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Consider testing the hypothesis:\n\n\[ \n{H}_{0} : \theta \in {\Theta }_{0} \leftrightarrow {H}_{1} : \theta \in {\Theta }_{1} \]\n\nLet \( \varphi \left( \mathbf{x}\right) \) be the UMP test with level of significance \( \alpha \) (suppose it exits) , then its power \( {E}_{\theta }\varphi \left( \mathbf{X}\right) \ge... | Proof. Let \( {\varphi }^{ * }\left( \mathbf{x}\right) \equiv \alpha \), then \( {\varphi }^{ * } \) is a test with level of significance \( \alpha \). Since \( \varphi \left( \mathbf{x}\right) \) is the UMP test, then\n\n\[ \n{E}_{\theta }\varphi \left( \mathbf{X}\right) \geq {E}_{\theta }{\varphi }^{ * }\left( \mathb... | Yes |
(i) Use the Neyman-Pearson Fundamental Lemma to derive the MP test for testing the hypothesis \( {H}_{0} : \theta = {\theta }_{0} \) against the alternative \( {H}_{A} : \theta = {\theta }_{1} \) at level of significance \( \alpha \) . | (i) \( {H}_{0} \) is rejected, if for some positive constant \( {C}^{ * } \):\n\n\[ \frac{{\theta }_{1}{x}^{{\theta }_{1} - 1}}{{\theta }_{0}{x}^{{\theta }_{0} - 1}} > {C}^{ * },\;\text{ or }\;{x}^{{\theta }_{1} - {\theta }_{0}} > \frac{{\theta }_{0}{C}^{ * }}{{\theta }_{1}},\;\text{ or }\;\left( {{\theta }_{1} - {\the... | Yes |
Example 8. On the basis of a random sample of size 1 from the p.d.f. \( f\left( {x;\theta }\right) = \) \( 1 + {\theta }^{2}\left( {\frac{1}{2} - x}\right) ,0 < x < 1, - 1 \leq \theta \leq 1 : \)\n\n(i) Use the Neyman-Pearson Fundamental Lemma to derive the MP test for testing the hypothesis \( {H}_{0} : \theta = 0 \) ... | (i) \( {H}_{0} \) is rejected whenever \( 1 + {\theta }_{1}^{2}\left( {\frac{1}{2} - x}\right) > {C}^{ * } \), or \( x < C \), where \( C = \frac{1}{2} - \left( {{C}^{ * } - }\right.\n\n1) \( /{\theta }_{1}^{2} \), and \( C \) is determined by \( {P}_{0}\left( {X < C}\right) = \alpha \), so that \( C = \alpha \), since... | Yes |
Example 6. On the basis of a random sample of size 1 from the p.d.f. \( f\left( {x;\theta }\right) = \) \( \theta {x}^{\theta - 1},0 < x < 1\left( {\theta > 1}\right) \) : | No UMP test for two-side test | No |
(i) Use the Neyman-Pearson Fundamental Lemma to derive the MP test for testing the hypothesis \( {H}_{0} : \theta = 0 \) (i.e., the p.d.f. is \( U\left( {0,1}\right) \) ) against the alternative \( {H}_{A} \) : \( \theta = {\theta }_{1} \) at level of significance \( \alpha \) . | (i) \( {H}_{0} \) is rejected whenever \( 1 + {\theta }_{1}^{2}\left( {\frac{1}{2} - x}\right) > {C}^{ * } \), or \( x < C \), where \( C = \frac{1}{2} - \left( {{C}^{ * } - }\right. \) 1) \( /{\theta }_{1}^{2} \), and \( C \) is determined by \( {P}_{0}\left( {X < C}\right) = \alpha \), so that \( C = \alpha \), since... | Yes |
