{ "cells": [ { "cell_type": "markdown", "id": "306d1152", "metadata": {}, "source": [ "# Exercise 1 #" ] }, { "cell_type": "markdown", "id": "5e7f658e", "metadata": {}, "source": [ "A fixed-coupond bond is a financial instrument characterised with maturity $T$-years, coupon $c$ (which is assumed to be paid out annually) and principal $P$. For such a bond, a bond holder gets $cP$ amound of money at the end of each year before maturity, and at the maturity they get $cP +P$. The value of the bond at time $t=0$ is equal to the present value of the underlying cashflows.\n", "\n", "1. Assume that the interest rate at time $t=0$ is flat and equal $r$. Write down a formula for calculating the value of the bond at time $t=0$ given the parameters $r, T, c, P$. (4p)" ] }, { "cell_type": "markdown", "id": "c6b90fed", "metadata": {}, "source": [ "## Solution exercise 1.1" ] }, { "cell_type": "markdown", "id": "7ce6a1b1", "metadata": {}, "source": [ "Blackboard" ] }, { "cell_type": "markdown", "id": "cb930eda", "metadata": {}, "source": [ "2. Write a function (script) which calculates the value of the bond, given parameters $r,c,P,T$. Run the function for $r=5\\%$, $c=4\\%$, $T=10$ and $P=1000$. Run the function again changing different parameters and plot the results (paramater agains the value of the bond), first with the interest rate ($r=1,2,3,4,5,6,7,8,9,10\\%$), then coupon ($c=1,2,3,4,5,6,7,8,9,10\\%$) and at last, the maturity of the bond ($T=2,4,6,8,10,12,14,16,18,20$). What does the results tell you about the relation between different level of parameters and the value of the bond? (8p)" ] }, { "cell_type": "markdown", "id": "7194049a", "metadata": {}, "source": [ "## Solution Exercise 1.2" ] }, { "cell_type": "code", "execution_count": 1, "id": "5b9be89e", "metadata": {}, "outputs": [], "source": [ "### INPUTS\n", "maturity<-10\n", "principal<-1000\n", "coupon<-0.04\n", "interest<-0.05" ] }, { "cell_type": "code", "execution_count": 3, "id": "4d57866c", "metadata": {}, "outputs": [ { "data": { "text/html": [ "922.782650708152" ], "text/latex": [ "922.782650708152" ], "text/markdown": [ "922.782650708152" ], "text/plain": [ "[1] 922.7827" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "#define\n", "calculate_bond_value<-function(maturity,principal,coupon,interest){\n", " #define the cashflows\n", " cashflow<-rep(coupon*principal,times=maturity)\n", " cashflow[maturity]<-cashflow[maturity]+principal\n", " cashflow <-cbind(1:maturity,cashflow)\n", " \n", " #discount the cashflows\n", " discounted_cashflow<-cashflow\n", " for (i in 1:maturity){\n", " discounted_cashflow[i,2]<-cashflow[i,2]/(1+interest)^cashflow[i,1]\n", " }\n", " \n", " #calculate value: OUTPUT\n", " value<-sum(discounted_cashflow[,2])\n", "\n", " return(value)\n", "}\n", "\n", "calculate_bond_value(maturity,principal,coupon,interest)" ] }, { "cell_type": "code", "execution_count": 6, "id": "b17235b6", "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "
A matrix: 10 × 2 of type dbl
0.011284.1391
0.021179.6517
0.031085.3020
0.041000.0000
0.05 922.7827
0.06 852.7983
0.07 789.2926
0.08 731.5967
0.09 679.1171
0.10 631.3260
\n" ], "text/latex": [ "A matrix: 10 × 2 of type dbl\n", "\\begin{tabular}{ll}\n", "\t 0.01 & 1284.1391\\\\\n", "\t 0.02 & 1179.6517\\\\\n", "\t 0.03 & 1085.3020\\\\\n", "\t 0.04 & 1000.0000\\\\\n", "\t 0.05 & 922.7827\\\\\n", "\t 0.06 & 852.7983\\\\\n", "\t 0.07 & 789.2926\\\\\n", "\t 0.08 & 731.5967\\\\\n", "\t 0.09 & 679.1171\\\\\n", "\t 0.10 & 631.3260\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "A matrix: 10 × 2 of type dbl\n", "\n", "| 0.01 | 1284.1391 |\n", "| 0.02 | 1179.6517 |\n", "| 0.03 | 1085.3020 |\n", "| 0.04 | 1000.0000 |\n", "| 0.05 | 922.7827 |\n", "| 0.06 | 852.7983 |\n", "| 0.07 | 789.2926 |\n", "| 0.08 | 731.5967 |\n", "| 0.09 | 679.1171 |\n", "| 0.10 | 631.3260 |\n", "\n" ], "text/plain": [ " [,1] [,2] \n", " [1,] 0.01 1284.1391\n", " [2,] 0.02 