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Proposition 3.3.17 (Size of a function with many vanishing partial derivatives). Let \( U \) be an open subset of \( {\mathbb{R}}^{n} \) and let \( g : U \rightarrow \mathbb{R} \) be a \( {C}^{k} \) function. If at \( \mathbf{a} \in U \) all partial derivatives of \( g \) up to order \( k \) vanish (including the 0th p... | Proof. The proof is by induction on \( k \), starting with \( k = 1 \) . The case\n\n--- \n\n\( k = 1 \) follows from Theorem 1.9.8: if \( g \) vanishes at \( \mathbf{a} \), and its first partials are continuous, \( g \) is differentiable at \( \mathbf{a} \), and its derivative is given by the Jacobian matrix. So if th... | Yes |
Proposition 3.4.4 (Chain rule for Taylor polynomials). Let \( U \subset {\mathbb{R}}^{n} \) and \( V \subset \mathbb{R} \) be open, and \( g : U \rightarrow V, f : V \rightarrow \mathbb{R} \) be of class \( {C}^{k} \). Then \( f \circ g : U \rightarrow \mathbb{R} \) is of class \( {C}^{k} \), and if \( g\left( \mathbf{... | \[ {P}_{f, b}^{k}\left( {{P}_{g,\mathbf{a}}^{k}\left( {\mathbf{a} + \overrightarrow{\mathbf{h}}}\right) }\right) \] and discarding the terms of degree \( > k \) . | Yes |
Proposition 3.9.16 (Frenet frame related to curvature and torsion). The Frenet frame satisfies the following equations, where \( \kappa \) is the curvature of the curve at \( \mathbf{a} = \delta \left( 0\right) \) and \( \tau \) is its torsion:\n\n\[ \n{\overrightarrow{\mathbf{t}}}^{\prime }\left( 0\right) = \;\kappa \... | Proof of Propositions 3.9.15 and 3.9.16. We may assume that the curve \( C \) is written in its adapted coordinates, i.e., as in equation 3.9.56, which we repeat here:\n\n---\n\nWhen equation A15.6 first appeared (as equation 3.9.56) we used dots \( \left( \ldots \right) \) to denote the terms that can be ignored. Here... | No |
Theorem 4.7.4 (Integrals using arbitrary pavings). Let \( X \subset {\mathbb{R}}^{n} \) be a bounded subset, and \( {\mathcal{P}}_{N} \) a nested partition of \( X \). If \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is integrable, then the limits \[ \mathop{\lim }\limits_{{N \rightarrow \infty }}{U}_{{\mathcal{P}}... | Proof. 1. The boundary \( \partial X \) has measure 0 since for any \( N \) it is contained in the union of the boundaries of the pieces of \( {\mathcal{P}}_{N} \), and this is a finite union of sets of volume 0 . The indicator function \( {\mathbf{1}}_{X} \) is integrable (why? \( {}^{4} \) ). By replacing \( f \) by ... | No |
Theorem 4.10.12 (Change of variables formula). Let \( X \) be a compact subset of \( {\mathbb{R}}^{n} \) with boundary \( \partial X \) of volume 0 ; let \( U \subset {\mathbb{R}}^{n} \) be an open set containing \( X \) . Let \( \Phi : U \rightarrow {\mathbb{R}}^{n} \) be a \( {C}^{1} \) mapping that is injective on \... | Proof. The proof is a (lengthy) matter of dotting the \( i \) ’s of the sketch in variables formula we will use the Section 4.10. As shown in Figure A19.1, we use the dyadic decomposition fact that \( \Phi \) is defined on \( U \), not of \( X \), and the image decomposition for \( Y \), whose paving blocks are the jus... | Yes |
Proposition 4.3.4 (Bounded part of graph has volume 0). Let \( f : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be an integrable function with graph \( \Gamma \left( f\right) \), and let \( {C}_{0} \subset {\mathbb{R}}^{n} \) be any dyadic cube. Then \[ {\operatorname{vol}}_{n + 1}\left( \underset{\text{bounded part of g... | Proof. The proof is not so very hard, but we have two types of dyadic | No |
