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In Example 1 we considered the polynomial which is -1 multiplied by\n\n\\[ \nf\\left( z\\right) = {z}^{3} + {z}^{2} - z - 1 = \\left( {{z}^{2} - 1}\\right) \\left( {z + 1}\\right) .\n\\]\n\n(7.4.44)\n\nThe test function of Corollary 7.33 is \\( t = f/g \\), where\n\n\\[ \ng\\left( z\\right) = {z}^{2} - 1\\text{.}\n\\]\... | We have included this example to emphasize the fact that for \\( f \\) to be stable, it is necessary not only for \\( t\\left( z\\right) \\) to be representable by a continued fraction of the form (7.4.46), but also for \\( m = n \\), where \\( n \\) is the degree of the polynomial \\( f\\left( z\\right) \\) . | Yes |
For \( a \in \mathbb{C} \) let \( {\mathcal{L}}_{a} \) be the family of continued fractions defined by\n\n\[ \n{\mathcal{C}}_{a} = \left\lbrack {K : K = \mathop{K}\limits_{{n = 1}}^{\infty }\left( {{a}_{n}/1}\right) ,\lim {a}_{n} = a}\right\rbrack .\n\]\n\nThus \( {\mathcal{C}}_{a} \) consists of all limit periodic con... | Clearly we have, for all \( K \in {\mathcal{L}}_{a} \) and \( n = \) \( 1,2,3,\ldots , \)\n\n\[ \n{\Psi }_{n}\left( {{\mathcal{L}}_{a}, K}\right) = \widehat{\mathbb{C}}\n\] | Yes |
Example 1. Bessel functions of the first kind. A continued fraction representation of ratios of Bessel functions was given in Theorem 5.16. Setting \( \nu = 0 \) and making an equivalence transformation of (5.4.32), we obtain for all \( m = 1,2,3,\ldots \) ,\n\n\[ \n{G}_{m}\left( z\right) = \frac{{J}_{m}\left( z\right)... | Then from Theorem 8.3 it can be shown that for all \( z \neq 0 \) in the disk\n\n\[ \nD\left( {r, s}\right) = \left\lbrack {z : \left| {z + \frac{2msi}{1 + {r}^{2} - {s}^{2}}}\right| \leq \frac{2mr}{1 + {r}^{2} - {s}^{2}}}\right\rbrack ,\n\]\n\n(8.3.10)\n\nwe have\n\n\[ \n\left| {{G}_{m}\left( z\right) - {f}_{m, n}\lef... | Yes |
Theorem 8.9. Let\n\n\\[ K = \\mathrm{K}\\left( \\frac{{a}_{n}z}{1}\\right) ,\\;{a}_{n} > 0, \\]\n\n(8.3.34)\n\nbe an \\( S \\) -fraction which converges, for each \\( z \\) in the cut plane \\( R = \\lbrack z : \\left| {\\arg z}\\right| < \\) \\( \\pi \\rbrack \\), to a function \\( f\\left( z\\right) \\) holomorphic i... | Proof. We recall first that necessary and sufficient conditions for convergence of (8.3.34) are given by Theorem 4.58 (Stieltjes), and that if the\n\n\n\nFigure 8.3.3. \\( n \\) th inclusion region \\( {\\Theta }_{n}... | Yes |
Example 2. Natural logarithm. After an equivalence transformation, (6.1.17) can be written as an \( S \) -fraction\n\n\[ \log \left( {1 + z}\right) = \frac{z}{1} + \frac{\frac{{1}^{2}}{1 \times {2}^{z}}z}{1} + \frac{\frac{{1}^{2}}{2 \times {3}^{z}}z}{1} + \frac{\frac{{2}^{2}}{3 \times {4}^{z}}z}{1} + \frac{\frac{{2}^{2... | Letting \( {f}_{n}\left( z\right) \) denote the \( n \) th approximant of the \( S \) -fraction and setting \( z = 1 \), we obtain from Theorem 8.9 the truncation-error bounds \( \left| {{f}_{n}\left( 1\right) - {f}_{n - 1}\left( 1\right) }\right| \) given in Table 8.3.2. Also given are actual truncation errors\n\nTabl... | Yes |
Example 3. Tangent function. After an equivalence transformation, (6.1.54) can be expressed as\n\n\\[ f\\left( z\\right) = - z\\tan z = \\frac{-{z}^{2}}{1} + \\frac{-\\frac{1}{1 \\times 3}{z}^{2}}{1} + \\frac{-\\frac{1}{3 \\times 5}{z}^{2}}{1} + \\frac{-\\frac{1}{5 \\times 7}{z}^{2}}{1} + \\cdots ,\\]\n\n(8.3.40)\n\nwh... | Letting \\( z = {e}^{{i\\pi }/4} \\), we obtain from Theorem 8.9 the truncation-error bounds \\( \\left| {{f}_{n}\\left( {e}^{{i\\pi }/4}\\right) - {f}_{n - 1}\\left( {e}^{{i\\pi }/4}\\right) }\\right| \\) given in Table 8.3.3. Here \\( {f}_{n}\\left( z\\right) \\) denotes the \\( n \\) th approximant of (8.3.40). Also... | Yes |
Example 4. Complementary error function. By (6.2.25) and (6.2.19) the function\n\n\\[ \nf\\left( w\\right) = \\frac{\\sqrt{\\pi }{e}^{{w}^{2}}}{w}\\operatorname{erfc}\\left( w\\right) = \\frac{2{e}^{{w}^{2}}}{w}{\\int }_{w}^{\\infty }{e}^{-{t}^{2}}{dt}\n\\]\n\n(8.3.42)\n\nis represented, for all \\( w \\) such that \\(... | Since (8.3.43) has the same form as (8.3.41d), it is equivalent to an \\( S \\) -fraction (8.3.41a), and hence Theorem 8.9 can be applied to estimate truncation errors. Let \\( {f}_{n}\\left( w\\right) \\) denote the \\( n \\) th approximant of (8.3.43). Then setting \\( w = {e}^{{i\\pi }/4} \\) and \\( {f}_{n} = {f}_{... | Yes |
Example 1. Incomplete gamma function. | \[ \Gamma \left( {a, z}\right) = {e}^{-z}{z}^{a}\left( {\frac{1}{z} + \frac{1 - a}{1} + \frac{1}{z} + \frac{2 - a}{1} + \frac{2}{z} + \frac{3 - a}{1} + \frac{3}{z} + \cdots }\right) \] is valid for all \( z \) in the cut plane \[ R = \left\lbrack {z : \left| {\arg z}\right| < \pi }\right\rbrack \] and all \( a \in \mat... | No |
Complementary error function \( \operatorname{erfc}\left( z\right) \) [see (6.2.19)]. | \[ \operatorname{erfc}\left( z\right) = \frac{1}{\sqrt{\pi }}\Gamma \left( {\frac{1}{2},{z}^{2}}\right) \] \[ = \frac{{e}^{-{z}^{2}}}{\sqrt{\pi }z}\left( {\frac{1}{1} + \frac{\frac{1}{2}}{{z}^{2}} + \frac{1}{1} + \frac{\frac{3}{2}}{{z}^{2}} + \frac{2}{1} + \frac{\frac{5}{2}}{{z}^{2}} + \frac{3}{1} + \cdots }\right) \] ... | Yes |
Gamma Function \( \Gamma \left( z\right) \) . We consider the logarithm of the gamma function,\n\n\[ \log \Gamma \left( z\right) = \left( {z - \frac{1}{2}}\right) \log z - z + \frac{1}{2}\log {2\pi } + J\left( z\right) . \] | Here \( z \in R = \left\lbrack {z : \left| {\arg z}\right| < \pi }\right\rbrack \), and \( \log z \) denotes the principal branch of the logarithm, which is real when \( z \) is real and positive. Binet’s function \( J\left( z\right) \) is holomorphic in \( R \) and tends to zero as \( z \rightarrow \infty \) in such a... | Yes |
Arctangent. From (6.1.14)\n\n\[ z\arctan z = \frac{{z}^{2}}{1} + \frac{{1}^{2}{z}^{2}}{3} + \frac{{2}^{2}{z}^{2}}{5} + \frac{{3}^{2}{z}^{2}}{7} + \cdots \]\n\n(10.3.17)\n\nis valid for all \( z \) in the cut plane with cuts along the imaginary axis from \( + i \) to \( + i\infty \) and from \( - i \) to \( - i\infty \)... | It follows from Corollary 10.8 that the relative error \( {\varepsilon }_{1}^{\left( n\right) } \) in computing the \( n \) th approximant of (10.3.17) by the BR algorithm is such that\n\n\[ {\varepsilon }_{1}^{\left( n\right) } = O\left( 1\right) \;\text{ for }\;\left| {\arg z}\right| \leq \pi /4. \]\n\n(10.3.18)\n\nT... | No |
From (6.1.53) we have, for \( m > - 1 \) ,\n\n\[ \n\frac{-z{J}_{m + 1}\left( z\right) }{{J}_{m}\left( z\right) } = \frac{-{z}^{2}}{2\left( {m + 1}\right) } + \frac{-{z}^{2}}{2\left( {m + 2}\right) } + \frac{-{z}^{2}}{2\left( {m + 3}\right) } + \cdots ,\n\]\n\n(10.3.19)\n\nwhich is valid for all \( z \in \mathbb{C} \) . | The continued fraction in (10.3.19) is equivalent to an \( S \) -fraction \( \mathrm{K}\left( {{a}_{n}w/1}\right) \) where \( w = - {z}^{2} \) and where \( \left\{ {a}_{n}\right\} \) is a bounded sequence of positive numbers. Hence by Corollary 10.8, the relative error \( {\varepsilon }_{1}^{\left( n\right) } \), in co... | Yes |
