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Theorem 17.1. Let Algorithm 17.1 be applied to a problem (17.1) consisting of floating point numbers on a computer satisfying (13.7). This algorithm is backward stable in the sense that the computed solution \( \widetilde{x} \in {\mathbb{C}}^{m} \) satisfies\n\n\[ \left( {R + {\delta R}}\right) \widetilde{x} = b \]\n\n... | To keep the ideas clear and interesting, our proof will be most leisurely.\n\n\( m = 1 \)\n\nAccording to (17.3), our task is to express every floating point error as a perturbation of the input. Let us begin with the simplest case, where \( R \) is of dimension \( 1 \times 1 \) . Back substitution in this case consist... | Yes |
Let the factorization \( {PA} = {LU} \) of a matrix \( A \in {\mathbb{C}}^{m \times m} \) be computed by Gaussian elimination with partial pivoting (Algorithm 21.1) on a computer satisfying the axioms (13.5) and (13.7). Then the computed matrices \( \widetilde{P},\widetilde{L} \), and \( \widetilde{U} \) satisfy\n\n\[ ... | Is Gaussian elimination backward stable? According to Theorem 22.2 and our definition (14.5) of backward stability, the answer is yes if \( \rho = O\left( 1\right) \) uniformly for all matrices of a given dimension \( m \), and otherwise no. | Yes |
Theorem 23.1. Every hermitian positive definite matrix \( A \in {\mathbb{C}}^{m \times m} \) has a unique Cholesky factorization (23.3). | Proof. Existence is what we just discussed; a factorization exists since the algorithm cannot break down. In fact, the algorithm also establishes uniqueness. At each step (23.2), the value \( \alpha = \sqrt{{a}_{11}} \) is determined by the form of the \( {R}^{ * }R \) factorization, and once \( \alpha \) is determined... | Yes |
Theorem 23.2. Let \( A \in {\mathbb{C}}^{m \times m} \) be hermitian positive definite, and let a Cholesky factorization of \( A \) be computed by Algorithm 23.1 on a computer satisfying (13.5) and (13.7). For all sufficiently small \( {\epsilon }_{\text{machine }} \), this process is guaranteed to run to completion (i... | Like so many algorithms of numerical linear algebra, this one would look much worse if we tried to carry out a forward error analysis rather than a backward one. If \( A \) is ill-conditioned, \( \widetilde{R} \) will not generally be close to \( R \) ; the best we can say is \( \parallel \widetilde{R} - R\parallel /\p... | Yes |
Theorem 24.1. \( \lambda \) is an eigenvalue of \( A \) if and only if \( {p}_{A}\left( \lambda \right) = 0 \) . | Proof. This follows from the definition of an eigenvalue:\n\n\( \lambda \) is an eigenvalue \( \Leftrightarrow \) there is a nonzero vector \( x \) such that \( {\lambda x} - {Ax} = 0 \)\n\n\[ \Leftrightarrow {\lambda I} - A\text{is singular} \]\n\n\[ \Leftrightarrow \det \left( {{\lambda I} - A}\right) = 0. \] | Yes |
Theorem 24.3. If \( X \) is nonsingular, then \( A \) and \( {X}^{-1}{AX} \) have the same characteristic polynomial, eigenvalues, and algebraic and geometric multiplicities. | Proof. The proof that the characteristic polynomials match is a straightforward computation:\n\n\[ \n{p}_{{X}^{-1}{AX}}\left( z\right) = \det \left( {{zI} - {X}^{-1}{AX}}\right) = \det \left( {{X}^{-1}\left( {{zI} - A}\right) X}\right) \n\]\n\n\[ \n= \det \left( {X}^{-1}\right) \det \left( {{zI} - A}\right) \det \left(... | Yes |
Theorem 24.4. The algebraic multiplicity of an eigenvalue \( \lambda \) is at least as great as its geometric multiplicity. | Proof. Let \( n \) be the geometric multiplicity of \( \lambda \) for the matrix \( A \) . Form an \( m \times n \) matrix \( \widehat{V} \) whose \( n \) columns constitute an orthonormal basis of the eigenspace \( \{ x : {Ax} = {\lambda x}\} \) . Then, extending \( \widehat{V} \) to a square unitary matrix \( V \) , ... | Yes |
Consider the matrices \[ A = \left\lbrack \begin{array}{lll} 2 & & \\ & 2 & \\ & & 2 \end{array}\right\rbrack ,\;B = \left\lbrack \begin{array}{lll} 2 & 1 & \\ & 2 & 1 \\ & & 2 \end{array}\right\rbrack . \] Both \( A \) and \( B \) have characteristic polynomial \( {\left( z - 2\right) }^{3} \), so there is a single ei... | An eigenvalue whose algebraic multiplicity exceeds its geometric multiplicity is a defective eigenvalue. A matrix that has one or more defective eigenvalues is a defective matrix. Any diagonal matrix is nondefective. For such a matrix, both the algebraic and the geometric multiplicities of an eigenvalue \( \lambda \) a... | Yes |
An \( m \times m \) matrix A is nondefective if and only if it has an eigenvalue decomposition \( A = {X\Lambda }{X}^{-1} \) . | Proof. \( \left( \Leftarrow \right) \) Given an eigenvalue decomposition \( A = {X\Lambda }{X}^{-1} \), we know by Theorem 24.3 that \( \Lambda \) is similar to \( A \), with the same eigenvalues and the same multiplicities. Since \( \Lambda \) is a diagonal matrix, it is nondefective, and thus the same holds for \( A ... | Yes |
Theorem 24.6. The determinant \( \det \left( A\right) \) and trace \( \operatorname{tr}\left( A\right) \) are equal to the product and the sum of the eigenvalues of \( A \), respectively, counted with algebraic multiplicity: | Proof. From (24.5) and (24.6), we compute\n\n\[ \det \left( A\right) = {\left( -1\right) }^{m}\det \left( {-A}\right) = {\left( -1\right) }^{m}{p}_{A}\left( 0\right) = \mathop{\prod }\limits_{{j = 1}}^{m}{\lambda }_{j}. \]\n\nThis establishes the first formula. As for the second, from (24.5), it follows that the coeffi... | Yes |
Theorem 24.9. Every square matrix \( A \) has a Schur factorization. | Proof. We proceed by induction on the dimension \( m \) of \( A \) . The case \( m = 1 \) is trivial, so suppose \( m \geq 2 \) . Let \( x \) be any eigenvector of \( A \), with corresponding eigenvalue \( \lambda \) . Take \( x \) to be normalized and let it be the first column of a unitary matrix \( U \) . Then, just... | Yes |
Theorem 27.1. Suppose \( \left| {\lambda }_{1}\right| > \left| {\lambda }_{2}\right| \geq \cdots \geq \left| {\lambda }_{m}\right| \geq 0 \) and \( {q}_{1}^{T}{v}^{\left( 0\right) } \neq 0 \) . Then the iterates of Algorithm 27.1 satisfy\n\n\[ \begin{Vmatrix}{{v}^{\left( k\right) } - \left( {\pm {q}_{1}}\right) }\end{V... | Proof. The first equation follows from (27.4), since \( {a}_{1} = {q}_{1}^{T}{v}^{\left( 0\right) } \neq 0 \) by assumption. The second follows from this and (27.3). If \( {\lambda }_{1} > 0 \), then the \( \pm \) signs are all + or all -, whereas if \( {\lambda }_{1} < 0 \), they alternate. | Yes |