Proposition 5.2.2 ( \( k \) -dimensional volume of a manifold). If integers \( m, k, n \) satisfy \( 0 \leq m < k \leq n \), and \( M \subset {\mathbb{R}}^{n} \) is a manifold of dimension \( m \) , any closed subset \( X \subset M \) has \( k \) -dimensional volume \( 0 \) . | Proof. By Definition 5.2.1, it is enough to show that for any \( R > 0 \) the set \( X \cap {\bar{B}}_{R}\left( \mathbf{0}\right) \) has \( k \) -dimensional volume 0 ; such an intersection is closed and bounded, hence compact. Denote by \( {Q}_{r}\left( \mathbf{x}\right) \subset {\mathbb{R}}^{n} \) the open box of sid... | No |
Proposition 4.11.5 (Convergence except on a set of measure 0). If \( {f}_{k} \) for \( k = 1,2,\ldots \) are Riemann-integrable functions on \( {\mathbb{R}}^{n} \) such that\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }{\int }_{{\mathbb{R}}^{n}}\left| {{f}_{k}\left( \mathbf{x}\right) }\right| \left| {{d}^{n}\mathbf{x... | Proof. Set\n\n\[ A\overset{\text{ def }}{ = }\mathop{\sum }\limits_{{k = 1}}^{\infty }{\int }_{{\mathbb{R}}^{n}}\left| {{f}_{k}\left( \mathbf{x}\right) }\right| \left| {{d}^{n}\mathbf{x}}\right| \]\n\nWe will take \( X \) to be the set of \( \mathbf{x} \in {\mathbb{R}}^{n} \) such that \( \mathop{\sum }\limits_{{k = 1}... | Yes |
Theorem 4.11.17 (A first limit theorem for Lebesgue integrals). Let \( k \mapsto {f}_{k} \) be a series of L-integrable functions such that\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{\infty }{\int }_{{\mathbb{R}}^{n}}\left| {{f}_{k}\left( \mathbf{x}\right) }\right| \left| {{d}^{n}\mathbf{x}}\right| < \infty \]\n\nThen \( f... | Proof. The idea is to write \( {f}_{k} = \mathop{\sum }\limits_{{i = 1}}^{\infty }{f}_{k, i} \), where the \( {f}_{k, i} \) are Riemann integrable with\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{\infty }{\int }_{{\mathbb{R}}^{n}}\left| {{f}_{k, i}\left( \mathbf{x}\right) }\right| \left| {{d}^{n}\mathbf{x}}\right| < \infty ... | Yes |
Theorem 4.11.19 (Dominated convergence for Lebesgue integrals). Let \( k \mapsto {f}_{k} \) be a sequence of L-integrable functions that converges pointwise to some function \( f \) almost everywhere. Suppose there is an L-integrable function \( F : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) such that \( \left| {{f}_{k... | Proof. We will make some preliminary simplifications. First, the numbers \( {A}_{k}\overset{\text{ def }}{ = }{\int }_{{\mathbb{R}}^{n}}{f}_{k}\left( \mathbf{x}\right) \left| {{d}^{n}\mathbf{x}}\right| \) are a bounded sequence of numbers, so by the Bolzano-Weierstrass theorem (Theorem 1.6.3) we can pass to a subsequen... | Yes |
Theorem 4.11.21 (Fubini’s theorem for the Lebesgue integral). Let \( f : {\mathbb{R}}^{n} \times {\mathbb{R}}^{m} \rightarrow \mathbb{R} \) be an L-integrable function. Then the function\n\n\[ \mathbf{y} \mapsto {\int }_{{\mathbb{R}}^{n}}f\left( {\mathbf{x},\mathbf{y}}\right) \left| {{d}^{n}\mathbf{x}}\right| \]\n\nis ... | Proof. Assume first that \( f \) is L-integrable. Then \( f = \mathop{\sum }\limits_{k}{\widetilde{f}}_{k} \), where the \( {\widetilde{f}}_{k} \) are R-integrable on \( {\mathbb{R}}^{n + m} \), with\n\n\[ \mathop{\sum }\limits_{k}{\int }_{{\mathbb{R}}^{n} \times {\mathbb{R}}^{m}}\left| {{\widetilde{f}}_{k}\left( {\mat... | Yes |