1179.6517\n", " [3,] 0.03 1085.3020\n", " [4,] 0.04 1000.0000\n", " [5,] 0.05 922.7827\n", " [6,] 0.06 852.7983\n", " [7,] 0.07 789.2926\n", " [8,] 0.08 731.5967\n", " [9,] 0.09 679.1171\n", "[10,] 0.10 631.3260" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "### INPUT \n", "interest_sen<-1:10/100\n", "\n", "#define the matrix with the bond_values for different interest rates\n", "values<-matrix(nrow=length(interest_sen),ncol=2)\n", "\n", "# calculate the bond value\n", "for (i in 1:length(interest_sen)){\n", " values[i,1]<-interest_sen[i]\n", " values[i,2]<-calculate_bond_value(maturity,principal,coupon,interest_sen[i])\n", "}\n", "\n", "###output\n", "values" ] }, { "cell_type": "code", "execution_count": 7, "id": "8acced40", "metadata": {}, "outputs": [ { "data": { "image/png": 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"text/plain": [ "plot without title" ] }, "metadata": { "image/png": { "height": 420, "width": 420 } }, "output_type": "display_data" } ], "source": [ "plot(values)" ] }, { "cell_type": "code", "execution_count": 8, "id": "8f3fdd0f", "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "
A matrix: 10 × 2 of type dbl
0.01 691.1306
0.02 768.3480
0.03 845.5653
0.04 922.7827
0.051000.0000
0.061077.2173
0.071154.4347
0.081231.6520
0.091308.8694
0.101386.0867
\n" ], "text/latex": [ "A matrix: 10 × 2 of type dbl\n", "\\begin{tabular}{ll}\n", "\t 0.01 & 691.1306\\\\\n", "\t 0.02 & 768.3480\\\\\n", "\t 0.03 & 845.5653\\\\\n", "\t 0.04 & 922.7827\\\\\n", "\t 0.05 & 1000.0000\\\\\n", "\t 0.06 & 1077.2173\\\\\n", "\t 0.07 & 1154.4347\\\\\n", "\t 0.08 & 1231.6520\\\\\n", "\t 0.09 & 1308.8694\\\\\n", "\t 0.10 & 1386.0867\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "A matrix: 10 × 2 of type dbl\n", "\n", "| 0.01 | 691.1306 |\n", "| 0.02 | 768.3480 |\n", "| 0.03 | 845.5653 |\n", "| 0.04 | 922.7827 |\n", "| 0.05 | 1000.0000 |\n", "| 0.06 | 1077.2173 |\n", "| 0.07 | 1154.4347 |\n", "| 0.08 | 1231.6520 |\n", "| 0.09 | 1308.8694 |\n", "| 0.10 | 1386.0867 |\n", "\n" ], "text/plain": [ " [,1] [,2] \n", " [1,] 0.01 691.1306\n", " [2,] 0.02 768.3480\n", " [3,] 0.03 845.5653\n", " [4,] 0.04 922.7827\n", " [5,] 0.05 1000.0000\n", " [6,] 0.06 1077.2173\n", " [7,] 0.07 1154.4347\n", " [8,] 0.08 1231.6520\n", " [9,] 0.09 1308.8694\n", "[10,] 0.10 1386.0867" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "### INPUT \n", "coupon_sen<-1:10/100\n", "\n", "#define the matrix with the coupon_sen for different interest rates\n", "values<-matrix(nrow=length(interest_sen),ncol=2)\n", "\n", "# calculate the bond value\n", "for (i in 1:length(coupon_sen)){\n", " values[i,1]<-coupon_sen[i]\n", " values[i,2]<-calculate_bond_value(maturity,principal,coupon_sen[i],interest)\n", "}\n", "\n", "###output\n", "values" ] }, { "cell_type": "code", "execution_count": 9, "id": "36255b9c", "metadata": {}, "outputs": [ { "data": { "image/png": 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zlfWdaGbj+1mUhIMIzWkDamK6W6b8jtdPbQ5KT3bCYSEgyj93Wkjyf8KP9T6+Mz\nfOr7/2s3j5BgGHc+aOy7Hfa3ExIMwyf2AQIICRBASIAAQgIEtNOQdm+P79dHe9MeQ6pZkKtU\nj1/siuNDoL1phyHVnN9rScnGZ0494Zv4PQbam3YY0pIeX4YO+wZNjN9joL1phyHlzYscV6Tw\nLpSQ0g5D6rgictytPojfg6CdaYchpb0SOZarD+P3IGhn2mFIZ9wROb6UVhm/B0E70w5DeqTr\n+tCh4qSfx+8x0N60w5AOjes67/W3lvQ9ze6jzoBWaYchWYGHBqYm9buDJ3aQ0x5DCjp4IL5f\nH+1NOw0JkEVIgABCAgQQEiCAkAABhAQIICRAACEBAggJEEBIgABCAgQQEiCAkAABhAQIICRA\nACEBAggJEEBIgABCAgQQEiCAkAABhAQIICRAACEBAggJEEBIgABCAgQQEiCAkAABhAQIICRA\nACEBAggJEEBIgADdIQU2rSwuXrUlyixCgmH0hlQ+s5cKy5m/324eIcEwWkMq66v65c8tKpo9\nIUv1L7eZSEgwjNaQpvifrxvVLvUV2EwkJBhGa0iZkxvH47NtJhISDKM1JP+CxvG8FJuJhATD\naA0pd1zjeEwfm4mEBMNoDanAt6gqMqqcowptJhISDKM1pIqBKn1U/vRpk0Z2UsPtUiEkGEbv\n60jVi/OSQi8j+Ycsq7WbR0gwjPYtQgc+LykprY4yiZBgGLYIAQLYIgQIYIsQIIAtQoAA72wR\nCqxe2aCAkGAW72wR2pSmmrD9EQrwGm9uEXpbRTtBDniKN7cIHS2kQ+uWL193qK0PD8SNN7cI\nHSWkklNUnz7qlJK2Pj4QL97cItRySJ9mTNxmWdsmZnzW5gUA8eHNLUIth3TZeYHQ4dB5lztZ\nABAHWkPavCvGiS2GVNPxT5HByx1r2roCID