Theorem 27.3. Rayleigh quotient iteration converges to an eigenvalue/eigenvector pair for all except a set of measure zero of starting vectors \( {v}^{\left( 0\right) } \) . When it converges, the convergence is ultimately cubic in the sense that if \( {\lambda }_{J} \) is an eigenvalue of \( A \) and \( {v}^{\left( 0\... | Proof. We shall not prove the assertion about convergence for almost all starting vectors. Here, however, is a proof that if convergence occurs, it is ultimately cubic. For simplicity, we assume that the eigenvalue \( {\lambda }_{J} \) is simple. By (27.3), if \( \begin{Vmatrix}{{v}^{\left( k\right) } - {q}_{J}}\end{Vm... | Yes |
Theorem 28.1. Suppose that the iteration (28.1)-(28.3) is carried out and that assumptions (28.4) and (28.5) are satisfied. Then as \( k \rightarrow \infty \), the columns of the matrices \( {\widehat{Q}}^{\left( k\right) } \) converge linearly to the eigenvectors of \( A \) : | Proof. Extend \( \widehat{Q} \) to a full \( m \times m \) orthogonal matrix \( Q \) of eigenvectors of \( A \), and let \( \Lambda \) be the corresponding diagonal matrix of eigenvalues; thus \( A = {Q\Lambda }{Q}^{T} \) . Just as \( \widehat{Q} \) is the leading \( m \times n \) section of \( Q \), define \( \widehat... | Yes |
Theorem 28.3. The processes (28.7)-(28.10) and (28.11)-(28.14) generate identical sequences of matrices \( {\underline{R}}^{\left( k\right) },{\underline{Q}}^{\left( k\right) } \), and \( {A}^{\left( k\right) } \), namely, those defined by the QR factorization of the \( k \) th power of \( A \) , \[ {A}^{k} = {\underli... | Proof. We proceed by induction in \( k \) . The base case \( k = 0 \) is trivial. For both simultaneous iteration and the QR algorithm, equations (28.7)-(28.15) imply \( {A}^{0} = {\underline{Q}}^{\left( 0\right) } = {\underline{R}}^{\left( 0\right) } = I \) and \( {A}^{\left( 0\right) } = A \), from which (28.16) and ... | Yes |
Theorem 34.1. As long as the Arnoldi iteration does not break down (i.e., \( {K}_{n} \) is of full rank \( n \) ),(34.3) has a unique solution \( {p}^{n} \), namely, the characteristic polynomial of \( {H}_{n} \) . | Proof. First we note that if \( p \in {P}^{n} \), then the vector \( p\left( A\right) b \) can be written \( p\left( A\right) b = {A}^{n}b - {Q}_{n}y \) for some \( y \in {\mathbb{C}}^{n} \), where \( {Q}_{n} \) is defined by (33.1). In other words, (34.3) is equivalent to a linear least squares problem: find the point... | Yes |
Here is a numerical example. Let \( A \) be a \( {200} \times {200} \) matrix whose entries are independent samples from the real normal distribution of mean 2 and standard deviation \( {0.5}/\sqrt{200} \) . In MATLAB, | \[ m = {200};A = 2 * \operatorname{eye}\left( m\right) + {0.5} * \operatorname{randn}\left( m\right) /\operatorname{sqrt}\left( m\right) \text{.} \] (35.17) Figure 35.2 shows the eigenvalues of \( A \), a set of points roughly uniformly dis- \( \bar{\text{tributed in the disk of radius 1/2 centered at }}z = 2 \) (Exerc... | No |
Theorem 36.1. The matrices \( {Q}_{n} \) of vectors \( {q}_{n} \) generated by the Lanczos iteration are reduced QR factors of the Krylov matrix (33.6), | \[ {K}_{n} = {Q}_{n}{R}_{n} \] (36.4) The tridiagonal matrices \( {T}_{n} \) are the corresponding projections \[ {T}_{n} = {Q}_{n}^{ * }A{Q}_{n} \] (36.5) and the successive iterates are related by the formula \[ A{Q}_{n} = {Q}_{n + 1}{\widetilde{T}}_{n} \] (36.6) which we can write in the form of a three-term recurre... | Yes |
Theorem 37.1. Let \( \left\{ {{q}_{n}\left( x\right) }\right\} \) be the sequence of orthogonal polynomials generated by Algorithm 37.1, let \( \left\{ {T}_{n}\right\} \) be the associated sequence of tridiagonal Jacobi matrices, and let \( {p}^{n} \) be the characteristic polynomial of \( {T}_{n} \) . Then for \( n = ... | \[ {p}^{n}\left( x\right) = {C}_{n}{q}_{n + 1}\left( x\right) \] (37.8) where \( {C}_{n} \) is a constant. In particular, the zeros of \( {q}_{n + 1}\left( x\right) \) are the eigenvalues of \( {T}_{n} \) . These \( n \) zeros are distinct and lie in the open interval \( \left( {-1,1}\right) \) . | Yes |
The n-point Gauss-Legendre quadrature formula has order of accuracy exactly \( {2n} - 1 \), and no quadrature formula (37.10) has order of accuracy higher than this. | Proof. Given any set of distinct points \( \left\{ {x}_{j}\right\} \), let \( f\left( x\right) \) be the polynomial \( \mathop{\prod }\limits_{{j = 1}}^{n}{\left( x - {x}_{j}\right) }^{2} \) of degree \( {2n} \) . Then \( I\left( f\right) > 0 \), but \( {I}_{n}\left( f\right) = 0 \) since \( f\left( {x}_{j}\right) = 0 ... | Yes |
Theorem 37.4. Let \( {T}_{n} \) be the \( n \times n \) Jacobi matrix (36.3) defined by Algorithm 37.1 or (36.5), with entries \( {\beta }_{1},\ldots ,{\beta }_{n - 1} \) given by (37.6). Let \( {T}_{n} = \) \( {VD}{V}^{T} \) be an orthogonal diagonalization of \( {T}_{n} \) with \( V = \left\lbrack {{v}_{1}\left| \cdo... | \[ {x}_{j} = {\lambda }_{j},\;{w}_{j} = 2{\left( {v}_{j}\right) }_{1}^{2},\;j = 1,\ldots, n. \] | Yes |
Theorem 38.2. Let the CG iteration be applied to a symmetric positive definite matrix problem \( {Ax} = b \) . If the iteration has not already converged (i.e., \( \left. {{r}_{n - 1} \neq 0}\right) \), then \( {x}_{n} \) is the unique point in \( {\mathcal{K}}_{n} \) that minimizes \( {\begin{Vmatrix}{e}_{n}\end{Vmatr... | Proof. From Theorem 38.1 we know that \( {x}_{n} \) belongs to \( {\mathcal{K}}_{n} \) . To show that it is the unique point in \( {\mathcal{K}}_{n} \) that minimizes \( \parallel e{\parallel }_{A} \), consider an arbitrary point \( x = {x}_{n} - {\Delta x} \in {\mathcal{K}}_{n} \), with error \( e = {x}_{ * } - x = {e... | Yes |