Theorem 4.11.20 (The change of variables formula for Lebesgue integrals). Let \( U, V \) be open subsets of \( {\mathbb{R}}^{n} \), and let \( \Phi : U \rightarrow V \) be bijective, of class \( {C}^{1} \), with inverse of class \( {C}^{1} \), such that both \( \Phi \) and \( {\Phi }^{-1} \) have Lipschitz derivatives.... | Proof. First we will show that if \( f \) is L-integrable, then \( \left| {\det \left\lbrack {\mathbf{D}\Phi }\right\rbrack }\right| \left( {f \circ \Phi }\right) \) is L-integrable and equation A21.63 is correct. Take all the cubes of \( {\mathcal{D}}_{1}\left( {\mathbb{R}}^{n}\right) \) with closures completely conta... | Yes |
Theorem 6.7.4 (Computing the exterior derivative of a \( k \) -form). Let\n\n\[ \n\varphi = \mathop{\sum }\limits_{{1 \leq {i}_{1} < \cdots < {i}_{k} \leq n}}{a}_{{i}_{1},\ldots ,{i}_{k}}d{x}_{{i}_{1}} \land \cdots \land d{x}_{{i}_{k}} \n\] \n\nbe a \( k \) -form of class \( {C}^{2} \) on an open subset \( U \subset {\... | Proof. Part 4 is proved in Example 6.7.3. We will first prove part 5, then part 1. Parts 2 and 3 will follow immediately.\n\nIt is enough to prove part 5 at the origin, which simplifies the notation; | No |
Proposition 6.10.10 (Trimming \( X \) to make \( {X}_{\epsilon } \) ). Let \( {B}_{r}\left( \mathbf{x}\right) \) be the open ball of radius \( r \) around \( \mathbf{x} \) . For all \( \epsilon > 0 \), there exist points \( {\mathbf{x}}_{1},\ldots ,{\mathbf{x}}_{p} \in X \) and \( {r}_{1} > 0,\ldots ,{r}_{p} > 0 \), su... | Proof. Proposition 6.10.10 will follow easily from the modified Whitney transversality theorem, Theorem A23.3. | No |
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 |
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}_{\infty }{\sigma... | 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}}^{... | If the return process is stationary with a constant variance \( {\\sigma }^{2} = \\operatorname{Var}\\left\\lbrack {R}_{t}\\right\\rbrack = {\\sigma }_{\\infty }^{2} \) and stationary correlation denoted by \( {\\rho }_{v - u} = \\operatorname{Corr}\\left( {{R}_{u},{R}_{v}}\\right) \), this relationship simplifies to\n... | Yes |
For the monthly returns of the CRSP value-weighted index (Jan., 1926 — Dec. 1997, \( T = {864} \) ), it is computed that \[ \text{AIC -5.807-5.805-5.817-5.816-5.819-5.821-5.819-5.820-5.821-5.818} \] \( {SE} = \frac{1}{\sqrt{T}} = {0.034},\;\mathbf{2}{SE} = \mathbf{0}\mathbf{.}\mathbf{{068}},\;\widehat{p} = 3 \) or \( 5... | 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), we compute the t-statistic \( {0.0103}/{0.002} = 5 \) and hence its associated P-value is \( {2\Phi }\left( {-5}\right) = 0 \) . We have strong evidence against \( {H}_{0} \), namely, the 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, \[ {f}^{\prime }\left( \mathbf{b}\right) = {\left( a,{b}_{0},{b}_{0},{b}_{0}\right) }^{T}/{a}^{2},\;a = 1 - {b}_{1} - {b}_{2} - {b}_{3}. \] Evaluation of the gradient at the estimates gives \[ {f}^{\prime }\left( \widehat{\mathbf{b}}\right) = {\left( {0.9746},{0.0098},{0.0098},{0.0098}\right) }^{T}. \] Suppose th... | Yes |
For the S&P 500 daily log-prices, \( T = {5348} \) . For testing against random-walk without a drift, it was computed that \[ \widehat{\rho } = {1.0006},\;T\left( {\widehat{\rho } - 1}\right) = {3.2088}. \] The random walk hypothesis can not be rejected. | Similarly, for random walk hypothesis with a drift (reasonable), \[ \widetilde{\rho } = {0.9997106},\;T = {5348},\;T\left( {\widetilde{\rho } - 1}\right) = - {1.5479}, \] which is bigger than the critical value -13.96 . We can not reject the random walk hypothesis. | 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). Fig 3.4 shows the ACFs and PACFs. Clearly, the returns are heteroscedastic. PACF suggests to fit an ARCH(3) model:\n\n\[ \n{r}_{t} = \mu + {X}_{t},\;{X}_{t} = {\sigma }_{t}{\varepsilon }_{t} \]\n\n\[ \n{\sigma }_{t}^{2} =... | ## Results:\n\n<table><thead><tr><th></th><th>\( \mu \)</th><th>\( {b}_{0} \)</th><th>\( {b}_{1} \)</th><th>\( {b}_{2} \)</th><th>\( {b}_{3} \)</th></tr></thead><tr><td>Est.