Added assignment_1 and missing-semester-lec6 files

Gowri/assignment_1/README.md

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+# Conversion of .jpeg, .png to .jpg
+
+#### Multiple government and exam application sites require image uploading, specifically in .jpg file format. Mutiple free file format conversion sites on the internet have a limit on the number of file conversion allowed at a time. To ease the conversion process I've written an elementary Bash script to perfom the conversion.
+
+#### I've demonstrated the same in a temporary directory on a few image files. sed command is used to remove Windows-style carriage  return characters (\r) from the script file before running.

Gowri/assignment_1/conversion.sh

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+#!/usr/bin/env bash
+
+for file in *.jpeg *.png; do
+    filename=${file%.*}
+    mv "$file" "$filename.jpg"
+done
+

Gowri/assignment_2/ProbStats1.ipynb

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+{
+  "nbformat": 4,
+  "nbformat_minor": 0,
+  "metadata": {
+    "colab": {
+      "provenance": []
+    },
+    "kernelspec": {
+      "name": "python3",
+      "display_name": "Python 3"
+    },
+    "language_info": {
+      "name": "python"
+    }
+  },
+  "cells": [
+    {
+      "cell_type": "markdown",
+      "source": [
+        "##Q1\n",
+        "\n",
+        "Given $Y=|Z|$\n",
+        "$$\n",
+        "\\begin{align}\n",
+        "& Z \\sim N(0,1) \\\\\n",
+        "& f_z(z)=\\frac{1}{\\sqrt{2 \\pi}} e^{-\\frac{z^2}{2}}\n",
+        "\\end{align}\n",
+        "$$\n",
+        "\n",
+        "Using CDF,\n",
+        "$$\n",
+        "\\begin{align}\n",
+        "F_Y(y) & =P(Y \\leq y) \\\\\n",
+        " &=P(|Z| \\leq y) \\\\\n",
+        " &=P(-y \\leq Z \\leq y)  \\\\\n",
+        "F_Y(y) & =F_z(y)-F_z(-y)\n",
+        "\\end{align}\n",
+        "$$\n",
+        "\n",
+        "Differentiating both sides with respect to y to get the PDF,\n",
+        "$$\n",
+        "\\begin{align}\n",
+        "\\frac{d F_Y(y)}{dy} = f_Y(y) \\\\\n",
+        "f_Y(y) = f_Z(y) + f_Y(-y) \\\\\n",
+        "\\end{align}\n",
+        "$$\n",
+        "\n",
+        "$$\n",
+        "\\begin{align}\n",
+        "= \\frac{2}{\\sqrt{2\\pi}} e^{-\\frac{y^2}{2}} \\\\\n",
+        "= {\\sqrt\\frac{2}{\\pi}} e^{-\\frac{y^2}{2}}, y \\geq 0\n",
+        "\\end{align}\n",
+        "$$\n",
+        "\n",
+        "Finding Expectation,\n",
+        "\n",
+        "$$\n",
+        "\\begin{align}\n",
+        "E[Y] = \\int_{-\\infty}^{+\\infty} yf_Y(y)dy \\\n",
+        "& = \\int_{-\\infty}^{+\\infty} y \\sqrt{\\frac{2}{\\pi}} e^{-\\frac{y^2}{2}} dy\n",
+        "\\end{align}\n",
+        "$$\n",
+        "\n",
+        "Substituting $\\frac{y^2}{2} = t$\n",
+        "\n",
+        "$$\n",
+        "\\frac{2y}{2} dy = dt\n",
+        "$$\n",
+        "\n",
+        "$$\n",
+        "\\begin{align}\n",
+        "E[Y] = \\sqrt{\\frac{2}{\\pi}} \\int_{0}^{\\infty} e^{-t} dt\n",
+        "= \\sqrt{\\frac{2}{\\pi}} \\left( -e^{-t} \\right) \\bigg|_{0}^{\\infty} = \\sqrt{\\frac{2}{\\pi}}\n",
+        "\\end{align}\n",
+        "$$\n",
+        "\n",
+        "Finding $E[Y^2]$,\n",
+        "$$\n",
+        "\\begin{align}\n",
+        "E[Y^2]=E[Z^2]\\\n",
+        "& =\\sigma_Z^2 +\\mu_z^2 =1\n",
+        "\\end{align}\n",
+        "$$\n",
+        "\n",
+        "Finding Variance,\n",