Example 3. Complementary error function. By (6.2.25)\n\n\\[ \n\\sqrt{\\pi }{e}^{{z}^{2}}\\operatorname{erfc}z = \\frac{1}{z} + \\frac{\\frac{1}{2}}{z} + \\frac{1}{z} + \\frac{\\frac{3}{2}}{z} + \\frac{2}{z} + \\frac{\\frac{5}{2}}{z} + \\frac{3}{z} + \\cdots \n\\]\n\n(10.3.21)\n\nis valid for \\( \\operatorname{Re}\\lef... | By making an equivalence transformation we see that\n\n\\[ \n\\frac{\\sqrt{\\pi }{e}^{{z}^{2}}}{z}\\operatorname{erfc}z = \\frac{1\\left( {1/{z}^{2}}\\right) }{1} + \\frac{\\frac{1}{2}\\left( {1/{z}^{2}}\\right) }{1} + \\frac{1\\left( {1/{z}^{2}}\\right) }{1}\n\\]\n\n\\[ \n+ \\frac{\\frac{3}{2}\\left( {1/{z}^{2}}\\righ... | Yes |
It is well known that \( {y}_{n}^{\left( 1\right) } = \cos {n\varphi } \) and \( {y}_{n}^{\left( 2\right) } = \sin {n\varphi } \) are solutions of the system of three-term recurrence relations \[ {y}_{n + 1} = 2\left( {\cos \varphi }\right) {y}_{n} - {y}_{n - 1},\;n = 1,2,3,\ldots \] | If \( \cos \varphi \) and \( \sin \varphi \) are known, then (B.2) provides a stable method for computing \( \cos {n\varphi } \) and \( \sin {n\varphi } \) . For \( \varphi = \pi /{10} \), starting with \( {y}_{0}^{\left( 1\right) } = 1,{y}_{1}^{\left( 1\right) } = \) \( \cos \left( {\pi /{10}}\right) = {0.951056516} \... | Yes |
The situation changes completely if one considers the recurrence relations\n\n\[ \n{y}_{n + 1} = 3{y}_{n} - {y}_{n - 1},\;n = 1,2,3,\cdots . \n\] | Two independent solutions of this system are\n\n\[ \n{y}_{n}^{\left( 1\right) } = {\left( \frac{3 - \sqrt{5}}{2}\right) }^{n}\text{ and }{y}_{n}^{\left( 2\right) } = {\left( \frac{3 + \sqrt{5}}{2}\right) }^{n}. \n\]\n\nStarting with \( {y}_{0}^{\left( 1\right) } = 1 \) and \( {y}_{1}^{\left( 1\right) } = \left( {3 - \s... | Yes |
Let \( \nu \) be real, \( x > 0 \), and let \( {J}_{n}\left( x\right) \) denote the Bessel function of the first kind of order \( n \). It can be seen from (5.2.13) that the numbers \[ {y}_{n} = {J}_{\nu + n}\left( x\right) \] satisfy the recurrence relations \[ {y}_{n + 1} = \frac{2\left( {\nu + n}\right) }{x}{y}_{n} ... | Recalling that the solutions \( \left\{ {y}_{n}\right\} \) of the recurrence relations (B.1) form a vector space of dimension 2 over \( \mathbb{C} \), we can give an explanation for the unfortunate results of the computation in Examples 2 and 3. The reason lies in the fact that in each case the difference equation has ... | Yes |
The Fresnel integrals are, for real \( \tau \), defined by\n\n\[ C\left( \tau \right) = {\int }_{0}^{\tau }\cos \left( {\frac{\pi }{2}{t}^{2}}\right) {dt},\;S\left( \tau \right) = {\int }_{0}^{\tau }\sin \left( {\frac{\pi }{2}{t}^{2}}\right) {dt} \] | For in-between values of \( \left| \tau \right| \) the expansions\n\n\[ C\left( \tau \right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }{J}_{\frac{1}{2} + {2k}}\left( {\frac{\pi }{2}{\tau }^{2}}\right) \]\n\n(B.16a)\n\n\[ S\left( \tau \right) = \mathop{\sum }\limits_{{k = 0}}^{\infty }{J}_{\frac{3}{2} + {2k}}\left( {\f... | Yes |
Corollary 1.5. Every convergent \( {c}_{i} = {p}_{i}/{q}_{i}, i \geq 1 \), of a simple continued fraction is in its lowest terms, that is, \( {p}_{i} \) and \( {q}_{i} \) have no common divisors other than +1 or -1 . | Proof. Since\n\n\[ \n{p}_{i}{q}_{i - 1} - {p}_{i - 1}{q}_{i} = {\left( -1\right) }^{i} \n\] \n\nit follows that any number which divides both \( {p}_{i} \) and \( {q}_{i} \) must be a divisor of \( {\left( -1\right) }^{i} \) . But the only divisors of \( {\left( -1\right) }^{i} \) are \( + 1 \) and -1 ; hence the numbe... | Yes |
Find integral solutions of the equation\n\n\\[ \n{205x} - {93y} = - 1\\text{.}\n\\] | Solution. The numbers 205 and 93 are relatively prime, hence the given equation has integral solutions. The continued fraction expansion for \\( \\frac{205}{93} \\) is\n\n\\[ \n\\frac{205}{93} = \\left\\lbrack {2,4,1,8,2}\\right\\rbrack\n\\]\n\nand has an odd number of partial quotients, so \\( {\\left( -1\\right) }^{n... | Yes |
Show that we can solve Example 2 if we have already solved Example 1. That is, solve the equation \( {205x} - {93y} = - 1 \), knowing that \( \left( {{x}_{0},{y}_{0}}\right) = \left( {{49},{108}}\right) \) is a particular solution of \( {205x} - {93y} = + 1 \) . | Solution. Using equations (2.17) we find that\n\n\[ \n{x}_{1} = b - {x}_{0} = {93} - {49} = {44}, \n\]\n\n\[ \n{y}_{1} = a - {y}_{0} = {205} - {108} = {97} \n\]\n\nis a particular solution of \( {205x} - {93y} = - 1 \). Hence the general solution, according to (2.18), is\n\n(2.19)\n\n\[ \n\begin{array}{l} x = {44} + {9... | Yes |
Solve the equation\n\n\[ \n{205x} - {93y} = 5. \n\] | Solution. From Example 1, Section 2.3, we know that \( \left( {{x}_{0},{y}_{0}}\right) = \left( {{49},{108}}\right) \) is a particular solution of the equation \( {205x} - {93y} = 1 \) , that is,\n\n\[ \n{205}\left( {49}\right) - {93}\left( {108}\right) = 1 \n\]\n\nMultiplying both sides by 5 we get\n\n\[ \n{205}\left(... | Yes |
Example 2. Solve the equation\n\n\\[ \n{205x} - {93y} = - 5. \n\\] | Solution. In Example 1 of this section we recalled that\n\n\\[ \n{205}\\left( {49}\\right) - {93}\\left( {108}\\right) = 1. \n\\]\n\nMultiplying through by -5 we get\n\n\\[ \n{205}\\left( {-5 \\cdot {49}}\\right) - {93}\\left( {-5 \\cdot {108}}\\right) = - 5 \n\\]\n\nor\n\n\\[ \n{205}\\left( {-{245}}\\right) - {93}\\le... | Yes |
Solve the indeterminate equation\n\n\[ \n{13x} + {17y} = {300}. \n\] | Solution. We find that \( \left( {{x}_{0},{y}_{0}}\right) = \left( {4,3}\right) \) is a particular solution of the\nequation\n\n\[ \n{13x} - {17y} = 1 \n\]\n\nor that \( {13}\left( 4\right) - {17}\left( 3\right) = 1 \), and so the given equation may be written in\nthe form\n\n\[ \n{13x} + {17y} = {300}\left( {{13} \cdo... | Yes |
Example 1. Solve the equation\n\n\\[ \n{410x} - {186y} = {10}. \n\\] | Solution. Since \( {410} = 2 \cdot 5 \cdot {41},{186} = 2 \cdot 3 \cdot {31} \), the g.c.d. of 410 and 186 is \( d = 2 \) . Since \( d = 2 \) divides 10, the equation can be solved. Divide the given equation by 2 to obtain\n\n\\[ \n{205x} - {93y} = 5 \n\\]\n\nwhere now 205 and 93 are relatively prime. This is the equat... | Yes |
Example 1. Expand \( \sqrt{2} \) into an infinite simple continued fraction. | Solution. The largest integer \( < \sqrt{2} = {1.414}\cdots \) is \( {a}_{1} = 1 \), so\n\n\[ \sqrt{2} = {a}_{1} + \frac{1}{{x}_{2}} = 1 + \frac{1}{{x}_{2}} \]\n\nSolving this equation for \( {x}_{2} \), we get\n\n\[ {x}_{2} = \frac{1}{\sqrt{2} - 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1} = \sqrt{2} + 1 \]\n\nConsequen... | Yes |
Find the infinite continued fraction expansion for\n\n\[ x = \frac{{25} + \sqrt{53}}{22}. \] | Solution. We proceed exactly as in Example 1. Since \( \sqrt{53} \) is between 7 and 8, the largest integer \( < x \) is \( {a}_{1} = 1 \) . Then\n\n\[ x = \frac{{25} + \sqrt{53}}{22} = {a}_{1} + \frac{1}{{x}_{2}} = 1 + \frac{1}{{x}_{2}} \]\n\nwhere\n\n\[ {x}_{2} = \frac{1}{x - 1} = \frac{22}{3 + \sqrt{53}}\frac{3 - \s... | Yes |
Show that the first few convergents to the number\n\n\\[ \ne = {2.718282}\\cdots \n\\]\n\ngive better and better approximations to this number. These convergents should be calculated by finding the first few convergents to 2.718282 , a decimal fraction which approximates \\( e \\) correctly to six decimal places. | Solution. Assuming the above expansion for \\( e \\), or being content with the approximation\n\n\\[ \ne = {2.718282} = \\frac{1359141}{500000}\n\\]\n\nwe find that\n\n\\[ \ne = \\left\\lbrack {2,1,2,1,1,4,1,\\cdots }\\right\\rbrack .\n\\]\n\nThe corresponding convergents are\n\n\\[ \n\\frac{2}{1},\\frac{3}{1},\\frac{8... | Yes |