</td><td>0.0196</td><td>0.0090</td><td>0.2973</td><td>0.0090</td><td>0.0626</td></tr><tr><td>SE</td><td>0.0062</td><td>0.0013</td><td>0.0887</td><t... | 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: (p + q ≤ 5) | 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\n\[ \n\left( {0.002}\right) \;\left( {0.004}\right) \;\left( {0.070}\right) \;\left( {0.083}\right) \;\left( {0.05... | 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 | \[ {r}_{t} = {0.0105} + {0.092}{r}_{t - 1} + {X}_{t},\;{X}_{t} = {\sigma }_{t}{\varepsilon }_{t} \] \[ {h}_{t} = - {0.794} + {0.856}{h}_{t - 1} + {\varepsilon }_{t - 1}^{ * } \] \[ {\varepsilon }_{t - 1}^{ * } = - {0.0795}{\varepsilon }_{t - 1} + {0.2647}\left\lbrack {\left| {\varepsilon }_{t - 1}\right| - \sqrt{2/\pi ... | 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 \( {\mu }_{p} = {18.48}\% \), optimal allocat... | \[ A = {6.4757},\;B = {0.7275},\;C = {92.5366},\;D = {25.3932} \] \[ \mathbf{g} = \left( \begin{matrix} {1.5052} \\ - {0.4244} \\ - {0.0807} \end{matrix}\right) \;\text{ and }\;\mathbf{h} = \left( \begin{matrix} - {4.7307} \\ {3.7627} \\ {0.9680} \end{matrix}\right) . \] \[ {\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 |
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\n4. a term s... | 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. | No |
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: \( \;{\widehat{\textbf{w}}}_{{}_{\mathrm{{opt}}}} = \underset{\textbf{argmin}}{\textbf{argmin}}{\iint }_{{\textbf{w}}^{T}\textbf{1} = 1}\widehat{R}\left( \textbf{w}\right) \). | 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 |
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\n\nConsider 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\% /... | Applying (3.8) twice to obtain the prices of these zeros and summing, we obtain the price of the zero-coupon bond:\n\n\[ \frac{40}{1.025} + \frac{1040}{{\left( {1.03}\right) }^{2}} = {1019.32}. \]\n\nThe yield to maturity on the coupon bond is the value of \( y \) that solves\n\n\[ \frac{40}{1 + y} + \frac{1040}{{\left... | 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 \( {y}_{1} = {0.06} \) . For a 2-year zero, the yield to maturity \( {y}_{2} \) solves\n\n\[ \frac{1000}{{\left( 1 + {y}_{2}\right) }^{2}} = {881.68} \]\n\nso that\n\n\[ {y}_{2} = \sqrt{\frac{1000}{881.68}} - 1 = {0.0650}. \]\n\nFor a 3-year zero, the yield to maturity \( {y}_{3} \) solves\n\n\[ \fra... | 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\[ \n{y}_{1} = \frac{1000}{920} - 1 = {0.087} \n\]\n\n\[ \n{y}_{2} = {\left\{ \frac{1000}{830}\right\} }^{1/2} - 1 = {0.097... | Then, using (3.15) and (3.16),\n\n\[ \n{r}_{1} = {y}_{1} = {0.087} \n\]\n\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\n\[ \n{r}_{3} = \frac{{\left( 1 + {y}_{3}\right) }^{3}}{{\left( 1 + {y}_{2}\ri... | 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 |
Continuously compounded forward rates and yields from prices | Using 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), \[ {r}_{1} = \log \left\{ \frac{1000}{920}\right\} = {0.083} \] \[ {r}_{2} = \log \left\{ \frac{920}{830}\right\} = {0.103} \] and \[ {r}_{3} = \log \left\{ \fra... | Yes |
Suppose 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\[ \n= {\left. {\left( {15}\right) }^{-1}\left( {0.03}t + {0.0005}{t}^{2}/2\right) \right| }_{0}^{15} = {0.033... | 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 |