Theorem 38.3. If the CG iteration has not already converged before step \( n \) (i.e., \( {r}_{n - 1} \neq 0 \) ), then (38.8) has a unique solution \( {p}_{n} \in {P}_{n} \), and the iterate \( {x}_{n} \) has error \( {e}_{n} = {p}_{n}\left( A\right) {e}_{0} \) for this same polynomial \( {p}_{n} \) . Consequently we ... | Proof. From Theorem 38.1 it follows that \( {e}_{n} = p\left( A\right) {e}_{0} \) for some \( p \in {P}_{n} \) . The equality in (38.9) is a consequence of this and Theorem 38.2. As for the inequality in (38.9), if \( {e}_{0} = \mathop{\sum }\limits_{{j = 1}}^{m}{a}_{j}{v}_{j} \) is an expansion of \( {e}_{0} \) in ort... | Yes |
Theorem 38.4. If \( A \) has only \( n \) distinct eigenvalues, then the \( {CG} \) iteration converges in at most \( n \) steps. | \( \textit{Proof. This is a corollary of }\left( {38.9}\right) , \) since a polynomial \( p\left( x\right) = \mathop{\prod }\limits_{{j = 1}}^{n}\left( {1 - x/{\lambda }_{j}}\right) \in \) \( {P}_{n} \) exists that is zero at any specified set of \( n \) points \( \left\{ {\lambda }_{j}\right\} \) . | No |
Theorem 38.5. Let the CG iteration be applied to a symmetric positive definite matrix problem \( {Ax} = b \), where \( A \) has 2-norm condition number \( \kappa \) . Then the \( A \) -norms of the errors satisfy\n\n\[ \left| \frac{{\begin{Vmatrix}{e}_{n}\end{Vmatrix}}_{A}}{{\begin{Vmatrix}{e}_{0}\end{Vmatrix}}_{A}}\ri... | Proof. By Theorem 38.3, it is enough to find a polynomial \( p \in {P}_{n} \) whose maximum value for \( \lambda \in \left\lbrack {{\lambda }_{\min },{\lambda }_{\max }}\right\rbrack \) is the middle expression of (38.10). The polynomial we choose is the scaled and shifted Chebyshev polynomial \( p\left( x\right) = \) ... | Yes |
Proposition 1. SR under normal i.i.d. returns assumption is asymptotically normal in \( n \) with standard deviation \( {\sigma }_{{IID},1} \) given by \[ {\sigma }_{{IID},1} = \sqrt{1 + \frac{S{R}_{\infty }^{2}}{2}} \] | ## Proof. Immediate using previous results and given in A.0.1 | No |
Proposition 2. under normal i.i.d. returns assumption, the estimator resulting from the empirical SR and the empirical variance is asymptotically efficient, meaning that it achieves the lower bound in terms of Cramer Rao bound given by\n\n\[ \n{CRB} = \frac{1}{n}\left( \begin{matrix} 1 + S{R}_{\infty }^{2}/2 & - S{R}_{... | Proof. Given in A.0.2 | No |
Proposition 3. The ratio between the \( q \) period returns \( {SR}\left( q\right) \) and the regular \( {SR} \) is the following:\n\n\[ \n\\frac{{SR}\\left( q\\right) }{SR} = \\frac{q{\\sigma }_{\\infty }}{\\sqrt{\\mathop{\\sum }\\limits_{{i = 0}}^{{q - 1}}{\\sigma }_{t - i}^{2} + 2\\mathop{\\sum }\\limits_{{k = 1}}^{... | ## Proof. Given in A.0.3 | No |
Example 3. For the monthly returns of the CRSP value-weighted index (Jan., 1926 — Dec. 1997, \( T = {864} \) ), it is computed that \[ \text{AIC -5.807-5.805-5.817-5.816-5.819-5.821-5.819-5.820-5.821-5.818} \] \( {SE} = \frac{1}{\sqrt{T}} = {0.034},\;\mathbf{2}{SE} = \mathbf{0}.\mathbf{{068}},\;\widehat{p} = 3 \) or \(... | Parameter estimation The AR coefficient is estimated by the least-squares: \[ \mathop{\min }\limits_{\mathbf{b}}\mathop{\sum }\limits_{{t = p + 1}}^{T}{\left( {X}_{t} - {b}_{0} - {b}_{1}{X}_{t - 1} - \cdots - {b}_{p}{X}_{t - p}\right) }^{2}. \] This is an auto-regression problem: \[ \mathop{\min }\limits_{\mathbf{b}}\m... | No |
For the CRSP index, an AR(3) fit results in\n\n\[ \left. {r}_{t}\right. = {0.0103} + {0.104}{r}_{t - 1} - {0.010}{r}_{t - 2} - {0.120}{r}_{t - 3} + {\varepsilon }_{t} \] | For example, to test whether \( {H}_{0} : {b}_{0} = 0 \) (or the mean return is zero),\n\nwe compute the t-statistic \( {0.0103}/{0.002} = 5 \) and hence its associated\n\nP-value is \( {2\Phi }\left( {-5}\right) = 0 \) . We have strong evidence against \( {H}_{0} \), namely,\n\nthe monthly returns are positive. | Yes |
Example 3 (Cont). \( \mu = f\left( \mathbf{b}\right) = \frac{{b}_{0}}{1 - {b}_{1} - {b}_{2} - {b}_{3}} \) . | Thus,\n\n\[ \n{f}^{\prime }\left( \mathbf{b}\right) = {\left( a,{b}_{0},{b}_{0},{b}_{0}\right) }^{T}/{a}^{2},\;a = 1 - {b}_{1} - {b}_{2} - {b}_{3}. \n\] \n\nEvaluation of the gradient at the estimates gives \n\n\[ \n{f}^{\prime }\left( \widehat{\mathbf{b}}\right) = {\left( {0.9746},{0.0098},{0.0098},{0.0098}\right) }^{... | Yes |
Consider the ARMA(1,1) model. For \( k > 1 \), we have\n\n\[ \n\gamma \left( k\right) = {b}_{1}\gamma \left( {k - 1}\right) = \cdots = {b}_{1}^{k - 1}\gamma \left( 1\right) ,\n\]\n\nwhich decays exponentially. | Using \( {X}_{t} = {b}_{1}{X}_{t - 1} + {\varepsilon }_{t} + {a}_{1}{\varepsilon }_{t - 1} \), we have\n\n\[ \n\gamma \left( 1\right) = \mathrm{{Cov}}\left( {{X}_{t},{X}_{t - 1}}\right) = \mathrm{{Cov}}\left( {{b}_{1}{X}_{t - 1} + {a}_{1}{\varepsilon }_{t - 1},{X}_{t - 1}}\right) = {b}_{1}\gamma \left( 0\right) + {a}_{... | Yes |
Consider an ARMA \( \\left( {1,1}\\right) \) model\n\n\[ \n{X}_{t} - b{X}_{t - 1} = {\\varepsilon }_{t} - a{\\varepsilon }_{t - 1}\n\]\nFor \( m \\geq 2 \), it is easy to see\n\n\[ \n{X}_{T}\\left( m\\right) - b{X}_{T}\\left( {m - 1}\\right) = 0 \\Rightarrow {\\widehat{X}}_{T + m} \\equiv {X}_{T}\\left( m\\right) = {b}... | Let us now consider the one-step forecasting. First of all, \( {X}_{T}\\left( 1\\right) = b{X}_{T} - a{\\varepsilon }_{T} \) . Note that\n\n\[ \n{\\varepsilon }_{t} = {\\left( 1 - aB\\right) }^{-1}\\left( {{X}_{t} - b{X}_{t - 1}}\\right)\n\]\n\n\[ \n= \\mathop{\\sum }\\limits_{{j = 0}}^{\\infty }{a}^{j}{B}^{j}\\left( {... | Yes |
Consider the monthly log-returns of the Intel stock from Jan. 1973 to Dec. 1997 (25 years, T=300). PACF suggests to fit an ARCH(3) model: | \[ {r}_{t} = \mu + {X}_{t},\;{X}_{t} = {\sigma }_{t}{\varepsilon }_{t} \] \[ {\sigma }_{t}^{2} = {b}_{0} + {b}_{1}{X}_{t - 1}^{2} + {b}_{2}{X}_{t - 2}^{2} + {b}_{3}{X}_{t - 3}^{2}. \] ## Results: <table><thead><tr><th></th><th>\( \mu \)</th><th>\( {b}_{0} \)</th><th>\( {b}_{1} \)</th><th>\( {b}_{2} \)</th><th>\( {b}_{3... | Yes |
Consider the stationary GARCH (1,1) process\n\n\\[ \n{\\sigma }_{t}^{2} = {b}_{0} + {b}_{1}{X}_{t - 1}^{2} + {a}_{1}{\\sigma }_{t - 1}^{2}\\;\\left( {{b}_{1} + {a}_{1} < 1}\\right) .\n\\]\n\nThen,\n\n\\[ \n{X}_{t}^{2} = {b}_{0} + \\left( {{a}_{1} + {b}_{1}}\\right) {X}_{t - 1}^{2} - {a}_{1}{\\eta }_{t - 1} + {\\eta }_{... | By using the ACF for ARMA model, we have\n\n\\[ \n\\operatorname{Corr}\\left( {{X}_{t}^{2},{X}_{t + k}^{2}}\\right) = \\frac{\\left( {1 - {a}_{1}^{2} - {a}_{1}{b}_{1}}\\right) {b}_{1}}{1 - {a}_{1}^{2} - 2{a}_{1}{b}_{1}}{\\left( {b}_{1} + {a}_{1}\\right) }^{k - 1}, k \\geq 1.\n\\] | Yes |