Given the fraction \( \frac{2065}{902} \), find a fraction with a smaller numerator and a smaller denominator whose value approximates that of the given fraction correctly to three decimal places. | Solution. Convert \( \frac{\mathbf{2065}}{\mathbf{902}} \) into a continued fraction and calculate the convergents. The table gives the numerical results:\n\n<table><thead><tr><th>\( i \)</th><th>\( - 1 \)</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th></tr></thead><tr><td>\( {a}_{i... | Yes |
Find a particular solution of the equation \( {x}^{2} - {21}{y}^{2} = 1 \) . | Solution. Here \( N = {21} \), and the continued fraction expansion given in Table 2 is\n\n\[ \sqrt{21} = \left\lbrack {4,\overline{1,1,2,1,1,8}}\right\rbrack = \left\lbrack {{a}_{1},\overline{{a}_{2},{a}_{3},{a}_{4},{a}_{5},{a}_{6},2{a}_{1}}}\right\rbrack ,\]\n\nwhich shows that \( {a}_{n} = {a}_{6} \), so that \( n =... | Yes |
Example 2. Find a particular solution of the equation \( {x}^{2} - {29}{y}^{2} = 1 \) . | Solution. The expansion of \( \sqrt{29} \) is\n\n\[ \sqrt{29} = \left\lbrack {5,\overline{2,1,1,2,{10}}}\right\rbrack = \left\lbrack {{a}_{1},\overline{{a}_{2},{a}_{3},{a}_{4},{a}_{5},2{a}_{1}}}\right\rbrack \]\n\nso that \( n = 5 \), an odd number. The first five convergents are\n\n\[ \frac{5}{1},\;\frac{11}{2},\;\fra... | Yes |
In Example 1 of Section 4.8 we found that \( {x}_{1} = {55} \) and \( {y}_{1} = {12} \) is a solution (minimal) of the equation \( {x}^{2} - {21}{y}^{2} = 1 \) . A second solution \( \left( {{x}_{2},{y}_{2}}\right) \) can be obtained by setting \( n = 2 \) in (4.42); this gives | \[ {x}_{2} + {y}_{2}\sqrt{21} = {\left( {55} + {12}\sqrt{21}\right) }^{2} \] \[ = {3025} + {1320}\sqrt{21} + {3024} \] \[ = {6049} + {1320}\sqrt{21} \] which implies that \( {x}_{2} = {6049},{y}_{2} = {1320} \) . These values satisfy the equation \( {x}^{2} - {21}{y}^{2} = 1 \), since \[ {\left( {6049}\right) }^{2} - {... | Yes |
Table 2 shows that \( {x}_{1} = 3,{y}_{1} = 1 \) is the minimal solution of \( {x}^{2} - {10}{y}^{2} = - 1 \) . A second solution is obtained from (4.44) by setting \( n = 3 \) . | We have\n\n\[ \n{x}_{3} + {y}_{3}\sqrt{10} = {\left( 3 + 1\sqrt{10}\right) }^{3} = {117} + {37}\sqrt{10} \n\]\n\nso that \( {x}_{3} = {117},{y}_{3} = {37} \) ; this is a solution since\n\n\[ \n{\left( {117}\right) }^{2} - {10}{\left( {37}\right) }^{2} = {13689} - {13690} = - 1. \n\] | Yes |
The payoff for the short side of the forward contract can therefore be expressed as | \[ F\left( t\right) - S\left( T\right) \] | Yes |
Example 1.3 (Exercising a Call Option) Let us revisit the example of Taf buying a call option on the Downhill, Inc., stock with maturity \( T = \) 6 months and strike price \( K = \$ \) 46.00. He pays \$1.00 for the option. | a. Suppose that at maturity the stock's market price is \( \$ {50.00} \) . Then Taf would exercise the option and buy the stock for \( \$ {46.00} \) . He could immediately sell the stock in the market for \$50.00, thereby cashing in the difference of \$4.00. His total profit is \$3.00, when accounting for the initial c... | Yes |
Example 1.4 (Using a Put Option for Hedging) Our conservative investor Taf has purchased one hundred shares of the stock of Big Blue Chip company as a large part of his portfolio, for the price of \$65.00 per share. He is concerned that the stock may go down during the next six months. As a hedge against that risk he b... | Taf has lost \( {100} \cdot {5.00} = {500} \) dollars in his stock position. However, by exercising the put options, he makes \( {100} \cdot {5.00} = {500} \) dollars. His total loss is the cost of put options equal to \( {100} \cdot {2.33} = {233} \) dollars. | Yes |
Example 1.5 Suppose that the Big Blue Chip stock has today's price of \$100. Imagine a weird stock market in which after one month there are only three possible prices of the stock: \( \$ {105},\$ {101} \), and \( \$ {98} \) . A European call option on that stock, with strike price \( K = {100} \) and maturity in one m... | \[ \left( {{105} - {100}}\right) /{100} = {0.05} = 5\% ,\;\left( {{101} - {100}}\right) /{100} = 1\% , \] \[ \text{or}\left( {{98} - {100}}\right) /{100} = - 2\% \] depending on the final price. However, the call option is increasing in the price of the stock, so Taf might decide to invest all his capital in the call o... | Yes |
Example 2.1 (Project Value) The powerful CEO (chief executive officer) of the Enterprise Search Company, a gentleman with the familar name Taf, has to make a decision between two different business projects. His reliable research group tells him that project Amazing is certain to result in the cash flow \( \left( {-{1.... | \[ - 1 + \frac{1.5}{1.06} + \frac{2}{{1.06}^{2}} = {2.1951} \] and of project Fabulous \[ - {1.2} + \frac{1.2}{1.06} + \frac{2.6}{{1.06}^{2}} = {2.2461} \] in millions. Taf concludes that project Fabulous might be a somewhat more profitable venture. | Yes |
Example 2.2 (Mortgage Payments Calculation) Our agent Taf has been doing well in the market and decides to buy a new house. Taf takes a 30-year loan on \( \$ {400},{000} \), at the fixed interest rate of \( 8\% \), compounded monthly. We want to compute the amount of Taf’s monthly payments. There are 12 months in a yea... | The interest rate per period is \( {0.08}/{12} = {0.0067} \) . The value of the loan is \( V\left( 0\right) = {400},{000} \) . Using formula (2.6) with \( r = {0.0067} \) and \( m = {360} \), we get approximately\n\n\( P = 2,{946} \)\n\nTaf will need to pay close to \( \$ 3,{000} \) in mortgage monthly payments. The wa... | Yes |
Example 2.4 (Annual Worth) A business venture is presented to Mr. Taf, a venture capitalist. It is projected that for the initial investment of $10 million the venture would return $2.2 million each year for the next 10 years. Mr. Taf thinks that he can find other risk-free investments returning 15% a year. He calculat... | \[ 2.2 - 1.9925 = 0.2075 \] million. This is a venture with a positive annual worth, and it should be explored further. | Yes |
Consider a default-free coupon bond that will pay a coupon of \$3.00 in six months, and will make a final payment of \$103.00 (the last coupon and the principal) in one year. This bond trades at \$101.505. The coupon bond is equivalent to a portfolio of a six-month pure discount bond with a nominal value of \$3.00, and... | \[ {101.505} - {0.03} \cdot {98.00} = {98.565} \] In six months the short position in the discount bond requires paying \( {0.03} \cdot {100.00} = {3.00} \) . Simultaneously, the coupon for the same amount is paid, resulting in zero profit/loss. In one year the principal and the final coupon of the coupon bond are paid... | Yes |
In the same setting as in Example 2.5, suppose that a one-year pure discount bond is also traded. Its price today (time zero) has to be\n\n\[ P\left( {0,1}\right) = \frac{100}{1 + {r}_{1y}} = \frac{100}{1.044996} = {95.6942} \] \nor else there are arbitrage opportunities. | In order to show this fact, suppose first that the price of the one-year pure discount bond is \$95.00, suggesting that the bond is underpriced. In order to take advantage of the arbitrage opportunity, we should buy the underpriced security and sell the equivalent portfolio. In this case we buy one unit of the one-year... | Yes |