+        "$$\n",
+        "\\text{Var}(Y) = E[Y^2] - \\left( E[Y] \\right)^2 \\\\\n",
+        "= 1 - \\frac{2}{\\pi}\n",
+        "$$\n",
+        "\n",
+        "Finding $E[Y^3]$,\n",
+        "$$\n",
+        "\\begin{align}\n",
+        "E[Y^3] = \\int_{-\\infty}^{+\\infty} y^3 \\, f_Y(y) \\, dy\n",
+        "= \\int_{-\\infty}^{0} y^3 \\times 0  dy + \\int_{0}^{\\infty} y^3 \\sqrt{\\frac{2}{\\pi}} e^{-\\frac{y^2}{2}} \\, dy\n",
+        "\\end{align}\n",
+        "$$\n",
+        "\n",
+        "Substituting  $\\frac{y^2}{2} = t$,  \n",
+        "$$\n",
+        "\\frac{2y}{2} dy = dt\n",
+        "$$\n",
+        "\n",
+        "$$\n",
+        "E[Y^3] = \\int_{0}^{\\infty} 2t \\sqrt{\\frac{2}{\\pi}} e^{-t} \\, dt  = 2 \\sqrt{\\frac{2}{\\pi}} \\int_{0}^{\\infty} t e^{-t} \\, dt\n",
+        "$$\n",
+        "\n",
+        "$$\n",
+        "E[Y^3] = 2 \\sqrt{\\frac{2}{\\pi}} \\times 1\n",
+        "$$\n"
+      ],
+      "metadata": {
+        "id": "ykfcZwlAqB1q"
+      }
+    },
+    {
+      "cell_type": "markdown",
+      "source": [
+        "##Q2\n",
+        "(a)  \n",
+        "By LOTUS,\n",
+        "\n",
+        "$E[g(X)] = \\sum_{x} g(x) f_X(x) $\n",
+        "\n",
+        "$$\n",
+        "E[X g(X)] = \\sum_{x} x g(X) \\frac{\\lambda^x e^{-\\lambda}}{x!}\n",
+        "= e^{-\\lambda} \\sum_{x=1}^\\infty  g(X) \\frac{\\lambda^x}{(x-1)!} + 0\n",
+        "$$\n",
+        "\n",
+        "$$\n",
+        "\\lambda E[g(X+1)] = \\sum_{x} g(X+1) \\frac{\\lambda^{x+1} e^{-\\lambda}}{x!}\n",
+        "= e^{-\\lambda} \\sum_{x=0}^\\infty  g(X+1) \\frac{\\lambda^{x+1}}{x!}\n",
+        "$$\n",
+        "\n",
+        "Both expansions give equivalent results. Hence proved.\n",
+        "\n",
+        "(b)\n",
+        "\n",
+        "$$\n",
+        "E(X) = \\lambda \\quad \\text{and} \\quad \\text{Var}(X) = \\lambda\n",
+        "$$\n",
+        "\n",
+        "For $g(X) = X^2$:\n",
+        "\n",
+        "$$\n",
+        "E(X^3) = \\lambda E[(X+1)^2] = \\lambda [E(X^2) + 1 + 2E(X)]\n",
+        "$$\n",
+        "\n",
+        "$Var(X)= E[X^2]-(E[X])^2$  \n",
+        "$E[X^2]= \\lambda+\\lambda^2$\n",
+        "\n",
+        "$$\n",
+        "E(X^3) = \\lambda [1 + 2\\lambda + \\lambda + \\lambda^2]\n",
+        "= \\lambda^3 + 3\\lambda^2 + \\lambda\n",
+        "$$"
+      ],
+      "metadata": {
+        "id": "Jx81KSD877va"
+      }
+    },
+    {
+      "cell_type": "markdown",
+      "source": [
+        "##Q3\n",
+        "\n",
+        "Given,\n",
+        "\n",
+        "$$\n",
+        "T_1 \\sim \\text{Exp}(\\lambda_1), \\quad T_2 \\sim \\text{Exp}(\\lambda_2)\n",
+        "$$\n",
+        "\n",
+        "The exponential distribution probability density function:\n",
+        "$$\n",
+        "f_X(x) =\n",
+        "\\begin{cases}\n",
+        "\\lambda e^{-\\lambda x}, & x > 0, \\lambda > 0 \\\\\n",
+        "0\\quad \\quad, & \\text{otherwise}\n",
+        "\\end{cases}\n",
+        "$$\n",
+        "\n",
+        "Calculating $P( T_1 \\lt T_2 )$,  \n",
+        "\n",
+        "Since the random variables are independent,\n",
+        "$$\n",
+        "P(T_1 < T_2) = \\int_{0}^{\\infty} \\int_{t_1}^{\\infty} f_{T_1}(t_1) f_{T_2}(t_2) \\, dt_2 \\, dt_1\n",
+        "$$\n",
+        "\n",
+        "$$\n",
+        "= \\lambda_1 \\lambda_2 \\int_{0}^{\\infty} \\int_{t_1}^{\\infty} e^{-\\lambda_1 t_1 - \\lambda_2 t_2} \\, dt_2 \\, dt_1\n",