Theorem 3. Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample with p.d.f. as in Theorem 2, and let \( V\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) be as in the same theorem. Consider the problem of testing the hypothesis \( {H}_{0} : \theta \leq {\theta }_{1} \) or \( \theta \geq {\theta }_{2} \) against the alternati... | (i) If \( Q \) is strictly increasing, the UMP test is given by:\n\n\[ \varphi \left( {{x}_{1},\ldots ,{x}_{n}}\right) = \left\{ \begin{array}{lll} 1 & \text{ if } & {C}_{1} < V\left( {{x}_{1},\ldots ,{x}_{n}}\right) < {C}_{2} \\ {\gamma }_{1} & \text{ if } & V\left( {{x}_{1},\ldots ,{x}_{n}}\right) = {C}_{1} \\ {\gamm... | Yes |
In the linear regression model\n\n\[ Y = {X\beta } + \epsilon \]\n\nsuppose that the design matrix is deterministic (not random) and the random\n\nerrors\n\n\[ {\epsilon }_{1},\cdots ,{\epsilon }_{n}\text{ i.i.d. } \sim N\left( {0,{\sigma }^{2}}\right) \]\n\nlet \( \widehat{\beta } = \left( {{\widehat{\beta }}_{0},{\wi... | ## Confidence Interval\n\nBy Theorem 1, we have, for \( j = 0,1,\cdots, p \) ,\n\n\[ {T}_{j} = \frac{{\widehat{\beta }}_{j} - {\beta }_{j}}{\widehat{\sigma }{\left\lbrack {\left( {X}^{T}X\right) }^{-1}\right\rbrack }_{jj}^{1/2}} \sim {t}_{n - p - 1} \]\n\nTherefore, a \( 1 - \alpha \) confidence interval for \( {\beta ... | Yes |
Example 5. Suppose we are interested in acquiring a fairly large number of equipments from among \( I \) brands entertained. The available workforce to use the equipments bought consists of \( J \) workers. Before a purchase decision is made, an experiment is carried out whereby each one of the \( J \) workers uses eac... | ## \(\text{Two-way ANOVA}\)\n\nGathering together the assumptions made so far, we have the following model.\n\n\[ \left. \begin{array}{l} {Y}_{ij} = \mu + {\alpha }_{i} + {\beta }_{j} + {e}_{ij},\mathop{\sum }\limits_{{i = 1}}^{I}{\alpha }_{i} = 0\text{ and }\mathop{\sum }\limits_{{j = 1}}^{J}{\beta }_{j} = 0\text{, th... | Yes |
Lemma 6. The unique minimizing values of \( \mu ,{\alpha }_{i} \), and \( {\beta }_{j} \) for expression (25) (i.e., the LSE’s of \( \mu ,{\alpha }_{i} \), and \( {\beta }_{j} \) ) are given by: | \[ \widehat{\mu } = {y}_{..},\;{\widehat{\alpha }}_{i} = {y}_{i.} - {y}_{..},\;i = 1,\ldots, I,\;{\widehat{\beta }}_{j} = {y}_{.j} - {y}_{..},\;j = 1,\ldots, J, \] (26) where \[ {y}_{i.} = \frac{1}{J}\mathop{\sum }\limits_{j}{y}_{ij},\;{y}_{.j} = \frac{1}{I}\mathop{\sum }\limits_{i}{y}_{ij},\;{y}_{..} = \frac{1}{IJ}\ma... | Yes |
Lemma 7. With \( \widehat{{\sigma }_{A}^{2}}, S{S}_{e} \) and \( S{S}_{A} \) defined by (34) and (36), it holds: \( {IJ}\widehat{{\sigma }_{A}^{2}} = {IJ}\widehat{{\sigma }^{2}} + \) \( S{S}_{A} = S{S}_{e} + S{S}_{A} \) | ## Proof. Deferred to Subsection 14.3.3. | No |
Lemma 11. Let \( S{S}_{e}, S{S}_{A} \), and \( S{S}_{B} \) be given by (36) and (43), and let \( S{S}_{T} \) be defined by:\n\n\[ S{S}_{T} = \mathop{\sum }\limits_{i}\mathop{\sum }\limits_{j}{\left( {y}_{ij} - {y}_{..}\right) }^{2} \]\n\n(46)\n\nThen:\n\n(i)\n\n(ii)\n\n\[ S{S}_{A} = J\mathop{\sum }\limits_{i}{y}_{i.}^{... | \[ S{S}_{A} = J\mathop{\sum }\limits_{i}{y}_{i.}^{2} - {IJ}{y}_{..}^{2},\;S{S}_{B} = I\mathop{\sum }\limits_{j}{y}_{.j}^{2} - {IJ}{y}_{..}^{2},\;S{S}_{T} = \mathop{\sum }\limits_{i}\mathop{\sum }\limits_{j}{y}_{ij}^{2} - {IJ}{y}_{..}^{2}. \]\n\n(47)\n\n\[ S{S}_{T} = S{S}_{e} + S{S}_{A} + S{S}_{B} \]\n\n(48) | No |