For \( \operatorname{GARCH}\left( {1,1}\right) \) model, \[ {\sigma }_{T}^{2}\left( m\right) = {b}_{0} + \left( {{b}_{1} + {a}_{1}}\right) {\sigma }_{T}^{2}\left( {m - 1}\right) \] | \[ = \frac{{b}_{0}\left\lbrack {1 - {\left( {a}_{1} + {b}_{1}\right) }^{m - 1}}\right\rbrack }{1 - {a}_{1} - {b}_{1}} + {\left( {a}_{1} + {b}_{1}\right) }^{m - 1}{\sigma }_{T}^{2}\left( 1\right) , \] and \( {\sigma }_{T}^{2}\left( 1\right) = {b}_{0} + {b}_{1}{X}_{T}^{2} + {a}_{1}{\sigma }_{T}^{2} \) . In particular, \[... | Yes |
GARCH model with Gaussian shocks: \( \left( {p + q \leq 5}\right) \) | Both AIC and BIC selected a GARCH(1,3) model\n\n\[ \n{\sigma }_{t}^{2} = {0.015} + {0.112}{X}_{t - 1}^{2} + {0.492}{\sigma }_{t - 1}^{2} - {0.034}{\sigma }_{t - 2}^{2} + {0.420}{\sigma }_{t - 3}^{2} \]\n\n\[ \left( {0.002}\right) \;\left( {0.004}\right) \;\left( {0.070}\right) \;\left( {0.083}\right) \;\left( {0.055}\r... | Yes |
Example 5. 864 monthly log-returns of IBM stock from Jan. 1926 to December 1997. An AR(1)-EGARCH(1,0) model is fitted to obtain\n\n\[ \n{r}_{t} = {0.0105} + {0.092}{r}_{t - 1} + {X}_{t},\;{X}_{t} = {\sigma }_{t}{\varepsilon }_{t} \]\n\n\[ \n{h}_{t} = - {0.794} + {0.856}{h}_{t - 1} + {\varepsilon }_{t - 1}^{ * } \]\n\n\... | Model checking: For \( {\widehat{\varepsilon }}_{t} \), the Ljung-Box \( Q\left( {10}\right) = {6.31}\left( {{79}\% }\right), Q\left( {20}\right) = 21.4(37%). For \( \left\{ {\widehat{\varepsilon }}_{t}^{2}\right\}, Q\left( {10}\right) = {4.13}\left( {{94}\% }\right), Q\left( {20}\right) = {15.96}\left( {{72}\% }\right... | Yes |
Theorem 4 In the decomposition (linear regression)\n\n\[ \n\\mathbf{Y} = \\mathbf{\\alpha } + \\mathbf{\\beta }{Y}^{m} + \\mathbf{\\varepsilon } \n\]\n\nwith \( \\mathrm{E}\\mathbf{\\varepsilon } = 0 \) and \( \\operatorname{Cov}\\left( {\\mathbf{\\varepsilon },{Y}^{m}}\\right) = 0 \), the intercept \( \\mathbf{\\alpha... | Proof: Note that from the decomposition\n\n\[ \n\\operatorname{Cov}\\left( {\\mathbf{Y},{Y}^{m}}\\right) = \\mathbf{\\beta }\\operatorname{Cov}\\left( {{Y}^{m},{Y}^{m}}\\right) \n\]\n\nIt follows that 不同维数一样求协方差阵\n\n\[ \n\\mathbf{\\beta } = \\frac{\\operatorname{Cov}\\left( {\\mathbf{Y},{Y}^{m}}\\right) }{\\operatornam... | Yes |
Example 4.3. Figure 4.3 illustrates how to compute the market betas for three stocks GS, IBM, GE in Jan. 2011. Market \( \beta \) is time varying. Suppose a hedge fund holds \$10 million, \$20 million, and \$30 million GS, IBM, GE stocks. Then, its market equivalent exposure is | \[ \$ {10} \times {1.37} + \$ {20} \times {0.74} + \$ {30} \times {1.66} = \$ {78.3}\text{ million. } \] | Yes |
Consider the three risky assets in Ex. 4.1. Then, and For any given \( {\mu }_{p} \), the optimal allocation is Its associated risk is given by This defines efficient frontiers in Fig. 4.5. For \( {\mu }_{p} = 0 \), the portfolio variance is 0.0287 and its SD is 16.94%. For \( {\mu }_{p} = {18.48}\% \) | \[ A = {6.4757},\;B = {0.7275},\;C = {92.5366},\;D = {25.3932} \] \[ \mathbf{g} = \left( \begin{matrix} {1.5052} \\ - {0.4244} \\ - {0.0807} \end{matrix}\right) \;\text{ and }\;\mathbf{h} = \left( \begin{matrix} - {4.7307} \\ {3.7627} \\ {0.9680} \end{matrix}\right) . \] \[ {\mathbf{\alpha }}^{ * } = \left( \begin{matr... | Yes |
Example 5.1 Fama-French model. Fama and French (1993) consider the following factors: excessive return of CRSP value-weighted stock index, | \[ \text{SMB} = \frac{1}{3}\left( {\text{Small Value} + \text{Small Neutral} + \text{Small Growth}}\right) - \frac{1}{3}\left( {\text{ Big Value } + \text{ Big Neutral } + \text{ Big Growth }}\right) , \] \[ \text{ HML } = \frac{1}{2}\left( {\text{ Small Value } + \text{ Big Value }}\right) - \frac{1}{2}\text{ (Small G... | No |
Example 5.3. Fama and French (1993) consider the following factors \( \left( {K = 2,3\text{and 5 factors}}\right) \) : 通常考虑的因子\n\n1. difference of returns between small and Big capitalization (SMB);\n\n2. difference of returns between high and low book-to-market ratios\n\n(HML);\n\n3. CRSP value-weighted stock index;\n... | Fama and French find\n\n— some improvement going from two factors to five factors;\n\n— three factors are necessary when testing portfolio consisting only\n\nof stocks;\n\n— five factors when bond portfolio is included. | Yes |
Example 5.4. Test Fama-French three factor models over 12 five-year periods (Jan 51 - Dec. 2010). | Testing Portfolios: 6 Fama-French portfolios \( \left( {2 \times 3}\right) \), and 25 Fama-French portfolios \( \left( {5 \times 5}\right) \) . Results: Comparing with the one-factor model (CAPM), the residual variances are smaller; hence multiple \( {R}^{2} \) is much larger. The better fits of 3-factor model are evid... | Yes |
Perceived minimum variance portfolio: \( \;{\widehat{\textbf{w}}}_{{}_{\mathrm{{opt}}}} = \underset{\textbf{argmin}}{\textbf{argmin}}{\pi }^{T}\textbf{1} = \widehat{R}\left( \textbf{w}\right) \). | Since the rank of \( \widehat{\mathbf{\sum }} \leq T - 1 \) , 维数比较大时大于T，而此时Σ是T个向 - the perceived risk \( \widehat{R}\left( {\widehat{\mathbf{w}}}_{\text{opt }}\right) = 0 \) , 量的样本协方差矩阵因此不满秩 - the actual risk \( R\left( {\widehat{\mathbf{w}}}_{\text{opt }}\right) \) is far from zero. - it behaves like a random portfoli... | No |
Estimate the volatility matrix of the returns of the SP500 and the changes of its VIX on the 10-year window from Jan. 29, 2001 to Feb. 28, 2011 presented in Chap 5. See Fig 5.2. | - \( \lambda = {0.94} \) and initial value \( {\widehat{\mathbf{\sum }}}_{0} = 0 \) . \n- The estimates at the initial 3 months (63 days) are influenced by the initial value since \( {\lambda }^{62} = {0.022} \), not much afterwards. \n- The correlation here is the measure of the leverage effect, which hovers around -0... | Yes |
Problem: \( {\widehat{\sum }}_{\lambda } \) is not necessarily positive definite. | - Method 1: Do SVD \( {\widehat{\mathbf{\sum }}}_{\lambda } = {\mathbf{\Gamma }}^{T}\operatorname{diag}\left( {{\lambda }_{1},\cdots ,{\lambda }_{p}}\right) \mathbf{\Gamma } \) . Set\n\n\[ \n{\widehat{\mathbf{\sum }}}_{\lambda }^{ + } = {\mathbf{\Gamma }}^{T}\operatorname{diag}\left( {{\lambda }_{1}^{ + },\cdots ,{\lam... | Yes |