Assume that today's date is 1/1/03. We want to determine as many points as possible in today's yield curve (or term structure of interest rates), given that we know that there are three risk-free bonds with the following characteristics: A pure discount bond that matures on 6/30/03 is selling today at \$98. A 6% bond (... | We first compute the six-month interest rate from\n\n\[ \n{98} = \frac{100}{1 + r} \n\]\n\nwith solution \( r = {2.041}\% \) . We want to construct the term structure in annual terms. We can use the simple interest rate (the most common approach when reporting the term structure), so that the annual rate \( {r}_{6m} \)... | Yes |
Example 2.8 (Speculating on Forward Rates) The one-year pure discount bond with nominal \$100 trades at \$95, and the two-year pure discount bond with nominal \$100 trades at \$89. The one-year spot rate \( r_{1y} \) is given by \( 95(1 + r_{1y}) = 100 \) with the solution \( r_{1y} = 5.2632\% \), and the two-year spot... | \[ 1.052632(1 + f_{1y,2y}) = (1.059998)^2 \] with the solution \( f_{1y,2y} = 6.7416\% \). If our fearless speculator Taf thinks that the rates will not change very much over the next year, he considers the forward rate to be relatively high. He wants to lock in the forward rate for his investment in the period between... | Yes |
Example 3.1 (Brownian Motion Squared) We want to find the dynamics of the process \( Y\left( t\right) \mathrel{\text{:=}} {W}^{2}\left( t\right) \) . We can think of process \( Y \) as a function \( Y\left( t\right) = f\left( {W\left( t\right) }\right) \) of Brownian motion, with \( f\left( x\right) = {x}^{2},{f}_{x}\l... | Since\n\n\[ \n{dW} = 0 \cdot {dt} + 1 \cdot {dW} \n\]\n\nBrownian motion is a diffusion process with the drift equal to zero and the diffusion term equal to one. Applying Itô's rule, we get\n\n\[ \nd\left( {{W}^{2}\left( t\right) }\right) = {2W}\left( t\right) {dW}\left( t\right) + {dt} \n\]\n\n\nThis expression is the... | Yes |
In this example we want to find the dynamics of the process\n\n\\[ Y\\left( t\\right) = {e}^{{aW}\\left( t\\right) + {bt}} \\] | We can think of process \\( Y \\) as a function \\( Y\\left( t\\right) = f\\left( {t, W\\left( t\\right) }\\right) \\) of Brownian motion and time, with\n\n\\[ f\\left( {t, x}\\right) = {e}^{{ax} + {bt}},\\;{f}_{t}\\left( {t, x}\\right) = {bf}\\left( {t, x}\\right) ,\\;{f}_{x}\\left( {t, x}\\right) = {af}\\left( {t, x}... | Yes |
We want to find the dynamics of the process\n\n\[ Y\left( t\right) = {W}^{2}\left( t\right) \cdot {e}^{{aW}\left( t\right) } \] | \[ {dY} = {e}^{aW} \cdot d\left( {W}^{2}\right) + {W}^{2} \cdot d\left( {e}^{aW}\right) + d\left( {W}^{2}\right) \cdot d\left( {e}^{aW}\right) \]\n\nUsing Examples 3.1 and 3.2, this gives us\n\n\[ {dY} = {e}^{aW}\left\lbrack {t + \frac{1}{2}{a}^{2}{W}^{2} + {2Wa}}\right\rbrack {dt} + a{e}^{aW}{W}^{2}{dW} \] | Yes |
Example 3.5 (Arbitrage with Arrow-Debreu Securities) Consider a single-period model with two states. There are two Arrow-Debreu securities of the type described in section 3.1.3, with prices 0.45 and 0.45 . There is a risk-free security that pays 1 in each of the two states and sells at moment \( t = 0 \) for 0.95 . | Here is an arbitrage strategy: buy each of the Arrow-Debreu securities and borrow (sell short) the risk-free security. Such a portfolio generates a revenue of \( - {0.45} - {0.45} + {0.95} = {0.05} \) at moment \( t = 0 \) . At moment \( t = 1 \) the portfolio has zero payoff in both states: a negative payment of 1.00 ... | Yes |
Consider a single-period model with \( r = {0.005}, S\left( 0\right) = {100},{s}_{1} = {101} \), and \( {s}_{2} = {99} \). The payoff Taf is trying to replicate is the European call option\n\n\[ g\left\lbrack {S\left( 1\right) }\right\rbrack = {\left\lbrack S\left( 1\right) - \widetilde{K}\right\rbrack }^{ + } = \max \... | The solution is \( {\delta }_{0} = - {49.2537},{\delta }_{1} = {0.5} \). So, Taf has to borrow 49.2537 from the bank (short-sell the bond or bank account) and buy half a share of the stock. We note that the cost of the replicating strategy (the amount of necessary initial investment) is\n\n\[ {\delta }_{0}B\left( 0\rig... | Yes |
Consider the single-period model in which the price of the stock today is \( S\left( 0\right) = \$ {100} \) . In the next period it can go up to \( \$ {120} \) , stay at \( \$ {100} \), or go down to \( \$ {90} \) . The price of the European call on this stock with strike price \( \$ {105} \) is \( \$ 5 \), and the pri... | \[ {120}{\delta }_{1} + {15}{\delta }_{2} + {25}{\delta }_{3} = 1 \] \[ {100}{\delta }_{1} + 5{\delta }_{3} = 1 \] \[ {90}{\delta }_{1} = 1 \] The solution to this system is \( {\delta }_{1} = 1/{90},{\delta }_{2} = 2/{135} \), and \( {\delta }_{3} = - 1/{45} \) . We note that the cost of this portfolio is \[ \frac{1}{... | Yes |
Consider the Merton-Black-Scholes model of section 3.3.6 with a stock with constant coefficients \( \mu \) and \( \sigma \), and a bond or bank account with constant interest rate \( r \). Suppose that the contingent claim follows the dynamics\n\n\[ \n{dC}\left( t\right) = a\left( t\right) {dt} + b\left( t\right) {dW}\... | This means that the amount \( {\pi }_{1}\left( t\right) \) invested in the stock has to be equal to \( \frac{b\left( t\right) }{\sigma } \) and, by the self-financing condition, the amount in the bank is \( {\pi }_{0}\left( t\right) = C\left( t\right) - {\pi }_{1}\left( t\right) \). | Yes |
Example 4.2 (Square-Root Utility) Our dear friend Taf has a choice between two investment opportunities. One pays $10,000 with certainty, while the other pays $22,500 with probability 0.3, $4,900 with probability 0.3, and $900 with probability 0.4. Taf decides that he likes the utility approach to establish his prefere... | In order to decide what value of \( \gamma \) to use, he decides he is pretty much indifferent between the choice of receiving $100 for sure, or $144 and $64 with fifty-fifty chance. He uses this equality in preferences to find his \( \gamma \) from\n\n\[ {\left( {100}\right) }^{\gamma } = {0.5} \cdot {\left( {144}\rig... | Yes |
For exponential utility \( U\left( x\right) = 1 - {e}^{-{\alpha x}} \), we have | \[ {U}^{\prime }\left( x\right) = \alpha {e}^{-{\alpha x}},\;{U}^{\prime \prime }\left( x\right) = - {\alpha }^{2}{e}^{-{\alpha x}},\;A\left( x\right) = \alpha \] | Yes |
One way to use the certainty equivalent for pricing a given claim is to require that the certainty equivalent of the profit or loss without buying the claim is the same as the certainty equivalent of profit or loss when buying the claim for a given amount. Such amount is then called the buying price of the claim. | As an example, consider a single period with a simple claim that pays the random value \( Y \) at time 1, equal to \( {Y}_{1} \) with probability \( {p}_{1} \), or \( {Y}_{2} \) with probability \( {p}_{2} = 1 - {p}_{1} \) . These amounts can also be negative, meaning that they represent a payment rather than a payoff.... | Yes |