Theorem 1. Let \( {x}_{1},\cdots ,{x}_{n} \) be any numbers and \( \bar{x} = \left( {{x}_{1} + \cdots + {x}_{n}}\right) /n \) . Then\n\na. \( \mathop{\min }\limits_{a}\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - a\right) }^{2} = \mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - \bar{x}\right) }^{2}, \)\n\nb.... | Proof. | No |
Theorem 3. If \( {X}_{1},\cdots ,{X}_{n} \) are mutually independent normal random variables with mean \( {\mu }_{1},\cdots ,{\mu }_{n} \) and variances \( {\sigma }_{1}^{2},\cdots ,{\sigma }_{n}^{2} \), then the linear combination\n\n\[ Y = \mathop{\sum }\limits_{{i = 1}}^{n}{c}_{i}{X}_{i} \sim N\left( {\mathop{\sum }... | Proof. | No |
Example 4. Uniform on \( \left\lbrack {0,\theta }\right\rbrack ,\theta > 0 \) is not exponential family | \[ f\left( {x;\theta }\right) = \frac{1}{\theta },\;x \in \left\lbrack {0,\theta }\right\rbrack ,\;\theta > 0 \] | Yes |
Example 5. The Cauchy distribution family is not exponential family | \[ f\left( {x;\theta }\right) = \frac{1}{\pi \left\lbrack {1 + {\left( x - \theta \right) }^{2}}\right\rbrack },\;x \in R \] | Yes |
1. All the distributions in the exponential family does not depend on \( \mathbf{\theta } \) . | Proof. | No |
We already known that if \( X \sim B\left( {n,\theta }\right) \) , then \( E\left( X\right) = {n\theta } \) . | \[ f\left( {x,\theta }\right) = {\left( 1 - \theta \right) }^{n}\exp \left\{ {x\log \frac{\theta }{1 - \theta }}\right\} {C}_{n}^{x} \] \[ \varphi = \log \left\lbrack {\theta /\left( {1 - \theta }\right) }\right\rbrack ;\;\theta = {e}^{\varphi }/\left( {1 + {e}^{\varphi }}\right) \] \[ f\left( {x;\varphi }\right) = {\l... | Yes |
Theorem 1. Let \( {x}_{1},\cdots ,{x}_{n} \) be any numbers and \( \bar{x} = \left( {{x}_{1} + \cdots + {x}_{n}}\right) /n \) . Then\n\na. \( \mathop{\min }\limits_{a}\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - a\right) }^{2} = \mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - \bar{x}\right) }^{2} \),\n\nb.... | Proof.  | No |
Theorem 2. Let \( {X}_{1},\cdots ,{X}_{n} \) be a random sample from a population with mean \( \mu \) and variance \( {\sigma }^{2} < \infty \) . Then\n\na. \( \mathrm{E}\left( \bar{X}\right) = \mu \)\n\nb. \( \operatorname{Var}\left( \bar{X}\right) = {\sigma }^{2}/n \)\n\nc. \( \mathrm{E}\left( {S}^{2}\right) = {\sigm... | Proof. 清华大学统计学研究中心 | No |
Theorem 3. If \( {X}_{1},\cdots ,{X}_{n} \) are mutually independent normal random variables with mean \( {\mu }_{1},\cdots ,{\mu }_{n} \) and variances \( {\sigma }_{1}^{2},\cdots ,{\sigma }_{n}^{2} \), then the linear combination | Proof.  | No |
Theorem 4. Suppose r.v.’s \( {X}_{1},\cdots ,{X}_{n} \) i.i.d. \( \sim N\left( {\mu ,{\sigma }^{2}}\right) \) . Let \( \mathbf{X} = {\left( {X}_{1},\cdots ,{X}_{n}\right) }^{T} \) , \( \mathbf{Y} = {\left( {Y}_{1},\cdots ,{Y}_{n}\right) }^{T} \) . Suppose \( \mathbf{A} = \left( {a}_{ij}\right) \) is an \( n \times n \)... | Then\n\n(1) \( {Y}_{1},\cdots ,{Y}_{n} \) are Normal r.v., and\n\n\[ E\left( {Y}_{i}\right) = \mu \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{ik},\;\operatorname{Var}\left( {Y}_{i}\right) = {\sigma }^{2}\mathop{\sum }\limits_{{k = 1}}^{n}{a}_{ik}^{2}, \]\n\n\[ \operatorname{Cov}\left( {{Y}_{i},{Y}_{j}}\right) = {\sigma }^{... | Yes |