Problem 5 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\( \operatorname{data}\left( {\text{ Capm,... | No |
Fitting an F-S skewed t-distribution to changes in risk-free returns | loglik_sstd = function(beta) \\(\\operatorname{sum}( - \\operatorname{dsstd}(x, mean = beta[1], sd = beta[2], nu = beta[3], xi = beta[4], log = TRUE))\\)\n\nstart = c\\(\\left( {\\operatorname{mean}\\left( x\\right) ,{sd}\\left( x\\right) ,5,1}\\right)\\)\n\nfit_sstd = optim(start, loglik_sstd, hessian = T,\nmethod = \ | No |
Example 5.5. Fitting a generalized error distribution to changes in risk-free returns | The fit of the generalized error distribution to diff rf was obtained using optim() similarly to the previous example. > fit_ged\$par [1] \\( - {0.00019493}\;{0.06883004}\;{1.00006805} \) > sd_ged [1] 0.0011470 0.0033032 0.0761374 \\( > \) AIC_ged [1] -1361.4 \\( > \) BIC_ged [1] -1344.4 The three parameters are the es... | 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{\ | No |
A \( {\mathrm{{BC}}}_{a} \) confidence interval for the ratio of quKurt for LSCC and CSGS is found with the following R program. | midcapD.ts = read.csv(\ | No |
Find a \( {90}\% \) confidence interval for \( \sigma \) . | To estimate the risk of a stock, a sample of 50 log returns was taken and \( s \) was 0.31 . To get a confidence interval for \( \sigma ,{10},{000} \) resamples were taken. Let \( {s}_{b\text{, boot }} \) be the sample standard deviation of the \( b \) th resample. The 10,000 values of \( {s}_{b,\text{ boot }}/s \) wer... | 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\[ \n{\mathbf{w}}_{1} = {\left( \begin{array}{lll} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{array}\right) }^{\mathsf{T}} \n\] \n\nand \n\n\[ \n{\mathbf{w}}_{2} = {\left( \begin{array}{lll} \frac{1}{2} & \frac{1}{2} & 0 \end{array}\right) }^{\top }. \n\] \n\nThen \n\n\( \operatorname{Cov}\left\{ {\frac{{Y}_{1... | Yes |
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. | Analysis 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 * * * {0.001} * * {... | 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 |
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 values of the predictor variables but not on \( {Y}_{1},\ldots ,{Y}_{n} \) such that\n\n\[ \n{\widehat{Y}}_{i} = \mathop{\sum }\limits_{{j = 1}}^{n}{H}_{ij}{Y}_{j} \n... | No |
Detecting nonlinearity: A simulated data example\n\nData were simulated to illustrate some of the techniques for diagnosing problems. In the example there are two predictor variables, \( {X}_{1} \) and \( {X}_{2} \). The assumed model is multiple linear regression, \( {Y}_{i} = {\beta }_{0} + {\beta }_{1}{X}_{i,1} + {\... | 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 |
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... | 7 data(bmw, package = \ | 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 |
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... | No |
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 |
Problem 13 The last three changes in \( \mathrm{r},\mathrm{y} \), and \( \mathrm{{pi}} \) are given next. What are the predicted values of the next set of changes in these series? | 10 tail(TbGdpPi, n = 4) \( \begin{array}{lll} \mathrm{r} & \mathrm{y} & \mathrm{{pi}} \end{array} \) [233,] \( {0.07} \) 9.7 1.38 [234,] 0.04 9.7 0.31 [235,] 0.02 9.7 0.28 [236,] 0.07 9.7 -0.47\n\nNow fit a VAR(1) using the following commands.\n\n11 var1 = ar(del_dat, order.max=1) | No |
