{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# 2019-03-06 Scratch\n", "\n", " " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## The Weak Instruments Problem\n", "\n", "Suppose: $y = {\\beta}x + u$; $x = {\\pi}z + v$; $u = {\\delta}v + w$\n", "\n", "then:\n", "\n", "${\\beta}_{IV} = \\frac{zy}{zx} = \\frac{{\\beta}xz + uz}{xz}$\n", "\n", "${\\beta}_{IV} = \\frac{{\\beta}{\\pi}zz + {\\beta}vz + uz}{{\\pi}zz + vz} = \\beta + \\frac{uz}{{\\pi}zz + vz}$\n", "\n", "${\\beta}_{IV} = \\beta + \\frac{{\\delta}vz + wz}{{\\pi}zz + vz} = \\beta + {\\delta}\\frac{1}{({\\pi}zz/(vz)) + 1} + \\frac{wz}{{\\pi}zz + vz}$\n", "\n", "And if $\\pi = 0$, then: \n", "\n", "${\\beta}_{IV0} = \\beta + \\frac{{\\delta}vz + wz}{vz} = \\beta + \\delta + \\frac{wz}{vz}$\n", "\n", "This is the weak instruments problem. As you get more and more data, $\\frac{wz}{vz}$ is not heading for zero, and even if it were your estimated ${\\beta}_{IV0}$ is not headed for $\\beta$ but is rather headed for $\\beta + \\delta$\n", "\n", "Now we would like to apply a file-drawer problem filter: $\\pi$ is zero, but in the sample you have the calculated $zv$ is large enough that you are happy to run and report the regression. What can we then say?\n", "\n", "----\n", "\n" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.6.6" } }, "nbformat": 4, "nbformat_minor": 2 }