Theorem 5. If \( {X}_{1},\cdots ,{X}_{n} \) are independent random sample from a \( \mathrm{N}\left( {\mu ,{\sigma }_{ \bot }^{2}}\right) \) population, then\na. \( \bar{X} \) and \( {S}^{2} \) are independent random variables\nb. \( \bar{X} \sim N\left( {\mu ,{\sigma }^{2}/n}\right) \)\nc. \( \left( {n - 1}\right) {S}... | Proof.\nA Simulation Study    \) be a random sample from \( \mathrm{N}\left( {\mu ,{\sigma }^{2}}\right) \), then the sample distribution family belongs to exponential family. | The joint p.d.f of \( \mathbf{X} \) is\n\n\[ f\left( {\mathbf{x};\mu ,{\sigma }^{2}}\right) = {\left( \sqrt{2\pi }\sigma \right) }^{-n}\exp \left\{ {-\frac{1}{2{\sigma }^{2}}\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - \mu \right) }^{2}}\right\} . \]\n\nLet \( \theta = \left( {\mu ,{\sigma }^{2}}\right) \), the... | Yes |
Example 3. Binomial distribution family \( \{ \mathsf{B}\left( {n,\theta }\right) \} \) belongs to exponential family. | Discussion. The joint p.d.f of \( X \sim B\left( {n,\theta }\right) \) is\n\n\[ f\left( {x;\theta }\right) = P\left( {X = x}\right) = {C}_{n}^{x}{\theta }^{x}{\left( 1 - \theta \right) }^{n - x} = {C}_{n}^{x}{\left( \frac{\theta }{1 - \theta }\right) }^{x}{\left( 1 - \theta \right) }^{n}, x = 0,1,2,\cdots, n, \]\n\nand... | Yes |
Example 4. Uniform on \( \left\lbrack {0,\theta }\right\rbrack ,\theta > 0 \) is not exponential family | \[ f\left( {x;\theta }\right) = \frac{1}{\theta },\;x \in \left\lbrack {0,\theta }\right\rbrack ,\;\theta > 0 \] | Yes |
Example 5. The Cauchy distribution family is not exponential family | \[ f\left( {x;\theta }\right) = \frac{1}{\pi \left\lbrack {1 + {\left( x - \theta \right) }^{2}}\right\rbrack },\;x \in R \] | Yes |
Normal distribution family | \[ f\left( {\mathbf{x},\theta }\right) = {\left( \sqrt{2\pi }\sigma \right) }^{-n}{e}^{-\frac{n{\mu }^{2}}{2{\sigma }^{2}}}\exp \left\{ {\frac{\mu }{{\sigma }^{2}}\mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i} - \frac{1}{2{\sigma }^{2}}\mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i}^{2}}\right\} \]\n\n\[ = C\left( \theta \right... | Yes |
Example 7 (discrete). Binomial distribution family\n\n\[ f\left( {x,\theta }\right) = \left( {{\left( 1 - \theta \right) }^{n}\exp \left\{ {x\log \frac{\theta }{1 - \theta }}\right\} {C}_{n}^{x}}\right. \] | Let \( \varphi = \log \left\lbrack {\theta /\left( {1 - \theta }\right) }\right\rbrack \), then \( - \infty < \varphi < + \infty \) and \( 1 - \theta = 1/\left( {1 + {e}^{\varphi }}\right) \) . Thus,\n\n\[ f\left( {x,\varphi }\right) = {\left( 1 + {e}^{\varphi }\right) }^{-n}\exp \{ {\varphi x}\} {C}_{n}^{x}, \]\nnatur... | Yes |
We already known that if \( X \sim B\left( {n,\theta }\right) \), then \( E\left( X\right) = {n\theta } \) | Solution.\n\[\nf\left( {x,\theta }\right) = {\left( 1 - \theta \right) }^{n}\exp \left\{ {x\log \frac{\theta }{1 - \theta }}\right\} {C}_{n}^{x}\n\]\n\n\[\n\varphi = \log \left\lbrack {\theta /\left( {1 - \theta }\right) }\right\rbrack ;\;\theta = {e}^{\varphi }/\left( {1 + {e}^{\varphi }}\right)\n\]\n\n\[\nf\left( {x;... | Yes |