Property 1 If the \( n \times n \) matrix \( A \) is symmetric, then \( A \) is positive-definite if and only if all of its eigenvalues are positive. | Proof. Theorem A. 5 says that, the set of unit eigenvectors is orthonormal and spans \( {R}^{n} \) . If \( A \) is positive-definite and \( {Av} = {\lambda v} \) for a nonzero vector \( v \), then \( 0 < {v}^{T}{Av} = \) \( {v}^{T}\left( {\lambda v}\right) = \lambda \parallel v{\parallel }_{2}^{2} \), so \( \lambda > 0... | Yes |
Property 2 If \( A \) is \( n \times n \) symmetric positive-definite and \( X \) is an \( n \times m \) matrix of full rank with \( n \geq m \) , then \( {X}^{T}{AX} \) is \( m \times m \) symmetric positive-definite. | Proof. The matrix is symmetric since \( {\left( {X}^{T}AX\right) }^{T} = {X}^{T}{AX} \) . To prove positive-definite, consider a nonzero \( m \) -vector \( v \) . Note that \( {v}^{T}\left( {{X}^{T}{AX}}\right) v = {\left( Xv\right) }^{T}A\left( {Xv}\right) \geq 0 \), with equality only if \( {Xv} = 0 \), due to the po... | Yes |
Property 3 Any principal submatrix of a symmetric positive-definite matrix is symmetric positive-definite. | Proof. Exercise 12. | No |
Property 1. There are \( n = 4 \) data points. We must check\n\n\[ \n{S}_{1}\left( 1\right) = 2\text{ and }{S}_{1}\left( 2\right) = 1 \]\n\n\[ \n{S}_{2}\left( 2\right) = 1\text{ and }{S}_{2}\left( 4\right) = 4 \]\n\n\[ \n{S}_{3}\left( 4\right) = 4\text{ and }{S}_{3}\left( 5\right) = 3. \]\n | These follow easily from the defining equations (3.16). | No |
We must check \( {S}_{1}^{\prime }\left( 2\right) = {S}_{2}^{\prime }\left( 2\right) \) and \( {S}_{2}^{\prime }\left( 4\right) = {S}_{3}^{\prime }\left( 4\right) \) . | The first is\n\n\[ - \frac{13}{8} + \frac{15}{8} = \frac{1}{4} \]\n\nand the second is\n\n\[ \frac{1}{4} + \frac{15}{4}\left( {4 - 2}\right) - \frac{15}{8}{\left( 4 - 2\right) }^{2} = \frac{1}{4} \]\n\nboth of which check out. | Yes |
We must check \( {S}_{1}^{\prime \prime }\left( 2\right) = {S}_{2}^{\prime \prime }\left( 2\right) \) and \( {S}_{2}^{\prime \prime }\left( 4\right) = {S}_{3}^{\prime \prime }\left( 4\right) \), both of which are true. Therefore,(3.16) is a cubic spline. | Property 3. The second derivatives are\n\n\[ \n{S}_{1}^{\prime \prime }\left( x\right) = \frac{15}{4}\left( {x - 1}\right) \n\]\n\n\[ \n{S}_{2}^{\prime \prime }\left( x\right) = \frac{15}{4} - \frac{15}{4}\left( {x - 2}\right) \n\]\n\n\[ \n{S}_{3}^{\prime \prime }\left( x\right) = - \frac{15}{4} + \frac{15}{4}\left( {x... | Yes |
Property 3 Brownian motion \( {B}_{t} \) can be represented by continuous paths. | The appearance of the normal distribution is a consequence of the Central Limit Theorem, a deep fact about probability.\n\nComputer simulation of Brownian motion is based on respecting these three properties. Establish a grid of steps\n\n\[ 0 = {t}_{0} \leq {t}_{1} \leq \cdots \leq {t}_{n} \]\non the \( t \) -axis, and... | No |
Property 1 The rank of the matrix \( A = {US}{V}^{T} \) is the number of nonzero entries in \( S \) . | Proof. Since \( U \) and \( {V}^{T} \) are invertible matrices, \( \operatorname{rank}\left( A\right) = \operatorname{rank}\left( S\right) \), and the latter is the number of nonzero diagonal entries. | Yes |
Property 2 If \( A \) is an \( n \times n \) matrix, \( \left| {\det \left( A\right) }\right| = {s}_{1}\cdots {s}_{n} \) . | Proof. Since \( {U}^{T}U = I \) and \( {V}^{T}V = I \), the determinants of \( U \) and \( {V}^{T} \) are 1 or -1, due to the fact that the determinant of a product equals the product of the determinants. Property 2 follows from the factorization \( A = {US}{V}^{T} \) . | "No" |
Property 3 If \( A \) is an invertible \( m \times m \) matrix, then \( {A}^{-1} = V{S}^{-1}{U}^{T} \) . | Proof. By Property \( 1, S \) is invertible, meaning all \( {s}_{i} > 0 \) . Now Property 3 follows from the fact that if \( {A}_{1},{A}_{2} \), and \( {A}_{3} \) are invertible matrices, then \( {\left( {A}_{1}{A}_{2}{A}_{3}\right) }^{-1} = \) \( {A}_{3}^{-1}{A}_{2}^{-1}{A}_{1}^{-1} \) | Yes |
Property 4 The \( m \times n \) matrix \( A \) can be written as the sum of rank-one matrices\n\n\[ A = \mathop{\sum }\limits_{{i = 1}}^{r}{s}_{i}{u}_{i}{v}_{i}^{T} \]\n\nwhere \( r \) is the rank of \( A \), and \( {u}_{i} \) and \( {v}_{i} \) are the \( i \) th columns of \( U \) and \( V \), respectively. | Proof.\n\n\[ A = {US}{V}^{T} = U\left\lbrack \begin{array}{lll} {s}_{1} & & \\ & \ddots & \\ & & {s}_{r} \\ & & \\ & & \\ & & \\ & & \\ & & \\ & & \end{array}\right\rbrack {V}^{T} \]\n\n\[ = U\left( {\left\lbrack \begin{array}{lll} {s}_{1} & & \\ & & \\ & & \\ & & \\ & & \end{array}\right\rbrack + \left\lbrack \begin{a... | Yes |
What is the probability that the value of the stock will be below \$950,000 at the close of at least one of the next 45 trading days? | niter \( = 1\mathrm{e}5\;\# \; \) number of iterations\nbelow \( = \operatorname{rep}\left( {0\text{, niter}}\right) \) #set up storage\nset.seed(2009)\nfor (i in 1:niter)\n\( r = \operatorname{rnorm}({45}, \) mean \( = {0.05}/{253} \) ,\nsd \( = {0.23}/\operatorname{sqrt}\left( {253}\right) ) \) #generate random numbe... | Yes |
Finding the price and yield to maturity of a coupon bond using spot rates | Consider the simple example of 1-year coupon bond with semiannual coupon payments of $40 and a par value of $1,000. Suppose that the one-half-year spot rate is 2.5% / half-year and the 1-year spot rate is 3% / half-year. Think of the coupon bond as being composed of two zero-coupon bonds, one with T = 1/2 and a par val... | Yes |
In this example, we first find the yields to maturity from the prices derived in Example 3.2 using the interest rates from Table 3.1. For a 1-year zero, the yield to maturity \( {y}_{1} \) solves\n\n\[ \frac{1000}{\left( 1 + {y}_{1}\right) } = {943.40} \] | which implies that \( {y}_{1} = {0.06} \) . | Yes |
Example 3.4. Finding yields and forward rates from prices\n\nSuppose that one-, two-, and three-year par $ 1,000 zeros are priced as given in Table 3.2. Using (3.12), the yields to maturity are\n\n\[ y_1 = \frac{1000}{920} - 1 = 0.087 \]\n\n\[ y_2 = \left\{ \frac{1000}{830} \right\}^{1/2} - 1 = 0.0976 \]\n\n\[ y_3 = \l... | Then, using (3.15) and (3.16),\n\n\[ r_1 = y_1 = 0.087 \]\n\n\[ r_2 = \frac{\left( 1 + y_2 \right)^2}{\left( 1 + y_1 \right)} - 1 = \frac{\left( 1.0976 \right)^2}{1.0876} - 1 = 0.108, \text{ and } \]\n\n\[ r_3 = \frac{\left( 1 + y_3 \right)^3}{\left( 1 + y_2 \right)^2} - 1 = \frac{\left( 1.096 \right)^3}{\left( 1.0976 ... | Yes |