60hqQ6/jq2s3EthlSm6p7SfarK2roCID70htQn\n6YdvxDKxxZC+U2sig3d8lW1dARAfekMqXDtAjXoz+sSWf0YaeEvkOGtgWxcAxInmkKza+45R\nI57cHWViyyH9PqU4dChOeaGtCwDiRHdIllW5sIdKOn3Kr5fYTDzK60gLk0bccsuIpIVtfXwg\nXvSHZFn7nr64s1J2X+VoOxs+Khw9uvCjtj48EDduhBR62A+e+63NRPbawTAuhRQFIcEwWkNK\nnR3jREKCYbz5QWOEBMMQEiCAkAABhAQIICRAACEBAggJEEBIgABCAgQQEiCAkAABhAQIICRA\nACEBArwZ0loFGGZtq/+axz8k68N18TXp5Kc9ZGaq2yto6mF1t9tLaOq4yW6voKkR5x/tr9SH\nrf9briGkeJt9jtsraOpPnd1eQVO7lKfeFeMHD7u9gqby8wW/GCEJIyQbhORlhHR0hGSDkA5H\nSEdHSDYI6XCEdHSEZIOQDkdIR0dINgjpcIR0dIRkg5AOR0hHR0g2COlwhHR0hGSDkA5HSEdH\nSDYI6XB3jXZ7BU291s3tFTS117fR7SU0derjbq+gqeuuE/xiCRBS5Tdur6CpQ5vdXsFhNrm9\ngMNs9dQvApSXC36xBAgJcB8hAQIICRBASIAAQgIEEBIggJAAAYQECCAkQAAhAQIICRBASIAA\nQgIEEBIggJAAAYQECDAwpIqCXP9xU8pavKJ8Zk5KnzFrPLKakJvVFK+s5pURXTLOfsMjq9l4\nVWZyz7HvaVxN8+VYNbd1GHS021rJvJCqB6rLF0z29y1v4YpdfdSFd16Z3PEfnlhNyNoknSHZ\nruZxdcLsWcektP6jf+KxmvXp3ec89avM5FXaVtN8OdaGgel1ITW/rbXMC2mx+s/g5e/UzBau\nmKYeCA7/oPS9i4PdaoIO5vXXGZLdarZ3GVBpWaVdbvDEaiaqvwaHH6mR2lbTfDl70gaXpg5q\n+bZWMy+kvPSq0OHEXoHmV8wYVRMcBdJyPbGaoHt8r+oMyW41i9SK0DBwlLtqXs2PVOj/lNW1\nj4vL2TWzxqoLqdltrWZcSAeSRoWP+WrTUa6wqvxnemQ1X6RNrdAYku1qfpJWY1Xt0baWKKuZ\npD4Ojr7tcIF7ywmLhNTyba1iXEifq8i7kc1VK49yhXV/+AmeF1Yz6rjdOkOyXU3uSe+f6VMn\nPOGN1Wzo1v/Nbe+P6vSue8sJi4TU8m2tYlxIJWpa+LhIFR/litUpww56YzVPqBcsnSHZriY9\n97iZL9yfo571xGqsT09SSuW8o2sxLSwnLBJSy7e1ioEhTQ8fi9SLLV/x36kDd3ljNdu7X2Rp\nDslmNanqydCzf8cAAAU+SURBVOCorEtmrRdWs6Fv9r0vP3ZyRpu/AzhfTlh9SC3d1irGhVSq\nJoWPs9XrLV0RmKPO3+uR1VzR5V96Q7JdTY+kfaHhvytdrw3YrmZIp6+Co329e9doWk3z5YRF\nQmr5tlYxLqTq5MgZ0wnqXy1cEZisbtT1L2601byi7ty6desnasJWXT/i2/63GZQU/jt7g9L1\nQpLdar7znR0e/kyt17Sa5ssJi4TU8m2tYlxI1o86hf5lPZSV3dIVBepuz6xmpqpX6IHVWNNV\n+Af789QWD6xmhxoavmacWqdrNc2XE1J3+rvF21rFvJCWqXnBy4fUXZZ14IMvDr/iD6rAO6vZ\n8HLIc+q8l7W9kb3df5t1vh9XWdbaDqfpWoztavr6PwsOK7p3rXJvOSF1ITW5rY3MC6l2uBpz\n1xW+U4P/hHysRh1+xQnqxsIwybdHb/NqwnT+jGS/mhkq765r01Le8MRqijv0+OXjC/qqpdpW\n03w5q4N/UZIygxc7m/1faz3zQrK+m5Xr7z0tdGYu8t+jyRUNT6Y2e2E1YVpDsl1N4OH+HTNG\n/90jq3ln7DHJ3c75s8bVNFvOwvq/LaXN/q+1noEhAd5DSIAAQgIEEBIggJAAAYQECCAkQAAh\nAQIICRBASIAAQgIEEBIggJAAAYQECCAkQAAhAQIICRBASIAAQgIEEBIggJAAAYQECCAkQAAh\nAQIICRBASIAAQgIEEBIggJAAAYQECCAkQAAhAQIICRBASGYYr7ZGnTNNKfVQs2t/oPUDDNst\nQjJDbCE99nLjx9vX3NYh/EHDb7x8CSHFHyGZIbaQNjf+YcPA9EhIllVASPFHSGZobUh70gaX\nphKSPoTkNWf6vg4dtvpGWNZ7Y3v4c6/abEVCulBVBEcHw5/I/c0NOf6eY0IfUV5VdFrXLqcW\nHTo8pF0zayxC0oiQvGapeiB0uE8ts9Z1zJq/7Lb0XjubhbQjN6Pw6buPT11tWdeoiQ89fKma\ndsRTuyBC0oiQvGZH8sjQYWhqhfXbgW8ERw+EyjoipKnJa4PDLemDLavT0ND0my+vJSQ3EZLn\nnJ+0PfTM7rLIn2oOrFIzjwwp0HPgtpCfqO+sjKztdXckJBcRkuc8qf4r9MyuODh8asT3VFDB\nkSF9o+p9Yt2vul79+FehOxKSiwjJc/amnRd8Ztet2rJuV4OfWL3m0eYhlaq8VyOC16wa21n5\nRv+TkFxFSN7z0+Tyrb7rLOtAWvZ3wT+uODykfeHvSHlN71C1cpLvxGpCchMheU+xevo+9X+W\ntVldGvrj7fUhjVU7gn9cHzrZ0LNjqClrR8N9pqr3CMlNhOQ9VRkTz8oNWNZ+34Dgnz7ora6P\nhDRV/S3451vDZ+3UHcHhjsyLrDVZT4buM029XxdS9QeldV+HkDQiJA+6pntyqBPrInX9/9zZ\n7ZXk4/+7MhTSGjXor+/ePjw9GNL2HHXN8rtz/K9ZB09JuXbpbyd3GBaoC6lUnRm8XF1YWJiU\nGbzYSUhaEJIHvabUhtBxx8RjMn78pnVXl8xt4S1Cy09KO/a63VnDgjdtm5qd/L1L3guOds04\noVNG/7u/sxpCGh68XFh/Wq+UkLQgpMRR9zPSY2OOuJ6QNCCkxFEX0uVFR1xPSBoQUuKYppa/\nusXaf1fFYde++epYQoo/Qkoc/IasiwgJEEBIgABCAgQQEiCAkAABhAQIICRAACEBAggJEEBI\ngABCAgQQEiCAkAABhAQIICRAACEBAggJEEBIgABCAgQQEiCAkAABhAQIICRAACEBAggJEEBI\ngABCAgT8Py7UuV+JzcAbAAAAAElFTkSuQmCC", "text/plain": [ "plot without title" ] }, "metadata": { "image/png": { "height": 420, "width": 420 } }, "output_type": "display_data" } ], "source": [ "plot(values)" ] }, { "cell_type": "code", "execution_count": 10, "id": "8a568511", "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "
A matrix: 10 × 2 of type dbl
2981.4059
4964.5405
6949.2431
8935.3679
10922.7827
12911.3675
14901.0136
16891.6223
18883.1041
20875.3779
