X(s) = Ee^{sX}$ and let it be finite at $s=0$,\n",
    "\n",
    "\n",
    "For $s \\geq 0$\n",
    "\n",
    "$ X \\geq a ⇔ sX \\geq sa ⇔ e^{sX} \\geq e^{sa}$\n",
    "\n",
    "Hence\n",
    "\n",
    "$P(X \\geq a) = P(e^{sX} \\geq e^{sa}) \\leq e^{-sa}Ee^{sX}$\n",
    "\n",
    "Similarly for $s < 0 $\n",
    "\n",
    "$P(X \\leq a ) \\leq e^{-sa}Ee^{sX}$\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "id": "BsCdNgaeUx3j"
   },
   "source": [
    "---\n",
    "Example contd. (some parts from\n",
    "\n",
    "https://sachdevasushant.github.io/courses/15s-cpsc665/notes/concentration.pdf\n",
    ")\n",
    "\n",
    "for all $s \\geq 0$ by the Chernoff bound\n",
    "\n",
    "$P(B_n - p \\leq ((1+\\delta) p)) \\leq e^{-s(1+\\delta) p)}Ee^{sB_n}$\n",
    "\n",
    "$Ee^{sB_n} = Ee^{\\frac{s}{n}\\sum_1^n X_i} = (Ee^{\\frac{s}{n}X})^n$ where $X\\ \\ \\tilde\\ \\ Ber(p)$\n",
    "\n",
    "$Ee^{sX/n} = pe^{s/n} + 1-p \\leq p(e^{s/n}-1) \\leq e^{p(e^{s/n}-1)}$\n",
    "\n",
    "Therefore. $Ee^{sB_n} \\leq e^{np(e^{s/n}-1)}$\n",
    "\n",
    "and\n",
    "\n",
    "$P(B_n - p \\leq ((1+\\delta) p)) \\leq e^{-s(1+\\delta) p)}e^{np(e^{s/n}-1)}$\n",
    "\n",
    "Above is true for all $s \\geq 0$\n",
    "\n",
    "In particular for $s/n = ln(1+\\delta)$\n",
    "\n",
    "$P((B_n - p) \\leq ((1+\\delta) p)) \\leq (\\frac{e^\\delta}{(1+\\delta)^{(1+\\delta)}})^{np}$\n",
    "\n",
    "We can show that $(1+\\delta)^{1+\\delta)} \\geq e^{\\delta + \\delta^2/3}$\n",
    "\n",
    "Hence\n",
    "\n",
    "$P((B_n - p) \\leq ((1+\\delta) p)) \\leq e^{-\\frac{\\delta^2np}{3}}$\n",
    "\n",
    "Similarly\n",
    "\n",
    "$P((B_n - p) \\geq ((1-\\delta) p)) \\leq e^{-\\frac{\\delta^2np}{2}}$\n",
    "\n",
    "Hence $P(|B_n-p| \\geq \\delta p) \\leq  e^{-\\frac{\\delta^2np}{3}} + e^{-\\frac{\\delta^2np}{2}} < 2e^{-\\frac{\\delta^2np}{3}}$ from Chernoff bound.\n",
    "\n",
    "Recall from Chebychev bound:\n",
    "\n",
    "$P(|B_n-p| \\geq \\delta p) \\leq \\frac{Var(B_n)}{(\\delta p)^2} = \\frac{1-p}{np\\delta^2 }$\n",
    "\n",
    "For example, if we know that $p$ is about 0.5 and we want to estimate it within 1% (i.e. $\\delta = 0.01$) and with a confidence exceeding 99.9%, then Chebychev bound says we need at least 10million samples, while Chernoff bounds tells that about 13.6Lakh samples is enough\n",
    "(see calculation below).\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/"
    },
    "id": "F9xCL-liSu7A",
    "outputId": "e97d2d19-3c90-44c9-af9d-15246ec621f0"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "From Chebychev's ineuality, n should be atleast 10000000.0\n",
      "From Chernoff bound, n should be at least 1368162.4427175745\n"
     ]
    }
   ],
   "source": [
    "q = 0.001 #(1-confidence)\n",
    "p = 0.5\n",
    "d = 0.01 #delta (accuracy)\n",
    "\n",
    "print(f\"From Chebychev's ineuality, n should be atleast {(1-p)/q/d**2/p}\")\n",
    "\n",
    "print(f\"From Chernoff bound, n should be at least {3*np.log(2/q)*3/p/d**2}\")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "colab": {
     "base_uri": "https://localhost:8080/",
     "height": 471
    },
    "id": "kL3guzhMK5L9",
    "outputId": "4ebec008-4c29-41da-ae3e-fde90970d97b"
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "EX = 9.994065465929562, Var(X)=20.282423928324594\n"
     ]
    },
    {
     "data": {
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",
      "text/plain": [
       "<Figure size 640x480 with 1 Axes>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# testing the bounds\n",
    "nSamples = 10000\n",
    "#draw samples from chisquare distribution\n",
    "#https://numpy.org/doc/stable/reference/random/generated/numpy.random.RandomState.chisquare.html\n",