Example 3.5. Forward rates from prices | Thus, using (3.17) and the prices in Table 3.2, the forward rates are\n\n\[ \n{r}_{1} = \frac{1000}{920} - 1 = {0.087} \]\n\n\[ \n{r}_{2} = \frac{920}{830} - 1 = {0.108} \]\n\nand\n\[ \n{r}_{3} = \frac{830}{760} - 1 = {0.092} \]\n | Yes |
Example 3.6. Continuously compounded forward rates and yields from prices\n\nUsing the prices in Table 3.2, we have \( P\left( 1\right) = {920}, P\left( 2\right) = {830} \), and \( P\left( 3\right) = {760} \) . Therefore, using (3.20), | \[ {r}_{1} = \log \left\{ \frac{1000}{920}\right\} = {0.083} \]\n\n\[ {r}_{2} = \log \left\{ \frac{920}{830}\right\} = {0.103} \]\n\nand\n\n\[ {r}_{3} = \log \left\{ \frac{830}{760}\right\} = {0.088} \]\n\nAlso, \( {y}_{1} = {r}_{1} = {0.083},{y}_{2} = \left( {{r}_{1} + {r}_{2}}\right) /2 = {0.093} \), and \( {y}_{3} =... | Yes |
Example 3.7. Finding continuous yield and discount functions from forward rates\n\nSuppose the forward rate is the linear function \( r\left( t\right) = {0.03} + {0.0005t} \) . Find \( r\left( {15}\right) ,{y}_{15} \), and \( D\left( {15}\right) \) . | Answer: \( r\left( {15}\right) = {0.03} + \left( {0.0005}\right) \left( {15}\right) = {0.0375} \) ,\n\n\[ \n{y}_{15} = {\left( {15}\right) }^{-1}{\int }_{0}^{15}\left( {{0.03} + {0.0005t}}\right) {dt} \]\n\n\[ \n= {\left. {\left( {15}\right) }^{-1}\left( {0.03}t + {0.0005}{t}^{2}/2\right) \right| }_{0}^{15} = {0.03375}... | Yes |
Problem 1 Use the plot to estimate graphically the yield to maturity. Does this estimate agree with that from spline interpolation? | As an alternative to interpolation, the yield to maturity can be found using a nonlinear root finder (equation solver) such as uniroot(), which is illustrated here:\n\n\[ \text{uniroot(function(r)}r \hat{} 2 - {.5}, c\left( {{0.7},{0.8}}\right) \text{)} \] | No |
Problem 1 Write a brief description of the time series plots of the four indices. Do the series look stationary? Do the fluctuations in the series seem to be of constant size? If not, describe how the volatility fluctuates. | Next, run the following R code to compute and plot the log returns on the indices.\n\nlogR \( = \operatorname{diff}\left( {\log \left( \text{ EuStockMarkets }\right) }\right) \)\n\nplot(logR) | No |
For the DAX index, state which choice of the degrees of freedom parameter gives the best-fitting \( t \) -distribution and explain why. | Run the next set of code to create a kernel density estimate and two parametric density estimates, \( t \) with df degrees of freedom and normal, for the DAX index. Here df equals 5, but you should vary df so that the \( t \) density agrees as closely as possible with the kernel density estimate.\n\nAt lines 5-6, a rob... | No |
Fisher information for a normal model mean\n\nSuppose that \( {Y}_{1},\ldots ,{Y}_{n} \) are i.i.d. \( N\left( {\mu ,{\sigma }^{2}}\right) \) with \( {\sigma }^{2} \) known. The log-likelihood for the unknown parameter \( \mu \) is\n\n\[ \log \{ L\left( \mu \right) \} = - \frac{n}{2}\left\{ {\log \left( {\sigma }^{2}\r... | Therefore,\n\n\[ \frac{d}{d\mu }\log \{ L\left( \mu \right) \} = \frac{1}{{\sigma }^{2}}\mathop{\sum }\limits_{{i = 1}}^{n}\left( {{Y}_{i} - \mu }\right) \]\n\nso that \( \bar{Y} \) is the MLE of \( \mu \) and\n\n\[ \frac{{d}^{2}}{d{\mu }^{2}}\log \{ L\left( \mu \right) \} = - \frac{\mathop{\sum }\limits_{{i = 1}}^{n}1... | Yes |
Fitting a t-distribution to changes in risk-free returns | This example uses one of the time series in Chap. 4, the changes in the risk-free returns that has been called diffrf. This time series will be used to illustrate several methods for fitting a \( t \) -distribution. The simplest method uses the \( R \) function fitdistr ().\n\n\( \operatorname{data}\left( {\text{ Capm,... | No |
Fitting an F-S skewed t-distribution to changes in risk-free returns | loglik_sstd = function(beta) \\(\\operatorname{sum}( - \\operatorname{dsstd}(x, mean = beta[1], sd = beta[2], nu = beta[3], xi = beta[4], log = TRUE))\\)\n\nstart = c\\(\\left( {\\operatorname{mean}\\left( x\\right) ,{sd}\\left( x\\right) ,5,1}\\right)\\)\n\nfit_sstd = optim(start, loglik_sstd, hessian = T,\n\nmethod =... | No |
Example 5.5. Fitting a generalized error distribution to changes in risk-free returns | The fit of the generalized error distribution to diff rf was obtained using optim() similarly to the previous example. > fit_ged$par [1] -0.00019493 0.06883004 1.00006805 > sd_ged [1] 0.0011470 0.0033032 0.0761374 \( > \) AIC_ged [1] -1361.4 \( > \) BIC_ged [1] -1344.4 The three parameters are the estimates of the mean... | Yes |
Finding a confidence interval for quKurt can be a daunting task without the bootstrap, but with the bootstrap it is simple. In this example, \( {\mathrm{{BC}}}_{a} \) confidence intervals will be found for quKurt. | The \( {\mathrm{{BC}}}_{a} \) intervals are found with the bcanon() function in the bootstrap package using \( B = 5,{000} \) . The seed of the random number generator was fixed so that these results can be reproduced.\n\n\( \mathrm{{bmw}} = \operatorname{read.csv}\left( \text{\ | Yes |
This example uses the data set midcapD.ts.csv of returns on midcap stocks. Two of the stocks in this data set are LSCC and CSGS. From Fig. 6.2, which is a QQ plot comparing the returns from these two companies, it appears that LSCC returns have lighter tails than CSGS returns. The values of quKurt are 2.91 and 4.13 for... | midcapD.ts = read.csv(\ | No |
Find a \( {90}\% \) confidence interval for \( \sigma \) . | To find a \( {90}\% \) confidence interval for \( \sigma \), we need to use the values of \( {s}_{b,\text{ boot }}/s \) from the table. Since we want a \( {90}\% \) confidence interval, we need to find the values at the 5th and 95th percentiles. These correspond to the ranks 500 and 9,500 respectively.\n\nFrom the tabl... | Yes |