Example 1. Let \( X \) be a Bernoulli population, i.e., \( X \sim B\left( {1,\theta }\right) \), and let \( X = \left( {{X}_{1},\cdots ,{X}_{n}}\right) \) be a random sample from \( X \). Goal: inference on \( \theta \). \[ P\left( {{X}_{i} = 1}\right) = \theta ,\;P\left( {{X}_{i} = 0}\right) = 1 - \theta ,\;0 < \theta... | Intuitively: The number of success contains all the information about \( \theta \) , while no information in the order, so \( T\left( \mathbf{X}\right) \) capture all the information Mathematically: Let \( t = \mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i} \) \( {f}_{\mathbf{X}}\left( {\mathbf{x};\theta }\right) = P\left( ... | Yes |
Example 6. Let \( \\mathbf{X} = \\left( {{X}_{1},\\cdots ,{X}_{n}}\\right) \) be a random sample from an exponential family, then \( T\\left( \\mathbf{X}\\right) = \\left( {{T}_{1}\\left( \\mathbf{X}\\right) ,\\cdots ,{T}_{k}\\left( \\mathbf{X}\\right) }\\right) \) is sufficient for \( \\theta \) . | \[ f\\left( {\\mathbf{x};\\theta }\\right) = C\\left( \\theta \\right) \\exp \\left\\{ {\\mathop{\\sum }\\limits_{{i = 1}}^{k}{Q}_{i}\\left( \\theta \\right) {t}_{i}\\left( \\mathbf{x}\\right) }\\right\\} h\\left( \\mathbf{x}\\right) = g\\left( {t\\left( \\mathbf{x}\\right) ;\\theta }\\right) h\\left( \\mathbf{x}\\righ... | Yes |
Example 3. Let \( \mathbf{X} = \left( {{X}_{1},\cdots ,{X}_{n}}\right) \) be a random sample from \( N\left( {\mu ,{\sigma }^{2}}\right) \) . The sample mean \( \bar{X} \) and the sample variance \( {S}^{2} \) are both unbiased estimators since, by Theorem 1,\n\n\[ E\left( \bar{X}\right) = \mu ,\;E\left( {S}^{2}\right)... | The MSE of these estimators are\n\n\[ E{\left( \bar{X} - \mu \right) }^{2} = \operatorname{Var}\left( \bar{X}\right) = \frac{{\sigma }^{2}}{n} \]\n\n\[ E{\left( {S}^{2} - {\sigma }^{2}\right) }^{2} = \operatorname{Var}\left( {S}^{2}\right) = \frac{2{\sigma }^{4}}{n - 1}. \]\n\nBy Theorem 3 in topic 2, \( \left( {n - 1}... | Yes |
An alternative estimator for \( {\sigma }^{2} \) is the maximum likelihood estimator \( {\widehat{\sigma }}^{2} = \frac{1}{n}\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {X}_{i} - \bar{X}\right) }^{2} = \frac{n - 1}{n}{S}^{2} \). It is straightforward to calculate | \[ E\left( {\widehat{\sigma }}^{2}\right) = E\left( {\frac{n - 1}{n}{S}^{2}}\right) = \frac{n - 1}{n}{\sigma }^{2} \] so \( {\widehat{\sigma }}^{2} \) is a biased estimator of \( {\sigma }^{2} \). The variance of \( {\widehat{\sigma }}^{2} \) can be calculated as \[ \operatorname{Var}\left( {\widehat{\sigma }}^{2}\righ... | Yes |
Example 23. Let \( \mathbf{X} = \left( {{X}_{1},\cdots ,{X}_{n}}\right) \) be a random sample from Bernoulli distribution \( \{ B\left( {1,\theta }\right) : 0 < \theta < 1\} \) . Find the moment estimate of \( \theta \) and \( g\left( \theta \right) = \theta \left( {1 - \theta }\right) \) . | Discussion. Here, \( {E}_{\theta }{X}_{1} = \theta \) . Thus, \( \bar{X} \) is the moment estimate of \( \theta \), and \( \bar{X}\left( {1 - \bar{X}}\right) \) is the moment estimate of \( g\left( \theta \right) = \theta \left( {1 - \theta }\right) \) . | No |