Problem 15 What do the elements of Phi_hat suggest about the relationships among the changes in \( r, y \), and \( {pi} \) ? | A VAR(1) process is stationary provided that the eigenvalues of \( \mathbf{\Phi } \) are less than one in magnitude. Compute the eigenvalues of \( \widehat{\mathbf{\Phi }} \):\n\n14 eigen.values \( = \) eigen(Phi_hat)\$values\n\n15 abs(eigen.values) | No |
Problem 16 Is the estimated process stationary? How does this result relate to the forecast calculations in Problem 14 above? | The dataset MacroVars.csv contains three US macroeconomic indicators from Quarter 1 of 1959 to Quarter 4 of 1997: Real Gross Domestic Product (a measure of economic activity), Consumer Price Index (a measure of inflation), and Federal Funds Rate (a proxy for monetary policy). Each series has been transformed to station... | No |
Problem 19 Plot DiffSqrtCpi and its ACF. Do you see any signs of long memory? If so, describe them. | Run the following code to estimate the amount of fractional differencing, fractionally difference DiffSqrtCpi appropriately, and check the ACF of the fractionally differenced series.\n\n4 library(\ | No |
Example 14.2. \( {AR}\left( 1\right) + {GARCH}\left( {1,1}\right) \) model fit to daily BMW stock log returns | This example uses the daily BMW stock log returns. The ugarchfit() function from R's rugarch package is used to fit an \( \mathrm{{AR}}\left( 1\right) + \mathrm{{GARCH}}\left( {1,1}\right) \) model to this series. Although ugarchfit () allows the white noise to have a nonGaussian distribution, we begin this example usi... | Yes |
In this example, an \( \mathrm{{AR}}\left( 1\right) + \mathrm{{APARCH}}\left( {1,1}\right) \) model with \( t \) -distributed errors is fit to the BMW log returns. The commands and abbreviated output from ugarchfit () is below. The estimate of \( \delta \) is 1.48 with a standard error of 0.14, so there is strong evide... | arma.aparch.t \( = \) ugarchspec \( \left( {\texttt{mean.model=list}\left( \texttt{armaOrder=c(1,0)}\right) \text{ }}\right) ,\n\nvariance.model=list (model=\ | No |
In Example 9.9, we saw that a parsimonious model for the yearly log returns on the stock index diff (log(sp)) used diff (log(ip)) and diff (bnd) as predictors. Figure 14.9 contains ACF plots of the residuals [panel (a)] and squared residuals [panel (b)]. Externally studentized residuals were used, but the plots for the... | The auto.arima() function from R's forecast package selected an MA(1) model [i.e., ARIMA(0,0,1)] for the residuals. Next an MA(1)+ARCH(1) model was fit to the regression model's raw residuals. Sample ACF plots of the standardized residuals from the \( \mathrm{{MA}}\left( 1\right) + \mathrm{{ARCH}}\left( 1\right) \) mod... | Yes |
Problem 1 Plot both Tbill and Tbill.diff. Use both time series and ACF plots. Also, perform ADF and KPSS tests on both series. Which series do you think are stationary? Why? What types of heteroskedasticity can you see in the Tbill.diff series? | In the following code, the variable Tbill can be used if you believe that series is stationary. Otherwise, replace Tbill by Tbill.diff. This code will fit an ARMA+GARCH model to the series.\n\n8 library(rugarch)\n\n9 arma.garch.norm = ugarchspec(mean.model=list(armaOrder=c(1,0)), variance.model=list(garch0rder=c(1,1))\... | No |
1. Let \( Z \) have an \( N\left( {0,1}\right) \) distribution. Show that\n\n\[ E\left( \left| Z\right| \right) = {\int }_{-\infty }^{\infty }\frac{1}{\sqrt{2\pi }}\left| z\right| {e}^{-{z}^{2}/2}{dz} = 2{\int }_{0}^{\infty }\frac{1}{\sqrt{2\pi }}z{e}^{-{z}^{2}/2}{dz} = \sqrt{\frac{2}{\pi }}. \] | \[ E\left( \left| Z\right| \right) = {\int }_{-\infty }^{\infty }\frac{1}{\sqrt{2\pi }}\left| z\right| {e}^{-{z}^{2}/2}{dz} = 2{\int }_{0}^{\infty }\frac{1}{\sqrt{2\pi }}z{e}^{-{z}^{2}/2}{dz} = \sqrt{\frac{2}{\pi }}. \] | Yes |