Suppose that \( \mathbf{Y} = {\left( {Y}_{1}{Y}_{2}{Y}_{3}\right) }^{\top },\operatorname{Var}\left( {Y}_{1}\right) = 2,\operatorname{Var}\left( {Y}_{2}\right) = 3,\operatorname{Var}\left( {Y}_{3}\right) = 5 \) , \( {\rho }_{{Y}_{1},{Y}_{2}} = {0.6} \), and that \( {Y}_{1} \) and \( {Y}_{2} \) are independent of \( {Y}... | The covariance between \( {Y}_{1} \) and \( {Y}_{3} \) is 0 by independence, and the same is true of \( {Y}_{2} \) and \( {Y}_{3} \) . The covariance between \( {Y}_{1} \) and \( {Y}_{2} \) is \( \left( {0.6}\right) \sqrt{\left( 2\right) \left( 3\right) } = \) 1.47. Therefore,\n\n\[ \operatorname{COV}\left( \mathbf{Y}\... | Yes |
Suppose that the random vector \( \mathbf{Y} = {\left( {Y}_{1},{Y}_{2},{Y}_{3}\right) }^{\top } \) has the mean vector and covariance matrix used in the previous example and contains the returns on three assets. Find the covariance between a portfolio that allocates \( 1/3 \) to each of the three assets and a second po... | Let\n\n\[ \mathbf{w}_{1} = {\left( \begin{array}{lll} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{array}\right) }^{\mathsf{T}} \]\n\nand\n\n\[ \mathbf{w}_{2} = {\left( \begin{array}{lll} \frac{1}{2} & \frac{1}{2} & 0 \end{array}\right) }^{\top }.\n\nThen\n\n\[ \operatorname{Cov}\left\{ {\frac{{Y}_{1} + {Y}_{2}}{2},\fr... | Yes |
(d) The nonzero correlations in the copula do not have the same values as the corresponding sample correlations. Do you think this is just due to random variation or is something else going on? If there is another cause besides random variation, what might that be? To help answer this question, you can get confidence i... | s cor.test(rand_t_cop[,1], rand_t_cop[,3]) will give a confidence interval (95 percent by default) for the correlation (Pearson by default) between components 1 and 3. Does this confidence interval include 0.75 ? | No |
Problem 7 Do you see any difference between the parametric estimates of the copula? If so, which seem closest to the empirical copula? Include the plot with your work. | A two-dimensional KDE of the copula's density will be compared with the parametric density estimates (PDFs).\n\n\\( \\operatorname{par}\\left( {\\mathrm{{mfrow}} = \\mathrm{c}\\left( {2,3}\\right) ,\\;\\mathrm{{mgp}} = \\mathrm{c}\\left( {{2.5},1,0}\\right) }\\right) \\)\n\ncontour(tCopula(param=ft$par[7], dim=2, df=ro... | Yes |
In this example, the null hypothesis is that, in the three-predictor model, the slopes for cm30_dif and ff_dif are zero. The \( F \) -test can be computed using R's anova function. | The output is\n\nAnalysis of Variance Table\n\nModel 1: aaa_dif ~ cm10_dif\n\nModel 2: aaa_dif ~ cm10_dif + cm30_dif + ff_dif\n\nRes.Df RSS Df Sum of Sq F Pr(>F)\n\n1 878 3.81\n\n\( \\begin{array}{llllllll} 2 & {876} & {3.66} & 2 & {0.15} & {18.0} & {2.1}\\mathrm{e} - {08} & * * * \\end{array} \)\n\nSignif. codes: \( 0... | Yes |
The question is whether VIF values of 14.4 and 14.1 are so large that the number of predictor variables should be reduced to 1 , that is, whether we should use only cm10_dif. | The answer is \ | No |
Problem 1 Describe any interesting features, such as outliers, seen in the scatterplot matrix. Keep in mind that the goal is to predict changes in consumption. Which variables seem best suited for that purpose? Do you think there will be collinearity problems? | Next, run the code below to fit a multiple linear regression model to consumption using the other four variables as predictors.\n\nfitLm1 = lm(consumption ~ dpi + cpi + government + unemp) summary(fitLm1) confint(fitLm1) | No |
Problem 3 For the purpose of variable selection, does the ANOVA table provide any useful information not already in the summary? | Upon examination of the \( p \) -values, we might be tempted to drop several variables from the regression model, but we will not do that since variables should be removed from a model one at a time. The reason is that, due to correlation between the predictors, when one is removed the significance of the others change... | No |
Problem 6 Was there much collinearity in the original four-variable model? Was the collinearity reduced much by dropping two variables? | Partial residual plots, which are also called component plus residual or cr plots, can be constructed using the function crPlot() in the car package. Run\n\n\\( \\operatorname{par}\\left( {{mfrow} = c\\left( {2,2}\\right) }\\right) \\)\n\n\\( \\mathrm{{sp}} = {0.8} \\)\n\ncrPlot(fitLm1, dpi, span = sp, col = \ | No |
2. Show that if \( {\epsilon }_{1},\ldots ,{\epsilon }_{n} \) are i.i.d. \( N\left( {0,{\sigma }_{\epsilon }^{2}}\right) \), then in straight-line regression the least-squares estimates of \( {\beta }_{0} \) and \( {\beta }_{1} \) are also the maximum likelihood estimates. | Hint: This problem is similar to the example in Sect. 5.9. The only difference is that in that section, \( {Y}_{1},\ldots ,{Y}_{n} \) are independent \( N\left( {\mu ,{\sigma }^{2}}\right) \), while in this exercise \( {Y}_{1},\ldots ,{Y}_{n} \) are independent \( N\left( {{\beta }_{0} + {\beta }_{1}{X}_{i},{\sigma }_{... | No |
The leverage of the \( i \) th observation, denoted by \( {H}_{ii} \), measures how much influence \( {Y}_{i} \) has on its own fitted value \( {\widehat{Y}}_{i} \) . We will not go into the algebraic details until Sect. 11.1. An important result in that section is that there are weights \( {H}_{ij} \) depending on the... | \[ {\widehat{Y}}_{i} = \mathop{\sum }\limits_{{j = 1}}^{n}{H}_{ij}{Y}_{j} \] (10.1) In particular, \( {H}_{ii} \) is the weight of \( {Y}_{i} \) in the determination of \( {\widehat{Y}}_{i} \) . It is a potential problem if \( {H}_{ii} \) is large since then \( {\widehat{Y}}_{i} \) is determined too much by \( {Y}_{i} ... | No |
Example 10.4. Cook's D for simulated data in Example 10.1 | The three columns of Fig. 10.4 show the values of square roots of Cook's D for the three simulated data examples in Fig. 10.1. In the top row, the square roots of Cook's D values are plotted versus case number (index). The bottom row contains half-normal plots of the square roots of the Cook's D values. In all panels, ... | Yes |
Detecting nonlinearity: A simulated data example | Figure 10.8a is a plot of the residuals versus \( {X}_{1} \) . The residuals appear to have a nonlinear trend. This is better revealed by adding a loess curve to the residuals. The curvature of the loess fit is evident and indicates that \( Y \) is not linear in \( {X}_{1} \) . A possible remedy is to add \( {X}_{1}^{2... | Yes |