On the basis of the random sample \( {X}_{1},\ldots ,{X}_{n} \) from the \( N\left( {\mu ,{\sigma }^{2}}\right) \) distribution with both \( \mu \) and \( {\sigma }^{2} \) unknown, determine the moment estimates of \( \mu \) and \( {\sigma }^{2} \) . |  | No |
Example 25. Let the random sample \( {X}_{1},\ldots ,{X}_{n} \) be from the \( U\left( {\alpha ,\beta }\right) \) distribution, where both \( \alpha \) and \( \beta \) are unknown. Determine their moment estimates. |  | No |
Example 5. Let \( {X}_{1},\cdots ,{X}_{10} \) be a random sample from \( B\left( {1,\theta }\right) \) distribution, \( 0 < \theta < 1 \), and let \( {x}_{1},\cdots ,{x}_{10} \) be the respective observed values. For convenience, set \( t = {x}_{1} + \cdots + {x}_{10} \) . Further, suppose that in the 10 trials, six re... | If these nine values were the only possible values for \( \theta \), one would reasonably enough choose the value of 0.6 as the value of \( \theta \) . This value maximizing the probability of attaining the six already observed successes.\n\nObserve that \( {0.6} = 6/{10} = \) \( t/n \) . Actually, the value \( t/n \) ... | Yes |
In terms of a random sample of size \( n,{X}_{1},\ldots ,{X}_{n} \) from the \( B\left( {1,\theta }\right) \) distribution with observed values \( {x}_{1},\ldots ,{x}_{n} \), determine the MLE \( \widehat{\theta } = \widehat{\theta }\left( \mathbf{x}\right) \) of \( \theta \in \) \( \left( {0,1}\right) ,\mathbf{x} = \l... | \[ L\left( {0 \mid x}\right) = \frac{h}{a}f\left( {x,0}\right) = {0}^{t}{\left( f0\right) }^{h - t}t = x + \cdots + h. \] \[ \frac{\partial }{\partial 0}\log L = \] | No |
Example 8. Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from the \( N\left( {\mu ,{\sigma }^{2}}\right) \) distribution, where only one of the parameters is known. Determine the MLE of the other (unknown) parameter. | \[ = - h\log \left( \sqrt{\ln b}\right) - \frac{1}{2{b}^{2}}\mathop{\sum }\limits_{{i = 1}}^{h}{\left( {x}_{i} - u\right) }^{2} \] | No |
Example 11. A Multinomial experiment is carried out independently \( n \) times, so that the likelihood function is\n\n\[ L\left( {{p}_{1},\ldots ,{p}_{r} \mid \mathbf{x}}\right) = \frac{n!}{{x}_{1}!\cdots {x}_{r}!}{p}_{1}^{{x}_{1}}\cdots {p}_{r}^{{x}_{r}}, \]\n\nwhere \( {x}_{i} \geq 0 \), integers, \( i = 1,\ldots, r... | \[ \Rightarrow {x}_{1}\frac{1}{{\rho }_{1}} - {x}_{r}\frac{1}{{\rho }_{r}} = 0.\;{P}_{r} = 1 - {\rho }_{1}\cdots - {\rho }_{r - 1} \]\n\n\[ \Rightarrow \frac{a}{b} = \frac{a!}{a}\;i = 1,\ldots r \] | No |
Theorem 5. Let \( {\widehat{\theta }}_{n} = {\widehat{\theta }}_{n}\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) be the MLE of \( \theta \in \Omega \subseteq \Re \) based on the random sample \( {X}_{1},\ldots ,{X}_{n} \) with p.d.f. \( f\left( {\cdot ,\theta }\right) \) . Then, under certain regularity conditions, \( \le... | Proof idea: Weak Law of Large Numbers or the Chebichev inequality \n\n\[ \Pr \left( {\left| {X - \mu }\right| \geq {k\sigma }}\right) \leq \frac{1}{{k}^{2}} \] | No |