Model (15.2)-(15.3) was simulated with \( {\phi }_{1} = {0.5},{\phi }_{2} = {0.55} \), and \( \lambda = 1 \) . A total of 5,000 observations were simulated, but, for visual clarity, only every 10th observation is plotted in Fig. 15.2. Neither \( {Y}_{1, t} \) nor \( {Y}_{2, t} \) is stationary, but \( {Y}_{1, t} - \lam... | 10 \( \mathrm{n} = {5000} \)\n\n11 set.seed(12345)\n\n12 a1 \( = {0.5} \)\n\na2 \( = {0.55} \)\n\nlambda \( = 1 \)\n\n\( \mathrm{{y1}} = \operatorname{rep}\left( {0,\mathrm{n}}\right) \)\n\ny2 = y1\n\n\( \mathrm{e}1 = \operatorname{rnorm}\left( \mathrm{n}\right) \)\n\n\( \mathrm{e}2 = \operatorname{rnorm}\left( \mathrm... | No |
Problem 5 Does this difference series appear stationary? Why? | Run the following commands to conduct Johansen's cointegration test.\n\n4 library(urca)\n\n5 summary(ca.jo(CokePepsi)) | No |
Problem 6 Are these two series cointegrated? Why? | Now consider the daily adjusted closing prices for 10 company stocks from January 2, 1987 to September 1, 2006 from the Stock_FX_Bond.csv dataset.\n\n6 Stock_FX_Bond = read.csv(\ | No |
Problem 7 Are these 10 stock price series cointegrated? If so, what is the rank of the cointegrating matrix, and what are the cointegrating vectors? | Rerun the Johansen’s cointegration test with \( \operatorname{lag}K = 8 \) . 10 summary(ca.jo(adjClose, K=8)) | No |
The expectation and variance of the return on a portfolio with two risky assets | Suppose that \( {\mu }_{1} = {0.14},{\mu }_{2} = {0.08},{\sigma }_{1} = {0.2},{\sigma }_{2} = {0.15} \), and \( {\rho }_{12} = 0 \) . Then\n\n\[ E\left( {R}_{P}\right) = {0.08} + {0.06w} \]\n\nand because \( {\rho }_{12} = 0 \) in this example,\n\n\[ {\sigma }_{{R}_{P}}^{2} = {\left( {0.2}\right) }^{2}{w}^{2} + {\left(... | Yes |
Example 16.3. The tangency portfolio with two risky assets\n\nSuppose as before that \( {\mu }_{1} = {0.14},{\mu }_{2} = {0.08},{\sigma }_{1} = {0.2},{\sigma }_{2} = {0.15} \), and \( {\rho }_{12} = 0 \) . Suppose as well that \( {\mu }_{f} = {0.06} \) . Then \( {V}_{1} = {0.14} - {0.06} = {0.08} \) and \( {V}_{2} = {0... | Therefore,\n\n\[ E\left( {R}_{T}\right) = \left( {0.693}\right) \left( {0.14}\right) + \left( {0.307}\right) \left( {0.08}\right) = {0.122}, \]\n\nand\n\n\[ {\sigma }_{T} = \sqrt{{\left( {0.693}\right) }^{2}{\left( {0.2}\right) }^{2} + {\left( {0.307}\right) }^{2}{\left( {0.15}\right) }^{2}} = {0.146}. \] | Yes |
In this example, we will find the optimal investment with \( {\sigma }_{{R}_{p}} = {0.05} \) . | The maximum expected return with \( {\sigma }_{{R}_{p}} = {0.05} \) mixes the tangency portfolio and the risk-free asset such that \( {\sigma }_{{R}_{p}} = {0.05} \) . Since \( {\sigma }_{T} = {0.146} \), we have that \( {0.05} = {\sigma }_{{R}_{p}} = \omega {\sigma }_{T} = {0.146\omega } \), so that \( \omega = {0.05}... | Yes |
Finding the efficient frontier, tangency portfolio, and minimum variance portfolio with no short selling using quadratic programming | In this example, Example 16.6 is modified so that short sales are not allowed. Only three lines of code need to be changed. When short sales are prohibited, the target expected return on the portfolio must lie between the smallest and largest expected returns on the stocks. To prevent numerical errors, the target expec... | Yes |