\n" ], "text/latex": [ "A matrix: 10 × 2 of type dbl\n", "\\begin{tabular}{ll}\n", "\t 2 & 981.4059\\\\\n", "\t 4 & 964.5405\\\\\n", "\t 6 & 949.2431\\\\\n", "\t 8 & 935.3679\\\\\n", "\t 10 & 922.7827\\\\\n", "\t 12 & 911.3675\\\\\n", "\t 14 & 901.0136\\\\\n", "\t 16 & 891.6223\\\\\n", "\t 18 & 883.1041\\\\\n", "\t 20 & 875.3779\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "A matrix: 10 × 2 of type dbl\n", "\n", "| 2 | 981.4059 |\n", "| 4 | 964.5405 |\n", "| 6 | 949.2431 |\n", "| 8 | 935.3679 |\n", "| 10 | 922.7827 |\n", "| 12 | 911.3675 |\n", "| 14 | 901.0136 |\n", "| 16 | 891.6223 |\n", "| 18 | 883.1041 |\n", "| 20 | 875.3779 |\n", "\n" ], "text/plain": [ " [,1] [,2] \n", " [1,] 2 981.4059\n", " [2,] 4 964.5405\n", " [3,] 6 949.2431\n", " [4,] 8 935.3679\n", " [5,] 10 922.7827\n", " [6,] 12 911.3675\n", " [7,] 14 901.0136\n", " [8,] 16 891.6223\n", " [9,] 18 883.1041\n", "[10,] 20 875.3779" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "### INPUT \n", "mat_sen<-1:10*2\n", "\n", "#define the matrix with the coupon_sen for different interest rates\n", "values<-matrix(nrow=length(mat_sen),ncol=2)\n", "\n", "# calculate the bond value\n", "for (i in 1:length(mat_sen)){\n", " values[i,1]<-mat_sen[i]\n", " values[i,2]<-calculate_bond_value(mat_sen[i],principal,coupon,interest)\n", "}\n", "\n", "###output\n", "values" ] }, { "cell_type": "code", "execution_count": 11, "id": "77c33a09", "metadata": {}, "outputs": [ { "data": { "image/png": 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"text/plain": [ "plot without title" ] }, "metadata": { "image/png": { "height": 420, "width": 420 } }, "output_type": "display_data" } ], "source": [ "plot(values)" ] }, { "cell_type": "markdown", "id": "adc5a498", "metadata": {}, "source": [ "3. Calculate the duration of the bond for $r=5\\%$, $c=4\\%$, $T=10$ and $P=1000$ and give an interpretation to it. Assume that the interest rate has increased from $r=5\\%$ to $r=6\\%$. How much would the price of the bond change, as prescribed by the duration? What is the real change in the value of the bond? Why those two value are different? (8p)" ] }, { "cell_type": "markdown", "id": "bc07c36d", "metadata": {}, "source": [ "## Solution Exercise 4" ] }, { "cell_type": "code", "execution_count": 13, "id": "d4f242b9", "metadata": {}, "outputs": [], "source": [ "calculate_bond_duration<-function(maturity,principal,coupon,interest){\n", " #define the cashflows\n", " cashflow<-rep(coupon*principal,times=maturity)\n", " cashflow[maturity]<-cashflow[maturity]+principal\n", " cashflow <-cbind(1:maturity,cashflow)\n", " \n", " #discount the cashflows\n", " discounted_cashflow<-cashflow\n", " for (i in 1:maturity){\n", " discounted_cashflow[i,2]<-cashflow[i,2]/(1+interest)^cashflow[i,1]\n", " }\n", " \n", " #calculate value: OUTPUT\n", " value<-sum(discounted_cashflow[,2]*cashflow[,1])/sum(discounted_cashflow[,2])\n", "\n", " return(value)\n", "}" ] }, { "cell_type": "code", "execution_count": 16, "id": "c870b5b9", "metadata": {}, "outputs": [ { "data": { "text/html": [ "8.35958916401051" ], "text/latex": [ "8.35958916401051" ], "text/markdown": [ "8.35958916401051" ], "text/plain": [ "[1] 8.359589" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "dur<-calculate_bond_duration(maturity,principal,coupon,interest)\n", "dur" ] }, { "cell_type": "markdown", "id": "ff3b3198", "metadata": {}, "source": [ "change approximated by duration" ] }, { "cell_type": "code", "execution_count": 17, "id": "9fb8d912", "metadata": {}, "outputs": [ { "data": { "text/html": [ "-73.4674652152072" ], "text/latex": [ "-73.4674652152072" ], "text/markdown": [ "-73.4674652152072" ], "text/plain": [ "[1] -73.46747" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "-dur * (0.06-0.05) * calculate_bond_value(maturity,principal,coupon,interest)/(1+interest)" ] }, { "cell_type": "markdown", "id": "55140bd6", "metadata": {}, "source": [ "Approximated value" ] }, { "cell_type": "code", "execution_count": 19, "id": "ee45cacc", "metadata": {}, "outputs": [ { "data": { "text/html": [ "849.315185492944" ], "text/latex": [ "849.315185492944" ], "text/markdown": [ "849.315185492944" ], "text/plain": [ "[1] 849.3152" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "-dur * (0.06-0.05) * calculate_bond_value(maturity,principal,coupon,interest)/(1+interest)+\n", " calculate_bond_value(maturity,principal,coupon,interest)" ] }, { "cell_type": "markdown", "id": "9dd14d5f", "metadata": {}, "source": [ "Real value" ] }, { "cell_type": "code", "execution_count": 20, "id": "ce1a332b", "metadata": {}, "outputs": [ { "data": { "text/html": [ "852.798258971706" ], "text/latex": [ "852.798258971706" ], "text/markdown": [ "852.798258971706" ], "text/plain": [ "[1] 852.7983" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "calculate_bond_value(maturity,principal,coupon,interest+0.01)" ] }, { "cell_type": "markdown", "id": "89149e49", "metadata": {}, "source": [ "# Exercise 2\n", "We assume that a random sample $Z_1,\\ldots,Z_n$ is Reyleigh-distributed. The density, expected value and variance of Reyleigh distribution is given by:\n", "$$f(x) =\\begin{cases}\n", " \\frac{x}{\\sigma^2} \\exp\\left(-\\frac{ x^2}{2\\sigma^2}\\right) & \\quad \\text{if } x>0\\\\\n", " 0 & \\quad \\text{otherwise } \n", " \\end{cases} $$\n", " $$ \\mathbb E X=\\sigma\\sqrt{\\pi/2}$$\n", " $$ \\mathbb{VAR}(X)=\\frac{4-\\pi}{2} \\sigma^2.$$\n", " \n", " " ] }, { "cell_type": "markdown", "id": "1e06f585", "metadata": {}, "source": [ "1. Derive an inverse sampler for a Reyleigh-distribution (Note: You are expected to compute first the CDF by integration by substitution, then the inverse of CDF). (5p)\n", "\n", "$F^{-1}(u)=\\sigma\\sqrt{-2\\ln(1-u)}$\n", "\n", "2. Derive the moment estimator of $\\sigma$. (5p)\n", "\n", "$ \\sigma_{mm}=\\bar X_n\\sqrt{\\frac{2}{\\pi}}$\n", "\n", "3. Write a log-likelihood function $\\mathcal L(\\sigma)$. Derive the maximum likelihood estimator of $\\sigma$. (5p)\n", "\n", "$ \\sigma_{ml}=\\sqrt{\\frac{1}{2n}\\sum_{i=1}^nx_i^2}$\n", "\n", "4. Write a script for a function which calculates the maximum likelihood estimatetors given a sample $Z_1,\\ldots,Z_n$. Function should be general enough to handle different $n$. " ] }, { "cell_type": "code", "execution_count": 24, "id": "1265efa0", "metadata": {}, "outputs": [], "source": [ "estimate_sigma_ml<-function(x){\n", " n<-length(x)\n", " sigma<- (sum(x^2)/(2*n))^0.5 \n", "}" ] }, { "cell_type": "markdown", "id": "594fd4da", "metadata": {}, "source": [ "5. Generate $n=10, 100, 1000, 10000, 1000000$ simulations using the inverse sampler (simulations for different $n$ should be drawn independently) for parameters $\\sigma=3$. Compute a moment estimatior $\\sigma_{MM}$ and maximum likelihood estimator $\\sigma_{ML}$ for each $n$ using a function from a previous points and prepare a table presenting the estimated parameters for different $n$. Compare it to the true value of parameter and comment on the estimation error. (5p)" ] }, { "cell_type": "code", "execution_count": 23, "id": "bd509137", "metadata": {}, "outputs": [], "source": [ "generate_ray<-function(n,sigma){\n", " u<-runif(n)\n", " x <- sigma*(-2*log(1-u))^0.5\n", " return(x)\n", "}" ] }, { "cell_type": "code", "execution_count": 30, "id": "7fff6732", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[1] \"N= 10\"\n", "[1] \"ML= 2.88897654850751\"\n", "[1] \"MM= 2.93817008695092\"\n", "[1] \"N= 100\"\n", "[1] \"ML= 2.89261395172873\"\n", "[1] \"MM= 2.89722725712971\"\n", "[1] \"N= 1000\"\n", "[1] \"ML= 3.05038940657277\"\n", "[1] \"MM= 3.06221336577764\"\n", "[1] \"N= 10000\"\n", "[1] \"ML= 3.00356565335221\"\n", "[1] \"MM= 3.00999659119662\"\n", "[1] \"N= 1e+05\"\n", "[1] \"ML= 3.00738545375832\"\n", "[1] \"MM= 3.00880345877067\"\n" ] } ], "source": [ "#### INPUT\n", "n=c(10,100,1000,10000,100000)\n", "sigma=3\n", "\n", "MM<-cbind(n,rep(0,times=length(n)))\n", "ML<-cbind(n,rep(0,times=length(n)))\n", "\n", "for (i in 1:length(n)){\n", " print(paste(\"N=\",n[i]))\n", " x=generate_ray(n[i],sigma)\n", " ML[i,2]=estimate_sigma_ml(x)\n", " print(paste(\"ML=\",ML[i,2]))\n", " MM[i,2]=mean(x)*(2/pi)^0.5\n", " print(paste(\"MM=\",MM[i,2]))\n", " \n", "}" ] }, { "cell_type": "markdown", "id": "3ef9ba11", "metadata": {}, "source": [ "6. Using the closed-form formula compute mean, standard deviation and $0,025$ and $0,975$-quantiles for \\textbf{the Reyleight-distribution} with estimated parameters $\\sigma_{MM}$ and $\\sigma_{ML}$ for each $n$ estimated in the previous point, as well as for true value of the parameter. Present the values in the table and comment on them.