Consider the single-period Cox-Ross-Rubinstein model, with stock \( S\\left( 0\\right) \) moving up to \( S\\left( 0\\right) u \) with probability \( p \), and down to \( S\\left( 0\\right) d \) with probability \( q \), \( d < 1 + r < u \). Assume the log utility, \( U\\left( {X\\left( T\\right) }\\right) = \\log \\le... | Note that we have already accounted for the budget constraint in this expression, and, therefore, we do not have to form a Lagrangian. Differentiating with respect to \( \\delta \) and setting the derivative equal to zero, we find the optimal number of shares \( \\widehat{\\delta } \):\n\n\\[ \\frac{\\widehat{\\delta }... | Yes |
Consider the multiperiod Cox-Ross-Rubinstein model, where the stock price \( S\left( t\right) \) can move up to \( S\left( t\right) u \) with probability \( p \), and down to \( S\left( t\right) d \) with probability \( q \), at each moment \( t \) . The bank account pays interest \( r \) each period. As before, we ass... | Introduce the notation\n\n\[ \widetilde{u} = u - \left( {1 + r}\right) ,\;\widetilde{d} = d - \left( {1 + r}\right) \]\n\nSuppose that the current wealth at time \( T - 1 \) is \( X\left( {T - 1}\right) = x \) and the current stock price is \( S\left( {T - 1}\right) = s \) . By the principle of dynamic programming, we ... | Yes |
Consider the same model as in the previous example, but our agent Taf now has the log utility \( U\left( {X\left( T\right) }\right) = \log \left( {X\left( T\right) }\right) \). We start the backward algorithm by setting \( V\left( {T, x}\right) = \log \left( x\right) \) and we want to find \( V\left( {T - 1, x}\right) ... | By the principle of dynamic programming, with \( S\left( {T - 1}\right) = s \), we have \( V\left( {T - 1, x}\right) = \mathop{\max }\limits_{\delta }{E}_{T - 1, x}\left\lbrack {\log \{ {\delta S}\left( T\right) + \left( {x - {\delta s}}\right) \left( {1 + r}\right) \} }\right\rbrack = \mathop{\max }\limits_{\delta }\l... | Yes |
The case of the logarithmic utility function\n\n\[ U\left( x\right) = \log \left( x\right) \] | We conjecture that the solution to the HJB PDE might be of the form\n\n\[ V\left( {t, x}\right) = \log \left( x\right) + K\left( {T - t}\right) \]\n\nfor some constant \( K \) . Indeed, using\n\n\[ {V}_{t} = - K,\;{V}_{x} = \frac{1}{x},\;{V}_{xx} = - \frac{1}{{x}^{2}} \]\n\nwe see that this function satisfies the HJB P... | Yes |
Example 4.9 (Log Utility) Let \( U\left( x\right) = \log \left( x\right) \) . Then \( I\left( z\right) = 1/z \) and \( \lambda = 1/x \) . Therefore, we need to have \( \widehat{X}\left( T\right) = x/\bar{Z}\left( T\right) \) . | Moreover, since \( \bar{Z}\left( t\right) \widehat{X}\left( t\right) \) is a martingale process,\n\n\[ \bar{Z}\left( t\right) \widehat{X}\left( t\right) = {E}_{t}\left\lbrack {\bar{Z}\left( T\right) \widehat{X}\left( T\right) }\right\rbrack = x \]\n\nIn particular, at time \( t = 1 \)\n\n\[ \widehat{X}\left( 1\right) =... | Yes |
Let us revisit the example with \( U\left( x\right) = \log \left( x\right) \) . We have | \[ {U}^{\prime }\left( x\right) = \frac{1}{x},\;I\left( z\right) = {\left( {U}^{\prime }\right) }^{-1}\left( z\right) = \frac{1}{z} \]\n\nFrom equation (4.72), the optimal wealth level is then\n\n\[ \widehat{X}\left( T\right) = \frac{1}{\widehat{\lambda }\bar{Z}\left( T\right) } \]\n\nThe budget constraint equation for... | Yes |
Example 4.11 (Logarithmic Utility) Let us again look at the log utility \( U\left( x\right) = \log \left( x\right) \) . We can use the same arguments as in the full-information case, that is, as in Example 4.10. Following the same steps as in that example, we get | \[ \widehat{\pi }\left( t\right) = {\sigma }^{-1}\widetilde{\theta }\left( t\right) {X}^{x,\widehat{\pi }}\left( t\right) \] (4.84) We see that the optimal portfolio is of the same form. The only difference is that we replace the unobserved market price of risk \( \theta \) (equivalently, the unobserved drift \( \mu \)... | Yes |
Example 4.12 (Power Utility) We consider power utility \( U\left( x\right) = {x}^{\gamma }/\gamma ,\gamma < 0 \), and the problem of maximizing expected terminal wealth \( E\left\lbrack {U(X\left( T\right) }\right. \) ]. We assume a particularly simple case in which \( \theta = \left( {\mu - r}\right) /\sigma \) is ass... | \[ \widehat{\Pi }\left( t\right) = \frac{{\sigma }^{-1}\widetilde{\theta }\left( t\right) }{1 - \gamma - {\gamma V}\left( t\right) \left( {T - t}\right) } \] (4.85) where \( V\left( t\right) \) is the conditional variance of \( \theta \), and \( \widetilde{\theta }\left( t\right) \) is the conditional expectation of \(... | Yes |
Example 5.1 (Deriving a Risk-Free Portfolio) Suppose that the stock of company A has an expected return of \( {10}\% \) and a standard deviation of \( 8\% \) . The stock of company B has an expected return of \( {14}\% \) and a standard deviation of \( {18}\% \) . The correlation between A and \( \mathrm{B} \) is 1 . W... | From equations (5.3) and (5.5), we get\n\n\[ \n{\sigma }_{\Pi }^{2} = {\Pi }_{A}^{2}{\sigma }_{A}^{2} + {\Pi }_{B}^{2}{\sigma }_{B}^{2} + 2{\Pi }_{A}{\Pi }_{B}{\sigma }_{AB} = {\left( {\Pi }_{A}{\sigma }_{A} + {\Pi }_{B}{\sigma }_{B}\right) }^{2} \n\] \n\nwhere we have used the fact that \( {\rho }_{AB} = 1 \) . We wan... | Yes |
Example 5.2 (Two Assets) In the case of only two assets, the problem becomes degenerate, in the sense that the portfolio weights are determined directly from the constraints (5.7). Indeed, suppose that we want to achieve a mean \( \mu \) with two assets. Then, denoting by \( \Pi \) the proportion in asset 1 [and, there... | Suppose for example that \( {\mu }_{1} = {0.15} \) and \( {\mu }_{2} = {0.10} \) . Then we have\n\n\[ {\Pi }_{1} = {20}\left( {\mu - {0.1}}\right) ,\;{\Pi }_{2} = 3 - {20\mu } \]\n\nThe portfolio variance is given by\n\n\[ {\sigma }^{2} = {400}{\left( \mu - {0.1}\right) }^{2}{\sigma }_{1}^{2} + {\left( 3 - {20}\mu \rig... | Yes |
Example 5.3 (Three Assets) Consider now the case of three uncorrelated assets, all having the same variance equal to 0.01 . Suppose also that\n\n\[ \n{\mu }_{1} = {0.1},\;{\mu }_{2} = {0.2},\;{\mu }_{3} = {0.3} \n\]\n\nThen, equations (5.8) become\n\n\[ \n{0.01}{\Pi }_{1} - {0.1}{\lambda }_{1} - {\lambda }_{2} = 0 \n\]... | To recap, equations (5.8) give us a way to find optimal weights for the assets in our portfolio (that is, the weights that yield the lowest possible variance for a given expected return) in a single-period model, assuming we know the variance-covariance structure, as well as the expected return of the assets. We will u... | No |
Example 5.4 (Optimal Mutual Fund) Taf can invest in asset 1 with \( {\mu }_{1} = {0.2},{\sigma }_{1} = {0.4} \) and asset 2 with \( {\mu }_{2} = {0.3},{\sigma }_{2} = {0.5} \), with correlation \( \rho = {0.2} \), and in the risk-free asset with return \( R = {0.05} \) . He wants to identify the weights of asset 1 and ... | Solving this system we get\n\n\[ \n{\Pi }_{1} = {0.7162}{\lambda }_{1},\;{\Pi }_{2} = {0.8854}{\lambda }_{1} \n\]\n\nFrom \( {\Pi }_{1} + {\Pi }_{2} = 1 \), we get \( {\lambda }_{1} = {0.6244} \) and\n\n\[ \n{\Pi }_{1} = {0.4472},\;{\Pi }_{2} = {0.5528} \n\]\n\nThis result means that first Taf should decide how much mo... | Yes |