Theorem 6. In the notation of Theorem 5, and under suitable regularity conditions, the MLE \( {\widehat{\theta }}_{n} \) is asymptotically Normal. More precisely, under \( {P}_{\theta } \) -probability, \[ \sqrt{n}\left( {{\widehat{\theta }}_{n} - \theta }\right) \overset{d}{ \rightarrow }N\left( {0,{\sigma }_{\theta }... | Using Delta method (Robert Dorfman 1938; Casella. G. and Bergerm R. L. 2002) \[ {\forall }^{J}\;\sqrt{n}\left\lbrack {g\left( {\widehat{\theta }}_{n}\right) - g\left( \theta \right) }\right\rbrack \overset{d}{ \rightarrow }N\left( {0,{\left( {g}^{\prime }\left( \theta \right) \right) }^{2}/I\left( \theta \right) }\righ... | Yes |
Lemma 1. Let \( T = T\left( \mathbf{X}\right) \) be a sufficient statistic of \( g\left( \theta \right) \) and \( \widehat{g}\left( \mathbf{X}\right) \) is an unbiased estimate of \( g\left( \theta \right) \), then\n\n\[ h\left( T\right) = E\left( {\widehat{g}\left( \mathbf{X}\right) \mid T}\right) \]\n\nis also an unb... | Proof. Since T is sufficient | No |
Example 15. Let \( \\mathbf{X} = \\left( {{X}_{1},\\cdots ,{X}_{n}}\\right) \) be a random sample from Bernoulli distribution \( \\{ B\\left( {1, p}\\right) : 0 < p < 1\\} \) . We known that \( {X}_{1} \) is an unbiased estimate of \( p \) and \( T = T\\left( \\mathbf{X}\\right) = \\mathop{\\sum }\\limits_{{i = 1}}^{n}... | \[ h\\left( t\\right) = E\\left( {{x}_{1} \\mid T = t}\\right) \\]\n\[ = {1.9}\\left( {{x}_{1} = 1 \\mid T = t}\\right) + {0.p}\\left( {{x}_{2} = 0}\\right) T = t \\]\n\[ = \\frac{{PC}{X}_{1} = {I}_{1}T = t)}{{PCT} = t)} = \\frac{P\\left( {{x}_{1} = 1,{x}_{2} + \\cdots + {x}_{n} = t - 1}\\right) }{P\\left( {T = t}\\rig... | No |
Example 16. Let \( \\mathbf{X} = \\left( {{X}_{1},\\cdots ,{X}_{n}}\\right) \) be a random sample from the distribution \( P\\left( \\theta \\right) \). Determine the UMVUEs of (1) \( {g}_{1}\\left( \\theta \\right) = \\theta \), (2) \( {g}_{2}\\left( \\theta \\right) = {\\theta }^{r}, r = 1,2,\\cdots \), and (3) \( {g... | Proof.\n\n① \( x = 7/h \) cm biased. estima 0\n\n, \( \\bar{x} \) is universal.\n\n(2) It is unbiased est-or \( {0}^{ \\vee }\)\n\n\( \\mathop{\\sum }\\limits_{{t = 0}}^{\\infty }\\delta \\left( t\\right) \\frac{{\\left( h\\theta \\right) }^{t}}{t!}{e}^{-{h\\theta }} = {\\theta }^{r},\\theta > 0 \)\n\n\\[ \\mathop{\\su... | No |
Example 17. Let \( \\mathbf{X} = \\left( {{X}_{1},\\cdots ,{X}_{n}}\\right) \) be a random sample from the normal distribution \( N\\left( {\\mu ,{\\sigma }^{2}}\\right) \) and let \( \\theta = \\left( {\\mu ,{\\sigma }^{2}}\\right) \). Determine the UMVUEs of \( \\left( 1\\right) \\mu \) and \( {\\bar{\\sigma }}^{2},\... | Proof.\n\\[ \nC\\left( {f, f, g}\\right) = \\left( {\\frac{\\mathop{\\sum }\\limits_{{i = 1}}^{n}\\frac{{x}_{i}}{n}}{n},\\mathop{\\sum }\\limits_{{i = 1}}^{n}{\\left( {x}_{i} - \\bar{x}\\right) }^{2}}\\right) \\text{is sufficient }R \n\\] | No |