(5p)" ] }, { "cell_type": "code", "execution_count": 31, "id": "4a30fef0", "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "
A matrix: 6 × 6 of type dbl
naveragestdevperc_2_5perc_97_5
1e+012.9381700.93524861.9249020.66115767.980666
1e+022.8972270.92221611.8980790.65194457.869457
1e+033.0622130.97473282.0061670.68907028.317593
1e+043.0099970.95811171.9719580.67732028.175762
1e+053.0088030.95773191.9711770.67705188.172521
0e+003.0000000.95492971.9654090.67507088.148609
\n" ], "text/latex": [ "A matrix: 6 × 6 of type dbl\n", "\\begin{tabular}{llllll}\n", " n & & average & stdev & perc\\_2\\_5 & perc\\_97\\_5\\\\\n", "\\hline\n", "\t 1e+01 & 2.938170 & 0.9352486 & 1.924902 & 0.6611576 & 7.980666\\\\\n", "\t 1e+02 & 2.897227 & 0.9222161 & 1.898079 & 0.6519445 & 7.869457\\\\\n", "\t 1e+03 & 3.062213 & 0.9747328 & 2.006167 & 0.6890702 & 8.317593\\\\\n", "\t 1e+04 & 3.009997 & 0.9581117 & 1.971958 & 0.6773202 & 8.175762\\\\\n", "\t 1e+05 & 3.008803 & 0.9577319 & 1.971177 & 0.6770518 & 8.172521\\\\\n", "\t 0e+00 & 3.000000 & 0.9549297 & 1.965409 & 0.6750708 & 8.148609\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "A matrix: 6 × 6 of type dbl\n", "\n", "| n | | average | stdev | perc_2_5 | perc_97_5 |\n", "|---|---|---|---|---|---|\n", "| 1e+01 | 2.938170 | 0.9352486 | 1.924902 | 0.6611576 | 7.980666 |\n", "| 1e+02 | 2.897227 | 0.9222161 | 1.898079 | 0.6519445 | 7.869457 |\n", "| 1e+03 | 3.062213 | 0.9747328 | 2.006167 | 0.6890702 | 8.317593 |\n", "| 1e+04 | 3.009997 | 0.9581117 | 1.971958 | 0.6773202 | 8.175762 |\n", "| 1e+05 | 3.008803 | 0.9577319 | 1.971177 | 0.6770518 | 8.172521 |\n", "| 0e+00 | 3.000000 | 0.9549297 | 1.965409 | 0.6750708 | 8.148609 |\n", "\n" ], "text/plain": [ " n average stdev perc_2_5 perc_97_5\n", "[1,] 1e+01 2.938170 0.9352486 1.924902 0.6611576 7.980666 \n", "[2,] 1e+02 2.897227 0.9222161 1.898079 0.6519445 7.869457 \n", "[3,] 1e+03 3.062213 0.9747328 2.006167 0.6890702 8.317593 \n", "[4,] 1e+04 3.009997 0.9581117 1.971958 0.6773202 8.175762 \n", "[5,] 1e+05 3.008803 0.9577319 1.971177 0.6770518 8.172521 \n", "[6,] 0e+00 3.000000 0.9549297 1.965409 0.6750708 8.148609 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "MM<-rbind(MM,c(0,sigma))\n", "\n", "average<-MM[,2]*(2/pi)*0.5\n", "stdev<-((4-pi)/2)^0.5 *MM[,2]\n", "perc_2_5<-MM[,2]*(-2*log(1-0.025))^0.5\n", "perc_97_5<-MM[,2]*(-2*log(1-0.975))^0.5\n", "\n", "MM<-cbind(MM,average,stdev,perc_2_5,perc_97_5)\n", "MM" ] }, { "cell_type": "code", "execution_count": 32, "id": "2fe18539", "metadata": {}, "outputs": [ { "data": { "text/html": [ "\n", "\n", "\n", "\t\n", "\n", "\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\t\n", "\n", "
A matrix: 6 × 6 of type dbl
naveragestdevperc_2_5perc_97_5
1e+012.8889770.91958981.8926740.65008797.847047
1e+022.8926140.92074761.8950570.65090647.856927
1e+033.0503890.97096911.9984210.68640968.285477
1e+043.0035660.95606461.9677450.67587318.158294
1e+053.0073850.95728051.9702480.67673278.168669
0e+003.0000000.95492971.9654090.67507088.148609