Suppose that the risk-free rate of interest is \( {\mu }_{f} = {0.06} \), the expected return on the market portfolio is \( {\mu }_{M} = {0.15} \), and the risk of the market portfolio is \( {\sigma }_{M} = {0.22} \). Then the slope of the CML is \( \left( {{0.15} - {0.06}}\right) /{0.22} = 9/{22} \). The CML of this e... | The CML is easy to derive. Consider an efficient portfolio that allocates a proportion \( w \) of its assets to the market portfolio and \( \left( {1 - w}\right) \) to the risk-free asset. Then\n\n\[ R = w{R}_{M} + \left( {1 - w}\right) {\mu }_{f} = {\mu }_{f} + w\left( {{R}_{M} - {\mu }_{f}}\right) .\n\]\n\n(17.2)\n\n... | Yes |
Why are the Dow Jones stocks behaving differently compared to the equity funds? | The Dow Jones stocks are similar to each other since they are all large companies in the United States. Thus, we can expect that their returns will be highly correlated with each other and a few principal components will explain most of the variation. | No |
Estimating the covariance matrix of GE, IBM, and Mobil excess returns | The estimate of \( {\mathbf{\sum }}_{F} \) is the sample covariance matrix of the factors:\n\nMkt.RF SMB HML\n\nMkt.RF 21.1507 4.2326 -5.1045\n\nSMB 4.2326 8.1811 -1.0760\n\nHML -5.1045 -1.0760 7.1797\n\nThe estimate of \( \mathbf{\beta } \) is the matrix of regression coefficients (without the intercepts):\n\nMkt.RF S... | Yes |
Factor analysis of equity funds using a 4-factor model. | The code for fitting a 4-factor model \( \left( {p = 4}\right) \) using factanal () is:\n\nequityFunds \( = \) read.csv(\ | No |
Problem 1 It is generally recommended that PCA be applied to time series that are stationary. Plot the first column of \( \mathtt{{yieldDat}} \) . (You can look at other columns as well. You will see that they are fairly similar.) Does the plot appear stationary? Why or why not? Include your plot with your work. | Another way to check for stationarity is to run the augmented Dickey-Fuller test. You can do that with the following code:\n\nlibrary(\ | No |
Based on the augmented Dickey-Fuller test, do you think the first column of yieldDat is stationary? Why or why not? | Run the following code to compute changes in the yield curves. Notice the use of \( \left\lbrack {-1,}\right\rbrack \) to delete the first row and similarly the use of \( \left\lbrack {-n,}\right\rbrack \) . \n\n\[ \n\text{n=dim(yieldDat)[1]} \n\] \n\n\[ \n\text{delta_yield = yieldDat[-1, ] - yieldDat[-n, ]} \n\] \n\nP... | No |
Problem 14 Does the likelihood ratio test suggest that two factors are enough? If not, what is the minimum number of factors that seems sufficient? | The following code will extract the loadings and uniquenesses.\n\nloadings \( = \operatorname{matrix}\left( \right. \) as.numeric \( \left( {\text{loadings}\left( \text{fact}\right) }\right) , \) ncol \( = 2) \)\n\nunique \( = \) as.numeric(fact\$unique) | No |
Parametric VaR and ES for a position in an S&P 500 index fund | This example uses the same data set as in Example 19.2 so that parametric and nonparametric estimates can be compared. We will assume that the returns are i.i.d. with a \( t \) -distribution. Under this assumption, VaR is\n\n\[ \n{\widehat{\operatorname{VaR}}}^{\mathrm{t}}\left( \alpha \right) = - S \times \left\{ {\wi... | Yes |
Bootstrap confidence intervals for VaR and ES for a position in an S&P 500 index fund | In this example, we continue Examples 19.2 and 19.3 and find an approximate confidence interval for \( \operatorname{VaR}\left( \alpha \right) \) and \( \operatorname{ES}\left( \alpha \right) \) . We use \( \alpha = {0.05} \) as before and \( \gamma = {0.1}.B = 5,{000} \) resamples were taken.\n\nThe basic percentile c... | Yes |
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