Example 6.1 (Deviation from the Put-Call Parity) Suppose that the today's price of stock \( S \) is \( S\\left( 0\\right) = \\$ {48} \) . The European call and put on \( S \) with strike price 45 and maturity \( T \) of three months are trading at \( \\$ {4.95} \) and \( \\$ {0.70} \), respectively. Three-month T-bills... | We can find the discount factor, call it \( d \), from the price of the T-bill:\n\n\[ \n{98.5} = {100d} \n\]\n\nfrom which \( d = {0.985} \) . We now compute the value of the left-hand side of the put-call parity,\n\n\[ \nc\\left( t\\right) + {Kd} = {49.275} \n\]\n\nand the value of the right-hand side,\n\n\[ \np\\left... | Yes |
Example 6.2 (Forward Pricing with Dividends) A given stock trades at \( \$ {100} \), and in six months it will pay a dividend of \$5.65. The one-year continuous interest rate is \( {10}\% \), and the six-month continuous interest rate is \( {7.41}\% \) . According to the analogue of expression (6.9) with different annu... | We take a long position in the forward contract (at zero initial cost) and sell the stock short for \( \$ {100} \) . We buy the six-month bond in the amount of \( {5.65}{e}^{-{0.0741} \times {0.5}} = {5.4445} \) . We invest the remaining balance, \( {100} - {5.4445} = {94.5555} \), in the one-year bond. In six months, ... | Yes |
Example 6.3 (Arbitrage by Replication) Recall Example 3.7: a single-period model with \( r = {0.005}, S\left( 0\right) = {100},{s}_{1} = {101} \), and \( {s}_{2} = {99} \), where \( {s}_{1} \) and \( {s}_{2} \) are the two possible prices of the stock at moment 1, that is, \( u = {1.01}, d = {0.99} \) . The payoff is a... | On the one hand, suppose, for example, that the price of the option is larger than the replication cost, say, equal to \( c = \$ {1.00} \) . Taf can then sell the option for \( \$ {1.00} \), deposit \$0.254 of that amount in the bank, and use the remaining \$0.746 to replicate the option's payoff. Suppose first that at... | Yes |
Consider again the setting of Example 6.3, with \( r = {0.005}, S\left( 0\right) = {100},{s}_{1} = {101},{s}_{2} = {99} \), and the price of the call option equal to the replication cost \( C\left( 0\right) = {0.746} \) . Let us compute the risk-neutral probabilities in this example. For the discounted stock to be a ma... | Solving for \( {p}^{ * } \), we obtain \( {p}^{ * } = {0.75} \) . Then, computing the risk-neutral expected value of the discounted payoff of the call option, we get\n\n\[ \n{E}^{ * }\left\lbrack {\bar{C}\left( 1\right) }\right\rbrack = {p}^{ * }\frac{{100} - 1}{1.005} = {0.746} \n\]\n\nwhich is the same amount we obta... | Yes |
Consider a two-period CRR model with continuously compounded interest rate \( r = {0.05}, S\left( 0\right) = {100}, u = {1.1} \), and \( d = {0.9} \) . The payoff is the European at-the-money call option with strike price \( K = S\left( 0\right) = {100} \) . We take \( {\Delta t} = 1 \) . We compute first the possible ... | We compute the risk-neutral probability\n\n\[ \n{p}^{ * } = \frac{{e}^{r\Delta t} - d}{u - d} = {0.7564} \n\]\n\nWe can then compute the option value at the previous node as\n\n\[ \n{e}^{-{r\Delta t}}\left\lbrack {{p}^{ * } \cdot {21} + \left( {1 - {p}^{ * }}\right) \cdot 0}\right\rbrack = {15.1088} \n\]\n\nSimilarly, ... | Yes |
Consider the same framework as in Example 7.1, that is, a two-period Cox-Ross-Rubinstein model with continuously compounded interest rate \( r = {0.05} \) and parameter values \( S\left( 0\right) = {100}, u = {1.1} \), and \( d = {0.9} \) . The security we want to price is an American at-the-money put option, so that t... | \[ \max \left\{ {{10},{e}^{-{r\Delta t}}\left\lbrack {{p}^{ * } \cdot 1 + \left( {1 - {p}^{ * }}\right) \cdot {19}}\right\rbrack }\right\} = \max \{ {10},{5.1229}\} = {10} \] | Yes |
Consider a two-period binomial setting with \( S\left( 0\right) = \$ {50}, r = 2\% \), and \( {\Delta t} = 1 \) . After one period the price of the stock can go up to \$55 or drop to \$47, and it will pay (in both cases) a dividend of \$3. If it goes up the first period, the second period it can go up to \$58 or down t... | In figure 7.8 each node in the tree shows the price of the stock and the price of the option, which we compute in this example. After one period the stock has two prices: before and after the dividend. After the dividend the price falls by the amount of the dividend. Because of dividends we cannot work here with the pr... | Yes |
Example 7.4 (Exchange Option) In the context of the model discussed in this section, consider the option to exchange one asset for another. This is an option that, at maturity, gives its holder the right to deliver asset \( {S}_{1} \) and receive in exchange asset \( {S}_{2} \). The payoff of this option is \[ g\left( ... | Since we have \[ {\left( {s}_{2} - {s}_{1}\right) }^{ + } = {s}_{1}{\left( \frac{{s}_{2}}{{s}_{1}} - 1\right) }^{ + } \] it is reasonable to expect that the option price will be of the form \[ C\left( {t,{s}_{1},{s}_{2}}\right) = {s}_{1}Q\left( {t, z}\right) \] for some function \( Q \) and a new variable \( z = {s}_{2... | No |
Consider a two-period setting, with dates \( t = 0,1,2 \) . We assume that the interest rate at the initial date \( t = 0 \) is \( 5\% \) . Suppose that the interest rate changes in the binomial tree according to an up factor \( u = {1.1} \) and a down factor \( d = {0.9} \) . Therefore, at \( t = 1 \) the interest rat... | This assumption gives us the following bond prices at time-1 nodes:\n\n\[ {P}^{u}\left( {1,2}\right) = \frac{1}{1.055}\left( {{0.5} \cdot 1 + {0.5} \cdot 1}\right) = {0.9479} \]\n\n\[ {P}^{d}\left( {1,2}\right) = \frac{1}{1.045}\left( {{0.5} \cdot 1 + {0.5} \cdot 1}\right) = {0.9569} \]\n\nThis result means that at \( ... | Yes |
Suppose that the one-year interest rate today is \( 4\% \) . In our model one year from now, the one-year interest rate can only have two values: it can either go up to \( 5\% \) or go down to \( 3\% \) . Our market consists of a two-year pure discount bond with nominal value \$100 that is trading at \$92.278. We want ... | The absence of arbitrage implies that the price of the current two-year bond has to be equal to the price of the contract that pays the price of this bond one year from now (at which moment it will be a one-year bond). If we denote by \( p \) the risk-neutral probability that the interest rate will be \( 5\% \) one yea... | Yes |
Consider a three-period setting with spot rates \( {r}_{0}\left( 1\right) = 4\% \) , \( {r}_{0}\left( 2\right) = 5\% \), and \( {r}_{0}\left( 3\right) = {5.5}\% \) . We assume that \( p = {0.5} \) and our estimate of \( \delta \) is \( \delta = {0.99} \) , so that \( {h}^{u}\left( 2\right) = {1.01005},{h}^{u}\left( 1\r... | After one period the one-period bond has a value of \( \$ 1 \), by definition. The two-period bond will have a value, after one period, in the upper node equal to\n\n\[ \n{P}_{1}^{u}\left( 1\right) = \frac{1}{1 + {f}_{0}\left( {1,2}\right) }{h}^{u}\left( 1\right) = {0.9481} \n\] \n\nand in the lower node \n\n\[ \n{P}_{... | Yes |
Example 8.5 (Bond Prices in the Vasicek Model) Consider the Vasicek model in the form\n\n\\[ \n{dr}\\left( t\\right) = \\left\\lbrack {b - {ar}\\left( t\\right) }\\right\\rbrack {dt} + {\\sigma dW}\\left( t\\right)\n\\]\n\nwith constant parameters \\( b, a \\), and \\( \\sigma \\) . The Ricatti ODE of equation (8.29) i... | The solution for \\( A \\) is\n\n\\[ \nA\\left( {t, T}\\right) = - b{\\int }_{t}^{T}B\\left( {u, T}\\right) {du} + \\frac{{\\sigma }^{2}}{2}{\\int }_{t}^{T}{B}^{2}\\left( {u, T}\\right) {du}\n\\]\n\nA computation of the previous expression yields\n\n\\[ \nA = \\frac{1}{{a}^{2}}\\left\\lbrack {\\left( {B - T + t}\\right... | Yes |