\n" ], "text/latex": [ "A matrix: 6 × 6 of type dbl\n", "\\begin{tabular}{llllll}\n", " n & & average & stdev & perc\\_2\\_5 & perc\\_97\\_5\\\\\n", "\\hline\n", "\t 1e+01 & 2.888977 & 0.9195898 & 1.892674 & 0.6500879 & 7.847047\\\\\n", "\t 1e+02 & 2.892614 & 0.9207476 & 1.895057 & 0.6509064 & 7.856927\\\\\n", "\t 1e+03 & 3.050389 & 0.9709691 & 1.998421 & 0.6864096 & 8.285477\\\\\n", "\t 1e+04 & 3.003566 & 0.9560646 & 1.967745 & 0.6758731 & 8.158294\\\\\n", "\t 1e+05 & 3.007385 & 0.9572805 & 1.970248 & 0.6767327 & 8.168669\\\\\n", "\t 0e+00 & 3.000000 & 0.9549297 & 1.965409 & 0.6750708 & 8.148609\\\\\n", "\\end{tabular}\n" ], "text/markdown": [ "\n", "A matrix: 6 × 6 of type dbl\n", "\n", "| n | | average | stdev | perc_2_5 | perc_97_5 |\n", "|---|---|---|---|---|---|\n", "| 1e+01 | 2.888977 | 0.9195898 | 1.892674 | 0.6500879 | 7.847047 |\n", "| 1e+02 | 2.892614 | 0.9207476 | 1.895057 | 0.6509064 | 7.856927 |\n", "| 1e+03 | 3.050389 | 0.9709691 | 1.998421 | 0.6864096 | 8.285477 |\n", "| 1e+04 | 3.003566 | 0.9560646 | 1.967745 | 0.6758731 | 8.158294 |\n", "| 1e+05 | 3.007385 | 0.9572805 | 1.970248 | 0.6767327 | 8.168669 |\n", "| 0e+00 | 3.000000 | 0.9549297 | 1.965409 | 0.6750708 | 8.148609 |\n", "\n" ], "text/plain": [ " n average stdev perc_2_5 perc_97_5\n", "[1,] 1e+01 2.888977 0.9195898 1.892674 0.6500879 7.847047 \n", "[2,] 1e+02 2.892614 0.9207476 1.895057 0.6509064 7.856927 \n", "[3,] 1e+03 3.050389 0.9709691 1.998421 0.6864096 8.285477 \n", "[4,] 1e+04 3.003566 0.9560646 1.967745 0.6758731 8.158294 \n", "[5,] 1e+05 3.007385 0.9572805 1.970248 0.6767327 8.168669 \n", "[6,] 0e+00 3.000000 0.9549297 1.965409 0.6750708 8.148609 " ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "ML<-rbind(ML,c(0,sigma))\n", "\n", "average<-ML[,2]*(2/pi)*0.5\n", "stdev<-((4-pi)/2)^0.5 *ML[,2]\n", "perc_2_5<-ML[,2]*(-2*log(1-0.025))^0.5\n", "perc_97_5<-ML[,2]*(-2*log(1-0.975))^0.5\n", "\n", "ML<-cbind(ML,average,stdev,perc_2_5,perc_97_5)\n", "ML" ] }, { "cell_type": "markdown", "id": "7ab477e6", "metadata": {}, "source": [ "# Exercise 3" ] }, { "cell_type": "markdown", "id": "174fd623", "metadata": {}, "source": [ "We assume the compound Poisson model. In this model, the claim number from all policies is modelled as a random variable $N\\sim Poiss(\\lambda)$ and the claim size $Z_i$ is distributed according to exponential distribution with a parameter $\\beta$ (defined as in the lectures). $Z_1,Z_2,\\ldots$ are i.i.d. and independent of $N$. Then the sum of all claims can be defined as $X=\\sum_{i=1}^N Z_i$. \n", "\n", "Assume that the insurer bought a reinsurance programme per event such that the reinsurer cover the part of the claim exceeding a fixed number $b>0$ ($H^{event}(x)=max(0,x-b)$).\n", "1. Define a sequence of random variables $U_i,\\ i=1,2,\\ldots$ as \n", "\t$$U_i =\\begin{cases}\n", " 1 & \\quad \\text{if } Z_i>b\\\\\n", " 0 & \\quad \\text{otherwise } \n", " \\end{cases} $$\n", " Prove that $(U_1,U_2,\\ldots)$ is an independent and identically distributed sequence of random variables. (5p)\n", "2. Define $N_b:=\\sum_{i=1}^N U_i$. Prove that $N_b\\sim Poiss(\\lambda \\mathbb P (Z_1>b))$. (15p)\n", "\tTips: 1. fix $k$. If $N_b=k$ it means that $N$ must be at least $k$. Use it together with the the law of total probability to express $\\mathbb P(N_b=k)=\\sum_{n=k}^{\\infty}\\mathbb P(N_b=k|N=n)\\mathbb P(N=n)$. 2. Note that for a fixed $n$ the random variable $\\sum_{i=1}^n U_i$ is binominally distributed. 3. Note that $e^x=\\sum_{n=0}^{\\infty} \\frac{x^k}{k!}$.\n", "3. Fix any $i=1,2,\\ldots$. Prove that $\\mathbb P(Z_i>x+b|Z_i>b)=\\mathbb P(Z_i>x)$. Argue that random variable $Z_i-b$ conditioned on $Z_i>b$ (i.e. using conditional probability) has an exponential distribution with the same parameter $\\beta$. (5p)\n", "4. For $n=1,2,3,...$ define the following random variable: $$\\tau_n=\\min\\{k\\in\\mathbb N: k>\\tau_{n-1}\\text{ and } U_k=1 \\}$$ where $\\tau_0=0$. $\\tau_n$ has the intrepretation as a (random) index identifying on which element of the sequence $U_1, U_2, \\ldots$ the value $1$ is taken for the n-th time. Argue that the following equality is true $$ \\sum_{i=1}^N H^{event}(Z_i)=\\sum_{k=1}^{N_b} (Z_{\\tau_k}-b).