+        "$$\n",
+        "\n",
+        "First integrating over $t_2$:\n",
+        "\n",
+        "$$\n",
+        "= \\lambda_1 \\int_{0}^{\\infty} e^{-\\lambda_1 t_1} \\left( \\int_{t_1}^{\\infty} \\lambda_2 e^{-\\lambda_2 t_2} \\, dt_2 \\right) dt_1 \\\\\n",
+        "= \\lambda_1 \\int_{0}^{\\infty} e^{-\\lambda_1 t_1} e^{-\\lambda_2 t_1} \\, dt_1 \\\\\n",
+        "= \\lambda_1 \\int_{0}^{\\infty} e^{-(\\lambda_1 + \\lambda_2) t_1} \\, dt_1 \\\\\n",
+        "= \\frac{\\lambda_1}{\\lambda_1 + \\lambda_2}\n",
+        "$$"
+      ],
+      "metadata": {
+        "id": "N5dPrGkV_Iph"
+      }
+    },
+    {
+      "cell_type": "markdown",
+      "source": [
+        "##Q4\n",
+        "(a)  \n",
+        "Let $X$ be a binary random variable representing the transmitted message:\n",
+        "$$\n",
+        "X = \\begin{cases}\n",
+        "1 & \\text{if message is } yes \\\\\n",
+        "0 & \\text{if message is } no\n",
+        "\\end{cases}\n",
+        "$$\n",
+        "\n",
+        "The noisy channel is modeled as additive Gaussian noise:\n",
+        "$Y \\sim \\mathcal{N}(0, \\sigma^2)$\n",
+        "\n",
+        "The received signal $Z$ is,\n",
+        "$Z = X + Y$  \n",
+        "\n",
+        "The probability of correct interpretation is:\n",
+        "$$\n",
+        "\\begin{align}\n",
+        "P(\\text{correct}) &= P(Z > \\tfrac{1}{2} \\mid X = 1)P(X = 1) \\\\\n",
+        "&\\quad + P(Z < \\tfrac{1}{2} \\mid X = 0)P(X = 0)\n",
+        "\\end{align}\n",
+        "$$\n",
+        "\n",
+        "$$\n",
+        "\\begin{align}\n",
+        "P(\\text{correct}) &= P(X = 1)P(Y > -\\tfrac{1}{2}) + P(X = 0)P(Y < \\tfrac{1}{2}) \\\\\n",
+        "&= P(X = 1)[1 - \\Phi(-\\tfrac{1}{2\\sigma})] + P(X = 0)\\Phi(\\tfrac{1}{2\\sigma}) \\\\\n",
+        "&= \\Phi(\\tfrac{1}{2\\sigma})[P(X = 1) + P(X = 0)] \\\\\n",
+        "&\\text{Since it is sure that the message is transmitted, } P(X = 1) + P(X = 0) = 1\\\\\n",
+        "&= \\Phi(\\tfrac{1}{2\\sigma})\n",
+        "\\end{align}\n",
+        "$$\n",
+        "where $\\Phi$ is the standard normal CDF.  \n",
+        "Hence, the probability that the received signal falls in the correct decision region is $\\Phi(\\frac{1}{2\\sigma})$  \n"
+      ],
+      "metadata": {
+        "id": "6tRmfWmVNuc1"
+      }
+    },
+    {
+      "cell_type": "markdown",
+      "source": [
+        "Behavior of the probability as noise varies extremely:  \n",
+        "Variation in noise can be interpreted using the variance of the distribution.  \n",
+        "As the noise becomes very small,  $\\sigma \\to 0$:  \n",
+        "$\\lim_{\\sigma \\to 0} \\Phi(\\frac{1}{2\\sigma}) = \\Phi(+\\infty) = 1$\n",
+        "\n",
+        "Intuitively, the probability of correctly interpreting the message tends to 1 as noise becomes negligible.\n",
+        "As noise becomes very large,  $\\sigma \\rightarrow \\infty$:  \n",
+        "$\\lim_{\\sigma \\to \\infty} \\Phi(\\frac{1}{2\\sigma}) = \\Phi(0) = \\frac{1}{2}$\n",
+        "\n",
+        "Intuitively, we understand that as $\\sigma \\rightarrow \\infty$ the spread of the data is very large and the actual message and noise conflate into being correctly interpreted only half the time. The probability degrades to $\\frac{1}{2}$ ."
+      ],
+      "metadata": {
+        "id": "v3e2SLMjWbjJ"
+      }
+    }
+  ]
+}