Example 8.6 (Flat Term Structure) Consider the case of a flat volatility structure\n\n\\[ \n\\sigma \\left( {t, T}\\right) \\equiv \\sigma \n\\]\n\nfor some positive constant \\( \\sigma \\) . This implies\n\n\\[ \n\\alpha \\left( {t, T}\\right) = \\sigma {\\int }_{t}^{T}{\\sigma du} = {\\sigma }^{2}\\left( {T - t}\\ri... | Integrating, we obtain\n\n\\[ \nf\\left( {t, T}\\right) = f\\left( {0, T}\\right) + {\\sigma }^{2}t\\left( {T - t/2}\\right) + {\\sigma W}\\left( t\\right) \n\\]\n\nHence,\n\n\\[ \nr\\left( t\\right) = f\\left( {t, t}\\right) = f\\left( {0, t}\\right) + {\\sigma }^{2}{t}^{2}/2 + {\\sigma W}\\left( t\\right) \n\\]\n\nor... | Yes |
Example 8.7 (Call Option Price in Affine Models) Consider a call option with maturity at time \( {T}_{1} \), written on a bond with maturity at time \( {T}_{2} \), with \( {T}_{1} \leq {T}_{2} \). The payoff of the option is\n\n\[ C = {\left\lbrack P\left( {T}_{1},{T}_{2}\right) - K\right\rbrack }^{ + }\n\]\nHere, the ... | Using Itô's rule we get\n\n\[ {dF}\left( t\right) = F\left( t\right) \left\lbrack \ldots \right\rbrack {dt} + F\left( t\right) {\sigma }_{F}\left( t\right) {dW}\left( t\right) \n\]\nwith\n\n\[ {\sigma }_{F}\left( t\right) = - \sigma \left\lbrack {B\left( {t,{T}_{2}}\right) - B\left( {t,{T}_{1}}\right) }\right\rbrack \n... | Yes |
We want to price the position in a swap of an investor who receives a \( {10}\% \) fixed rate in semiannually paid coupons and pays the six-month LIBOR. The swap still has nine months left to maturity. At the last resetting date, the six-month LIBOR was \( 6\% \). The continuous three-month and nine-month rates are \( ... | \[ {C}_{1} = \frac{1}{P\left( {{T}_{0},{T}_{1}}\right) } - \left( {1 + {R\Delta T}}\right) \] and nine months from now it is \[ {C}_{2} = \frac{1}{P\left( {{T}_{1},{T}_{2}}\right) } - \left( {1 + {R\Delta T}}\right) \] The first payoff is already known, and its price is simply obtained by multiplying by the price of th... | Yes |
Example 9.1 (Cross-Hedging Currency Risk) Consider a U.S. firm that has a contract with a company in Surfland, a country whose official currency is called the val. The Surfland company will pay \( 1\mathrm{{million}} \) val to the U.S. firm six months from now. The U.S. firm wants to hedge its exposure to the exchange-... | formula (9.3) for the minimum variance hedge ratio. Since \( U = T \), the value \( {F}_{2}\left( {T, T}\right) \) is simply the value of the playa exchange rate at time \( T \) (rather than the value of a futures contract on playa). We get\n\n\[ \delta = {0.00054}/{0.0004} = {1.35} \]\n\nTherefore, for each val the U.... | Yes |
Denote by \( \Gamma \) the gamma of portfolio \( X \), and by \( {\Gamma }_{C} \) the gamma of a contingent claim \( C \) . Taf wants to buy or sell \( n \) contracts of \( C \) in order to make the portfolio gamma-neutral. That is, Taf wants\n\n\[ \Gamma + n{\Gamma }_{C} = 0 \] | hence\n\n\[ n = - \frac{\Gamma }{{\Gamma }_{C}} \]\n\nHowever, taking this additional position in \( C \) will change the delta of the portfolio. Taf can then buy or sell some shares of the underlying asset in order to make the portfolio delta-neutral. This action does not change the gamma, because the underlying asset... | Yes |
Consider a bond with a nominal value of \$100 that pays a coupon of \$8 at the end of each year and has five years left until maturity. Suppose that the yield of this bond is \( 5\% \) . The price of the bond is | \[ P = \mathop{\sum }\limits_{{i = 1}}^{5}\frac{8}{{1.05}^{i}} + \frac{100}{{1.05}^{5}} = {112.99} \] The duration of the bond is \[ D = \mathop{\sum }\limits_{{i = 1}}^{5}i\frac{\frac{8}{{1.05}^{i}}}{112.99} + 5\frac{\frac{100}{{1.05}^{5}}}{112.99} = {4.36} \] | Yes |
Example 10.2 (Different Yields) Consider a bond with a nominal value of \$100 that pays a coupon of \$8 at the end of each year and has 30 years left until maturity. We first compute the duration for the case in which the yield of the bond is \( 5\% \) . The price of the bond is | \[ P = \mathop{\sum }\limits_{{i = 1}}^{{30}}i\frac{8}{{1.05}^{i}} + \frac{100}{{1.05}^{30}} = {146.12} \] The duration of the bond is \[ D = \mathop{\sum }\limits_{{i = 1}}^{{30}}i\frac{\frac{8}{{1.05}^{i}}}{146.12} + 5\frac{\frac{100}{{1.05}^{30}}}{146.12} = {14.82} \] We now compute the duration for a yield of \( {1... | Yes |
Example 10.3 (Duration and Change in Bond Price) Suppose that the duration of a bond is 5 and its yield is \( 6\% \) . The investor holding this bond is worried about what will happen to its price if the yield increases to \( 7\% \), so that \( 1 + y \) increases from 1.06 to 1.07, which is close to 1%. From equation (... | Since interest rates in general have relatively low values, an increase of \( y \) by 0.01 usually means an increase in \( 1 + y \) of approximately \( 1\% \) . Therefore, the duration tells us the percentage drop in price we should expect as a result of an increase in interest rates of \( 1\% \) .\n\nAs we have emphas... | Yes |
Suppose that the term structure is flat and the annual interest rate is \( 5\% \) . We have a liability with a nominal value of \( \$ {100} \), and the payment will take place in two years. In the market there are only two pure discount bonds: a one-year pure discount bond with nominal value \( \$ {100} \) and a four-y... | Suppose that immediately after investing in such a portfolio, interest rates go up to \( 6\% \) (and the term structure remains flat). The present values of both the liability and the portfolio of bonds drop. The new values of the liability and the portfolio of bonds are\n\n\[ \n\frac{100}{{1.06}^{2}} = {89},\;{0.63}\f... | Yes |
Consider the liability of the previous example and the portfolio of bonds formed for immunization purposes. Suppose that right after the portfolio of bonds is formed, the interest rates go down to \( 4\% \) . As we computed in Example 10.4, the increase in the value of the portfolio will match the increase in the value... | As we saw in Example 10.4, the new price of the portfolio of bonds is 92.20. The new weights of the one-year bond and the four-year bond are\n\n\[ \frac{{0.63}\frac{100}{1.04}}{92.20} = {0.66},\;\frac{{0.37}\frac{100}{{1.04}^{4}}}{92.20} = {0.34} \]\n\nThe duration of the portfolio is\n\n\[ D = {0.66} \cdot 1 + {0.34} ... | Yes |
Example 11.1 (Pricing in a Tree) As we did in Example 7.2, consider a Cox-Ross-Rubinstein model with continuously compounded interest rate \( r = {0.05}, S\\left( 0\\right) = {100} \) , and \( \\sigma = {0.30} \) . The security we consider is an American at-the-money put option with strike price \( K = S\\left( 0\\righ... | We have here\n\n\[ u = {e}^{\\sigma \\Delta t} = {1.0779},\;d = 1/u = {0.9277} \]\n\nThe risk-neutral probability is\n\n\[ {p}^{ * } = \\frac{{e}^{r\\Delta t} - d}{u - d} = {0.5650} \]\n\nWe compute first the possible values of the stock prices at times 1 and 2 , and payoff values \( g\\left( {S\\left( 2\\right) }\\rig... | No |