$$ The equality should be understood $\\omega$-wise (i.e. those two objects are really the same). (5p)\n", "5. It can be proven that $Z_{\\tau_1},Z_{\\tau_2},\\ldots$ are i.i.d. and that $(Z_{\\tau_n}-b)\\sim Exp(\\beta)$. Moreover, is independent of $N_b$. Use those facts (without proving them) and the previous point to prove the following explicit formulas for the pure reinsurance premium for our contract as well for the standard deviation of the reinsurance recoverables from that contract $X^{re}:= \\sum_{i=1}^N H^{event}(Z_i)$. (10p)\n", "\t$$ \\pi^{reinsurance,pure}=\\frac{\\lambda}{\\beta}\\exp(-\\beta b)$$\n", "\t$$ \\mathbb{VAR}(X^{re})=\\frac{2\\lambda\\exp(-\\beta b)}{\\beta^2}$$" ] }, { "cell_type": "markdown", "id": "e2b9f611", "metadata": {}, "source": [ "## Exercise 4\n", "We are in the setup of Assignment 3. Write a program which simulates the sum of reinsurance recoverables $X^{re}$ from that exercise and simulate $n=10, 100,1000, 10000$ simulations and for parameters $\\lambda=10$, $\\beta=0.25$ and $b=5$. Compute the estimate of the pure reinsurance premium and standard deviation of the reinsurance recoverables from the simulations for different $n$. Moreover, compute the pure reinsurance premium and standard deviation by the explicit formulas given in the Assignment 3(5) and compare it with the simulated ones. Comment on simulation error. \n", "\t\n", "Note: you can either use the inverse sampler to simulate from the exponential distribtion or you can use the fact that exponential distribution is a special case of gamma distribution and use a ready gamma sampler. In the second case make sure that you use the right parametrisation!" ] }, { "cell_type": "code", "execution_count": 7, "id": "6a97223e", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[1] \"Number of simulations: 10\"\n", "[1] \"Simulated average: 12.7645367169462\"\n", "[1] \"Simulated sd: 9.3294132139151\"\n", "[1] \"True average: 11.4601918744076\"\n", "[1] \"True sd: 9.57504751921685\"\n", "[1] \"-----\"\n", "[1] \"Number of simulations: 100\"\n", "[1] \"Simulated average: 12.5470140009292\"\n", "[1] \"Simulated sd: 11.0940703765532\"\n", "[1] \"True average: 11.4601918744076\"\n", "[1] \"True sd: 9.57504751921685\"\n", "[1] \"-----\"\n", "[1] \"Number of simulations: 1000\"\n", "[1] \"Simulated average: 11.098532377864\"\n", "[1] \"Simulated sd: 9.37508244487934\"\n", "[1] \"True average: 11.4601918744076\"\n", "[1] \"True sd: 9.57504751921685\"\n", "[1] \"-----\"\n", "[1] \"Number of simulations: 10000\"\n", "[1] \"Simulated average: 11.4646351404571\"\n", "[1] \"Simulated sd: 9.472376627215\"\n", "[1] \"True average: 11.4601918744076\"\n", "[1] \"True sd: 9.57504751921685\"\n", "[1] \"-----\"\n" ] } ], "source": [ "#input\n", "m=c(10,100,1000,10000)\n", "lambda=10\n", "beta=0.25\n", "b=5\n", "\n", "#draw exponential\n", "draw_exponential<-function(beta,n){\n", " u<-runif(n)\n", " x<-(log(1-u)/(-beta))\n", " return(x)\n", "}\n", "\n", "#program\n", "generate_compound_poiss<-function(lambda,beta,b,m){\n", " N <- rpois(m,lambda)\n", " Xre <- rep(0,m)\n", " for(i in 1:m)\n", " {\n", " Z <- draw_exponential(beta,N[i])\n", " Xre[i] <- sum(pmax(Z-b,0))\n", " }\n", " return(Xre)\n", "}\n", "\n", "for (i in 1:4){\n", " Xre<-generate_compound_poiss(lambda,beta,b,m[i])\n", " print(paste(\"Number of simulations: \",m[i]))\n", " print(paste(\"Simulated average:\", mean(Xre)))\n", " print(paste(\"Simulated sd:\", sd(Xre)))\n", " print(paste(\"True average:\", lambda/beta*exp(-beta*b)))\n", " print(paste(\"True sd:\", (2*lambda/beta^2*exp(-beta*b))^0.5))\n", " print(\"-----\")\n", "}\n", "\n" ] }, { "cell_type": "code", "execution_count": null, "id": "1857f5bc", "metadata": {}, "outputs": [], "source": [ "\n", "\n" ] } ], "metadata": { "kernelspec": { "display_name": "R", "language": "R", "name": "ir" }, "language_info": { "codemirror_mode": "r", "file_extension": ".r", "mimetype": "text/x-r-source", "name": "R", "pygments_lexer": "r", "version": "4.0.5" } }, "nbformat": 4, "nbformat_minor": 5 }