    "df = 10 #Number of degrees of freedom, obtained by adding squares of 'df' std normal variables\n",
    "X_D = rng.chisquare(df,nSamples)\n",
    "\n",
    "nbins = 20\n",
    "amin=0\n",
    "amax=40\n",
    "x_axis = np.arange(amin, amax, (amax-amin)/nbins)\n",
    "#print(x_axis)\n",
    "\n",
    "X_D_Hist,_ = np.histogram(X_D,nbins,(amin,amax))\n",
    "\n",
    "#convert to pdf\n",
    "X_pdf = X_D_Hist/nSamples\n",
    "\n",
    "print(f\"EX = {np.mean(X_D)}, Var(X)={np.var(X_D)}\")\n",
    "\n",
    "plt.plot(x_axis,X_pdf)\n",
    "plt.legend(['Chisquare density'])\n",
    "plt.ylabel(\"pdf f(x)\")\n",
    "plt.xlabel(\"X\")\n",
    "plt.show()\n"
   ]
  },
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    "id": "DpdX3BT_hIOP"
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   "source": [
    "---\n",
    "*$\\underline{Weak\\ Law\\ of\\ Large\\ Numbers}$*\n",
    "\n",
    "Let $X_1, X_2, ..., X_n$ be i.i.d with mean $\\mu$ and variance $\\sigma^2$\n",
    "\n",
    "Let $M_n=\\frac{\\sum_{i=1}^n X_i}{n}$\n",
    "\n",
    "Then $EM_n = \\mu$ and $Var(M) = \\frac{\\sigma^2}{n}$\n",
    "\n",
    "By Chebychev's inequality:\n",
    "\n",
    "So $P(|M_n - \\mu| \\geq \\epsilon ) \\leq \\frac{\\sigma^2}{n\\epsilon^2}$\n",
    "\n",
    "Hence $lim_{n→∞} P(|M_n - \\mu| \\geq \\epsilon) = 0$\n",
    "\n",
    "This is also stated $M_n$ converges to $\\mu$ in probability.\n",
    "\n",
    "*$\\underline{Convergence\\ in\\ Probability}$*\n",
    "\n",
    "Let $Y_1, Y_2, ..... Y_n$ be a sequence of r.v. and let $a$ be a real number. Then\n",
    "\n",
    "$Y_n$ converges to $a$ in probability if:\n",
    "\n",
    "$lim_{n→∞} P(|Y_n-a| \\geq \\epsilon) =0$ for all $\\epsilon > 0 $\n",
    "\n",
    "What this means is that the pdf of $Y_n$ gets increasingly concentrated around $a$ with increasing $n$.\n",
    "\n",
    "\n",
    "*$\\underline{Central\\ Limit\\ Theorem}$*\n",
    "\n",
    "Let $X_1, X_2, .., X_n$ be i.i.d with mean $\\mu$ and variance $\\sigma^2$\n",
    "\n",
    "$Z_n = \\frac{\\sum_{i=1}^n X_i - n\\mu}{\\sqrt{n}\\sigma}$.\n",
    "\n",
    "We know that $EZ_n=0$ and $Var(Z_n)=1$\n",
    "\n",
    "Then the CDF $F_{Z_n}(z)$ converges to CDF of standard normal. This is more precisely expressed as:\n",
    "\n",
    "$lim_{n\\rightarrow ∞} P(Z_n \\leq z) = \\Phi(z)$ for all $z$, where $\\Phi(z)$ is CDF of standard normal.\n",
    "\n",
    "Note that this holds regardless of pdf of $X_i$\n",
    "\n",
    "\n",
    "*$\\underline{Almost\\ Sure\\ Convergence}$*\n",
    "\n",
    "If $P(lim_{n→∞} Y_n = a) =1$, it is called a.s. convergence.\n",
    "\n",
    "The way to interpret this is at in the sample space of infinitely long sequences drawn from $Y_i$, all the outcomes are sequences with all $a$ except for a finite sub-sequence in the beginnnig.\n",
    "\n",
    "\n",
    "*Example*\n",
    "\n",
    "As an example of a sequence which converges in probability but not almost surely,\n",
    "\n",
    "Let $Y_i = 0$ with probability $\\frac{n-1}{n}$\n",
    "\n",
    "$\\ \\ = 1$ with probability $\\frac{1}{n}$\n",
    "\n",
    "Then $\\lim_{n→∞} P(Y_n=0)=1$ and converges in probability to the 0.\n",
    "\n",
    "However, it doesnt almost surely converge to 0, i.e. $P(lim_{n→∞} Y_n =0) \\neq 1$. To see this, assume that $lim_{n→∞} Y_n =0$.\n",
    "\n",
    "Then for every $\\delta$, there exists $N(\\delta)$ such that\n",
    "\n",
    "$Y_i = 0$ for all $n > N(\\delta)$.\n",
    "\n",
    "However the probability of this happening is $lim_{n→∞}(\\frac{n-1}{n})^{n-N(\\delta)} = e^{-1} < 1$\n",
    "\n",
    "\n",
    "\n"
   ]
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