Example 11.7. Who gets a credit card? | First, a logistic regression model is fit with all seven predictors using the glm() function. The R code is:\n\nlibrary(\ | No |
Problem 6 What is the MLE of \( \alpha \) ? (Hint: Type ?boxcox to learn what is returned by this function.) | Next, fit a linear model with price transformed by \( \widehat{\alpha } \) (the MLE). Here the function bcPower () in the AER package computes a Box-Cox transformation for a given value of \( \alpha \) and must be distinguished from boxcox (), which computes the profile log-likelihood for \( \alpha \) . In the followin... | No |
Show that their solution is\n\[ \n{\beta }_{1} = \frac{{\sigma }_{XY}}{{\sigma }_{X}^{2}} \n\]\nand\n\[ \n{\beta }_{0} = E\left( Y\right) - {\beta }_{1}E\left( X\right) = E\left( Y\right) - \frac{{\sigma }_{XY}}{{\sigma }_{X}^{2}}E\left( X\right) .\n\] | When we were finding the best linear predictor of \( Y \) given \( X \), we derived the equations\n\n\[ \n0 = - E\left( Y\right) + {\beta }_{0} + {\beta }_{1}E\left( X\right) \n\]\n\n\[ \n0 = - E\left( {XY}\right) + {\beta }_{0}E\left( X\right) + {\beta }_{1}E\left( {X}^{2}\right) .\n\] | Yes |
The daily log returns for BMW stock between January 1973 and July 1996 from the bmw data set in R's evir package are shown in Fig. 12.6a. Their sample ACF and quantiles are shown in Fig. 12.6b and c, respectively. The estimated autocorrelation coefficient at lag 1 is well outside the test bounds, so the series has some... | Next, an AR(1) model was fit using the arima() command in R.\n\n9 fitAR1 = arima(bmw, order = c(1,0,0))\n\n10 print(fitAR1)\n\nThe order parameter will be explained later, but for an AR(1) process it should be \( \mathrm{c}\left( {1,0,0}\right) \) . A summary of the output is below.\n\n![9e2cddb4-a886-43fa-8e39-a68ab2d... | No |
Example 12.6. Changes in the inflation rate-AR(p) models | Figure 12.10 is a plot of AIC and BIC versus \( p \) for \( \operatorname{AR}\left( p\right) \) fits to the changes in the inflation rate. Both criteria suggest that \( p \) should be large. AIC decreases steadily as \( p \) increases from 1 to 19, though there is a local minimum at 8 . Even the conservative BIC criter... | No |
Fitting an ARIMA model to CPI data | This example uses the CPI.dat.csv data set. CPI is a seasonally adjusted U.S. Consumer Price Index. The data are monthly. Only data from January 1977 to December 1987 are used in this example. Figure 12.15 shows time series plots of \( \log \left( \mathrm{{CPI}}\right) \) and the first and second differences of this se... | Yes |
Problem 5 Do you think that there is residual autocorrelation? If so, describe this autocorrelation and suggest a more appropriate model for the \( T \) -bill series. | GARCH effects, that is, volatility clustering, can be detected by looking for auto-correlation in the mean-centered squared residuals. Another possibility is that some quarters are more variable than others. This can be detected for quarterly data by autocorrelation in the squared residuals at time lags that are a mult... | Yes |
(a) Show that the ACF of \( {Y}_{t} \) satisfies the equation\n\n\[ \rho \left( k\right) = {\phi }_{1}\rho \left( {k - 1}\right) + {\phi }_{2}\rho \left( {k - 2}\right) \]\n\nfor all values of \( k > 0 \) . (These are a special case of the Yule-Walker equations.) | \n\n\( \lbrack \) Hint: \( \gamma \left( k\right) = \mathrm{{Cov}}\left( {{Y}_{t},{Y}_{t - k}}\right) = \mathrm{{Cov}}\{ {\phi }_{1}\left( {{Y}_{t - 1} - \mu }\right) + {\phi }_{2}\left( {{Y}_{t - 2} - \mu }\right) + \)\n\n\( \left. {{\epsilon }_{t},{Y}_{t - k}}\right\} \) and \( {\epsilon }_{t} \) and \( {Y}_{t - k} \... | No |
Suppose that \( {\epsilon }_{1},\ldots ,{\epsilon }_{n} \) is a stationary \( \operatorname{AR}\left( 1\right) \) process so that \( {\epsilon }_{t} = \phi {\epsilon }_{t - 1} + \) \( {u}_{t} \), where \( \left| \phi \right| < 1 \) and \( {u}_{1},{u}_{2},\ldots \) is weak \( \operatorname{WN}\left( {0,{\sigma }_{u}^{2}... | \[ {\mathbf{\sum }}_{\mathbf{\epsilon }} = {\sigma }_{\epsilon }^{2}\left( \begin{matrix} 1 & \phi & {\phi }^{2} & \cdots & {\phi }^{n - 1} \\ \phi & 1 & \phi & \cdots & {\phi }^{n - 2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ {\phi }^{n - 1} & {\phi }^{n - 2} & {\phi }^{n - 3} & \cdots & 1 \end{matrix}\right) ... | No |
In Sect. 13.3.1 a regression of aaa_dif on cm10_dif and cm30_dif produced residuals that exhibited minor autocorrelation; AIC suggested an MA(3) model for the residuals while BIC selected ARIMA \( \\left( {0,0,0}\\right) \), i.e., white noise. We now consider whether ignoring the small autocorrelations has a practical ... | The HC and HAC covariance matrix estimates can be computed using the NeweyWest () function from the R package sandwich. The first argument is a fitted model object, in this case fit. In both cases we set prewhite \( = F \) . For the HAC estimate, the argument lag corresponds to the maximal lag \( L \) used in the Bartl... | Yes |
Cross-correlation between changes in CPI (consumer price index) and IP (industrial production) | The largest absolute cross-correlations are at negative lags and these correlations are negative. This means that an above-average (below-average) change in \( {cpi} \) predicts a future change in \( {ip} \) that is below (above) average. As just emphasized, correlation does not imply causation, so we cannot say that c... | Yes |
Problem 5 Forecast log(consumption) for the next eight quarters using the models you found in Problems 2 and 4. Plot the two sets of forecasts in side-by-side plots with the same limits on the \( x \) - and \( y \) -axes. Describe any differences between the two sets of forecasts. | Note: To predict an arima object (an object returned by the arima() function), use the predict function. To learn how the predict() function works on an arima object, use ?predict.Arima. To forecast an object returned by auto.arima(), use the forecast() function in the forecast package. For example, the following code ... | No |
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