{
"cells": [
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"source": [
"# Problem Set 1: Hodrick-Prescott Filter [](https://mybinder.org/v2/gh/tobiasraabe/time_series/master?filepath=docs%2Fproblem_sets%2Fproblem_set_1.ipynb)\n",
"\n",
"## Introduction\n",
"\n",
"A commonly used approach to decompose a time series into a permanent component $y^p$ and a transitory (cyclical) component $y^c$ goes back to Hodrick and Prescott (1997). Suppose you have a sequence of data, $y_t$ with $t = 1,...,T$ (i.e. there are $T$ total observations). Our objective is to find a trend $\\{y^p_t\\}^T_{t=1}$ to minimize the following objective function:\n",
"\n",
"$$\n",
"f = \\underset{y^p_t}{\\min} \\sum^T_{t=1} (y_t - y^p_t)^2 + \\lambda \\sum^{T-1}_{t=2} [(y^p_{t+1} - y^p_t) - (y^p_t - y^p_{t-1})]^2\n",
"$$\n",
"\n",
"where the parameter $\\lambda \\geq 0$."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exercise 1\n",
"\n",
"**Question**: Provide a verbal interpretation of this objective function. What are the trade-offs?*\n",
"\n",
"**Answer**: The first term of the function is the sum of quadratic differences between the realization $y_t$ and the trend component $y^p_t$. If the function would only consist of this component, the minimization problem would yield $y^p_t = y_t$. Therefore, $y^p_t$ has also to minimize the squared difference to its neighbours $y^p_{t+1}$ and $y^p_{t-1}$ so that all trend components are similar to each other. The weighting parameter $\\lambda$ controls how important it is that the trend values are similar to each other. A higher $\\lambda$ will lead to more equal trend components meaning a smoother trend, but it might also wash out important differences in the trend."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exercise 2\n",
"\n",
"**Question**: Prove that if $\\lambda = 0$, the solution is $y^p_t = y_t \\forall t$, i.e. the trend and the actual series are identical.\n",
"\n",
"**Answer**: If $\\lambda = 0$ the former minimization problem simplifies to\n",
"\n",
"$$\n",
"\\underset{y^p_t}{\\min} \\sum^T_{t=1} (y_t - y^p_t)^2\n",
"$$\n",
"\n",
"The first derivative with respect to $y^p_t$ is given by\n",
"\n",
"$$\n",
"\\begin{align}\n",
"\\frac{d f}{d y^p_t} &= 2 (y_t - y^p_t)\\\\\n",
" &= y_t - y^p_t\\\\\n",
" &= 0\n",
"\\end{align}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exercise 3\n",
"\n",
"**Question**: Prove that as $\\lambda \\to \\infty$, the filtered series $\\{y^p_t\\}^T_{t=1}$ is linear (i.e. $y^p_t = \\beta t$ for some $\\beta$)\n",
"\n",
"**Answer**: If the formula is divided by $\\lambda$ and $\\lambda \\to \\infty$ the first term of the equation approaches 0."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exercise 4\n",
"\n",
"**Question**: For the more general case in which $0 < \\lambda < \\infty$, derive analytical conditions which implicitly define the trend. To do this, take the derivative with respect to $y^p_t$ for each $t = 1, \\dots, T$ and set them equal to zero, yielding $T$ first order conditions.\n",
"\n",
"**Answer**: Here are the three different derivations:\n",
"\n",
"For $y^p_1$:\n",
"$$\n",
"\\begin{align}\n",
" - (y_1 - y^p_1) + \\lambda (y^p_1 - 2 y^p_2 + y^p_3) &= 0\\\\\n",
" (1 + \\lambda) y^p_1 - 2 \\lambda y^p_2 + \\lambda y^p_3 &= y_1\n",
"\\end{align}\n",
"$$\n",
"\n",
"For $y^p_2$:\n",
"$$\n",
"\\begin{align}\n",
" - (y_2 - y^p_2) + \\lambda [-2(y^p_1 - 2 y^p_2 + y^p_3) + (y^p_2 - 2 y^p_3 + y^p_4)] &= 0\\\\\n",
" -2 \\lambda y^p_1 + (1 - 5\\lambda) y^p_2 - 4 \\lambda y^p_3 + \\lambda y^p_4 &= y_2\n",
"\\end{align}\n",
"$$\n",
"\n",
"For $y^p_t$ for $3 \\leq t \\leq T-2$:\n",
"$$\n",
"\\begin{align}\n",
" - (y_t - y^p_t) + \\lambda [(y^p_t - 2y^p_{t-1} + y^p_{t-2}) - 2(y^p_{t+1} - 2 y^p_t + y^p_{t-1}) + (y^p_{t+2} - 2y^p_{t+1} + y^p_{t}) &= 0\\\\\n",
" \\lambda y^p_{t-2} - 4 \\lambda y^p_{t-1} + (1 + 6 \\lambda) y^p_t - 4 \\lambda y^p_{t+1} + \\lambda y^p_{t+2} &= y_t\n",
"\\end{align}\n",
"$$\n",
"\n",
"for $y^p_{T-1}$:\n",
"$$\n",
"\\begin{align}\n",
" - (y_{T-1} - y^p_{T-1}) + \\lambda [2(y^p_{T-3} - 2y^p_{T-2} + y^p_{T-1}) - 2(y^p_T - 2y^p_{T-1} + y^p_{T-2})] &= 0\\\\\n",
" \\lambda y^p_{T-3} - 4 \\lambda y^p_{T-2} + (1 + 5\\lambda) y^p_{T-1} - 2\\lambda y^p_T&= y_{T-1}\n",