Example 11.2 (Option Pricing by Monte Carlo Simulation) Consider the Merton-Black-Scholes model and the European call option with\n\n\[ \nS\\left( 0\\right) = {65}, T = {0.08}, K = {55}, r = 0,\\sigma = {0.317} \n\] \n\nThe SDE for the price \( S \) (in the risk-neutral world) is \n\n\[ \n{dS} = {\\sigma Sd}{W}^{ * } \... | We can simply simulate the values \( {W}^{ * }\\left( T\\right) \) as \( {z}_{i}\\sqrt{T} \), where \( {z}_{i} \) ’s are independent standard normal random values. We use Microsoft Excel's Visual Basic to write and execute a program to perform the simulation. Some simulation results are presented in figure 11.3. For ex... | Yes |
Example 12.1 (Aggregation Property) Suppose that the endowments of the individuals satisfy\n\n\[ \n\\frac{1}{{e}^{A}\\left( 0\\right) } = p\\frac{1}{{e}_{u}^{A}\\left( 1\\right) } + \\left( {1 - p}\\right) \\frac{1}{{e}_{d}^{A}\\left( 1\\right) }\n\]\n\n\[ \n{\\left\\lbrack {e}^{B}\\left( 0\\right) \\right\\rbrack }^{\... | Then it is straightforward to check (see Problem 4) that an equilibrium is obtained if\n\n\[ \n{\\widehat{c}}^{i}\\left( j\\right) = {e}^{i}\\left( j\\right) ,\\;i = A, B, j = 0,1\n\]\n\n\[ \n1 + r = \\frac{1}{\\beta }\n\]\n\nMoreover, there is a representative agent with power utility with coefficient \\( \\alpha < 1 ... | No |
Consider a representative agent who chooses an investment amount \( \pi \) in a single risky security. The objective of the agent is\n\n\[ \mathop{\max }\limits_{{\{ \pi \} }}\left\{ {-E\left\lbrack {e}^{-{\alpha X}\left( 1\right) }\right\rbrack }\right\} \] \n\nOne dollar invested in the risk-free security at time 0 r... | In this setting, a possible equilibrium problem is the following: Suppose that the standard deviation \( \sigma \) of the stochastic return is given. What would be the expected return \( \mu \) at which the individual would optimally invest all of \( X\left( 0\right) \) in the risky security? That is, we want \( \wideh... | Yes |
Example 12.3 (Log Utility) The representative agent chooses optimal consumption according to\n\n\[ \mathop{\max }\limits_{{\{ c\} }}E\left\lbrack {{\int }_{0}^{T}{e}^{-{\beta t}}\log c\left( t\right) {dt}}\right\rbrack \] | From equation (12.22) the optimal consumption is\n\n\[ \widehat{c}\left( t\right) = \lambda {e}^{{\int }_{0}^{t}\left\lbrack {r\left( u\right) - \beta + \frac{1}{2}{\theta }^{2}\left( u\right) }\right\rbrack {du} + {\int }_{0}^{t}\theta \left( u\right) {dW}\left( u\right) } \]\n\n(12.26)\n\nSince, in equilibrium, \( \w... | Yes |
Assume that both agents have logarithmic utility, \( {U}_{i}\left( c\right) = \log \left( c\right) \). Then the representative agent’s utility function \( U \) is obtained by maximizing\n\n\[ \n{\lambda }_{A}\log \left( {c}^{A}\right) + {\lambda }_{B}\log \left( {c - {c}^{A}}\right) \n\] | which gives\n\n\[ \n{\widehat{c}}^{i} = \frac{{\lambda }_{i}}{{\lambda }_{A} + {\lambda }_{B}}c \n\]\n\nand\n\n\[ \nU\left( c\right) = \left( {{\lambda }_{A} + {\lambda }_{B}}\right) \log \left( c\right) + {\lambda }_{A}\log \left( \frac{{\lambda }_{A}}{{\lambda }_{A} + {\lambda }_{B}}\right) + {\lambda }_{B}\log \left... | Yes |
Ideally, we would like to use equilibrium models for predicting the difference in return between the risk-free asset and the risky asset, if not in precise quantitative terms, then at least in a theoretically sound qualitative way. We see from \( \theta = {\sigma }^{-1}\left( {\mu - r}\right) \) and equation (12.47) th... | In practice, a broad market index such as the S&P 500 can be used as a proxy for \( S \), and there are also ways to estimate the aggregate consumption in the economy. Thus, in principle, we can test the CCAPM using available economic data, and use it for future predictions. However, there is one big difficulty: the es... | Yes |
Example 13.1 (Market Portfolio) Consider a very simple market that consists of only two risky assets, A and B. Suppose that there are 1,000 shares of asset A with a market value of \$4.00 per share, and 2,000 shares of asset B with a market value of \$3.00 per share. Then the total capitalization of asset A is \$4,000,... | Let us repeat again the CAPM equilibrium idea: if a security is traded, it has to be one of the securities that form portfolio \( M \), because this is the only portfolio the investors will consider. Then, and this is the corollary of our analysis, the portfolio \( M \) has to be the market portfolio, which includes al... | Yes |
Example 13.2 (Capital Market Line) Our CAPM investor Taf considers a venture that is expected to return \( {16}\% \) after a year, with the uncertainty measured by a standard deviation of \( {50}\% \) . The risk-free rate is \( 5\% \) and the market expected return is \( {9.5}\% \), with the standard deviation \( {15}\... | \[ {0.05} + \frac{{0.095} - {0.05}}{0.15}{0.50} = {0.2} \] or \( {20}\% \) . We conclude that this venture is not on the capital market line. (Our conclusion does not necessarily mean it is a bad investment if combined with other investments.) | Yes |
Example 13.3 (Diversifying Specific Risk) Suppose, for simplicity, that all securities have the same specific risk \( {\sigma }_{{\epsilon }_{i}}^{2} = {\sigma }_{\epsilon }^{2} \) and that \( {\epsilon }_{i} \) are independent of each other. Consider a portfolio that invests the same proportion \( 1/N \) in the \( N \... | The variance \( {\sigma }_{P}^{2} \) of our portfolio is given by\n\n\[ \n{\sigma }_{P}^{2} = \operatorname{Var}\left\lbrack {\frac{1}{N}\mathop{\sum }\limits_{{j = 1}}^{N}{R}_{j}}\right\rbrack = \frac{1}{{N}^{2}}\mathop{\sum }\limits_{{j = 1}}^{N}\mathop{\sum }\limits_{{i = 1}}^{N}{\sigma }_{ji} \]\n\nUsing the fact t... | Yes |
Example 13.4 (Using the CAPM for Estimation) Suppose that the expected return and standard deviation of the market portfolio are \( {11}\% \) and \( {20}\% \), respectively. The expected return of security 1 is \( 6\% \) . The expected return of security 2 is \( {10}\% \), and its standard deviation is \( {18}\% \) . A... | In order to compute this risk, we first need to find the beta of security 2 . We know that we have \[ {0.5}{\beta }_{1} + {0.5}{\beta }_{2} = {0.5} \] from which we get \( {\beta }_{1} = 1 - {\beta }_{2} \) . We can now solve the system of two CAPM equations: \[ {0.06} - r = \left( {1 - {\beta }_{2}}\right) \left( {{0.... | Yes |
Example 13.5 (CAPM Pricing) Suppose that our investor Taf considers investing in a mutual fund whose beta is reported as 0.8 . The fund has been returning an average excess return of \( {15}\% \) in the last five years. Thus, for lack of better information, Taf estimates the expected annual return to be \( {15}\% \) ab... | \[ S\left( 0\right) = \frac{{50} \cdot {1.2}}{1 + {0.05} + {0.8} \cdot {0.15}} = {51.28} \] | Yes |
Proposition 1.1 Suppose \( \mathbf{u} = \left( {{u}_{1},{u}_{2},\ldots ,{u}_{r}}\right) \) and \( \mathbf{v} = \left( {{v}_{1},\ldots ,{v}_{s}}\right) \) are lists of positive integers. Let \( \mathbf{{uv}} \) denote the concatenation \( \left( {{u}_{1},\ldots ,{u}_{r},{v}_{1},\ldots ,{v}_{s}}\right) \) of \( \mathbf{u... | The proof is by induction. The identity \( {q}_{j + 1} = {u}_{j + 1}{q}_{j} + {q}_{j - 1} \) determines the successive denominators, together with \( {q}_{0} \mathrel{\text{:=}} 1 \) and \( {q}_{-1} \mathrel{\text{:=}} 0 \) . Clearly, the determinant obeys this same recursion, and agrees with \( {q}_{r} \) for \( r = 1... | Yes |
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