"\\end{align}\n",
"$$\n",
"\n",
"for $y^p_{T}$:\n",
"$$\n",
"\\begin{align}\n",
" - (y_T - y^p_T) + \\lambda(y^p_T - 2y^p_{T-1} + y^p_{T-2}) &= 0\\\\\n",
" \\lambda y^p_{T-2} - 2\\lambda y^p_{T-1} + 2\\lambda y^p_T &= y_T\n",
"\\end{align}\n",
"$$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exercise 5\n",
"\n",
"**Question**: Write a *Julia* function to find the trend taking the actual data series $y_t$ and parameter $\\lambda$ as inputs. To do this, express the first order conditions in Exercise 4 in matrix form as follows:\n",
"\n",
"$$\n",
"\\begin{align}\n",
"\\Lambda Y^t &= Y\\\\\n",
"Y^t &= \\Lambda^{-1}Y\n",
"\\end{align}\n",
"$$\n",
"\n",
"where $\\Lambda$ is a $T \\times T$ matrix whose elements are functions of $\\lambda$, $Y^t$ is a $T \\times 1$ vector equal to $[y^t_1, y^t_2, \\dots, y^t_T]'$ and $Y$ is a $T \\times 1$ vector equal to $[y_1, y_2, \\dots, y_T]'$."
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"function hodrick_prescott_filter(y, lambda)\n",
" T = size(y, 1)\n",
" matrix = zeros(T, T)\n",
" \n",
" matrix[1, 1:3] = [1 + lambda, -2 * lambda, lambda]\n",
" matrix[2, 1:4] = [-2 * lambda, 1 + 5 * lambda, -4 * lambda, lambda]\n",
" \n",
" for i = 3 : T - 2\n",
" matrix[i, i-2 : i+2] = [lambda, -4*lambda, 1 + 6 * lambda, -4 * lambda, lambda]\n",
" end\n",
" \n",
" matrix[T-1, T-3:T] = [lambda, -4 * lambda, 1 + 5 * lambda, -2 * lambda]\n",
" matrix[T, T-2:T] = [lambda, -2 * lambda, 1 + lambda]\n",
" \n",
" trend = matrix \\ y\n",
" cycle = y - trend\n",
" \n",
" return trend, cycle\n",
"end;"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Exercise 6"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {},
"outputs": [],
"source": [
"using DataFrames, Plots, SparseArrays, XLSX"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
"outputs": [],
"source": [
"y = float.(XLSX.readdata(\"problem_set_1_data/us_real_gdp.xlsx\", \"FRED Graph\", \"B12:B295\"));"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [],
"source": [
"trend, cycle = hodrick_prescott_filter(y, 1600);"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {},
"outputs": [],
"source": [
"timeline = 1947:0.25:2017.75;"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {},
"outputs": [
{
"data": {
"image/svg+xml": [
"\n",
"\n"
]
},
"execution_count": 13,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"plot(\n",
" plot(timeline, cycle, label=\"Cycle\", color=:blue),\n",
" plot(timeline, trend, label=\"Trend\", color=:red),\n",
" link=:x, layout=(2, 1),\n",
" legend=:topleft,\n",
")"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"To validate the method, we take another implementation of the HP-Filter from http://www.econforge.org/posts/2014/juil./28/cef2014-julia/ (note the API change introduced by https://github.com/JuliaLang/julia/pull/23757 and the new form explained in https://github.com/JuliaLang/julia/pull/23757/files#diff-7904f4ddd9158030529e0ed5ee8707eeR1771)."
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {},
"outputs": [],
"source": [
"function hp_filter(y, lambda)\n",
" n = length(y)\n",
" @assert n >= 4\n",
"\n",
" diag2 = lambda*ones(n-2)\n",
" diag1 = [ -2lambda; -4lambda*ones(n-3); -2lambda ]\n",
" diag0 = [ 1+lambda; 1+5lambda; (1+6lambda)*ones(n-4); 1+5lambda; 1+lambda ]\n",
"\n",
" D = spdiagm(-2 => diag2, -1 => diag1, 0 => diag0, 1 => diag1, 2 => diag2)\n",
"\n",
" trend = D \\ y\n",
" cycle = y - trend\n",
" \n",
" return trend, cycle\n",
"end;"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [],
"source": [
"trend, cycle = hp_filter(y, 1600);"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {},
"outputs": [
{
"data": {
"image/svg+xml": [
"\n",
"\n"
]
},
"execution_count": 9,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"plot(\n",
" plot(timeline, cycle, label=\"Cycle\", color=:blue),\n",
" plot(timeline, trend, label=\"Trend\", color=:red),\n",
" link=:x, layout=(2, 1), legend=:topleft,\n",
")"
]
}
],
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"display_name": "Julia 1.0.0",
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"file_extension": ".jl",
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