Theorem 9. Let \( \mathcal{F} = \{ f\left( {x;\theta }\right) ,\theta \in \Theta \} \) be a C-R regularity distribution family, \( g\left( \theta \right) \) is a differentiable function defined on the parameter space. Let \( \mathbf{X} = \) \( \left( {{X}_{1},\cdots ,{X}_{n}}\right) \) be a random sample from the distr... | Proof. \( {x}_{1},\cdots {x}_{n}, f\left( {x,\theta }\right) = \frac{h}{n}f\left( {x,\theta }\right) \)\n\n\n\n\[ \n\text{Score}\left( {\overrightarrow{x},\theta }\right) = \frac{\partial \log f\left( {\overrightarrow{... | No |
Example 18. Let \( \\mathbf{X} = \\left( {{X}_{1},\\cdots ,{X}_{n}}\\right) \) be a random sample from Bernoulli distribution \( B\\left( {1,\\theta }\\right) \). Using C-R inequality to show that \( \\bar{X} \) is UMVUE of \( \\theta \). | Proof. \( f\\left( {X;\\theta }\\right) = {\\theta }^{x}{\\left( 1 - \\theta \\right) }^{1 - x} \) \( \\therefore \) JULEALF bond. \( \\because B\\left( {1,0}\\right) \) belongs to \( {Expo} + a{m}^{\\prime }/g \) . 指数族\n\n\( I\\left( \\theta \\right) = {\\left\\lbrack 0,0 + \\frac{\\partial \\log f\\left( {{x}_{i},\\t... | No |
Example 19. Let \( \mathbf{X} = \left( {{X}_{1},\cdots ,{X}_{n}}\right) \) be a random sample from \( P\left( \theta \right) \) . Using C-R inequality to show that \( \bar{X} \) is UMVUE of \( \theta \) . | Proof.\n\[ \text{Asher Information C.-R band.} \] | No |
Let \( \\mathbf{X} = \\left( {{X}_{1},\\cdots ,{X}_{n}}\\right) \) be a random sample from \( N\\left( {\\mu ,{\\sigma }^{2}}\\right) \) . Suppose \( {\\sigma }^{2} \) is known, using C-R inequality to show that \( \\bar{X} \) is UMVUE of \( \\mu \) . | Proof. | No |
Let \( \mathbf{X} = \left( {{X}_{1},\cdots ,{X}_{n}}\right) \) be a random sample from \( N\left( {\mu ,{\sigma }^{2}}\right) \) . Suppose both \( \mu \) and \( {\sigma }^{2} \) are unknown. Let \( \theta = \left( {{\theta }_{1},{\theta }_{2}}\right) = \left( {\mu ,{\sigma }^{2}}\right) \), derive the CR lower bounds o... | \[ \text{let}\theta = \left( {0,{\theta }_{L}}\right) = \left( {u,{\sigma }^{2}}\right) \] \[ f\left( {x,0}\right) = \frac{1}{\sqrt{{2\pi }{\theta }_{2}}}\exp \left\{ \frac{-{\left( x - {\theta }_{1}\right) }^{2}}{2{\theta }^{2}}\right) \] \[ \frac{2\left( g\right) f}{2\left( g\right) } = \frac{x - 0}{{\theta }_{2}};\f... | No |
Suppose an insurance company pays the amount of \$1000 for lost luggage on an airplane trip. From past experience, it is known that the company pays this amount in 1 out of 200 policies it sells. What premium should the company charge? | Define the r.v. \( X \) as follows: \( X = 0 \) if no loss occurs, which happens with probability \( 1 - \left( {1/{200}}\right) = {0.995} \), and \( X = - {1000} \) with probability \( \frac{1}{200} = \) 0.005 . Then the expected loss to the company is: \( {EX} = - {1000} \times {0.005} = - 5 \) . Thus, the company mu... | Yes |
A roulette wheel consists of 18 black slots, 18 red slots, and 2 green slots. If a gambler bets \( \$ {10} \) on red, what is the gambler’s expected gain or loss? | Define the r.v. \( X \) by: \( X = {10} \) with probability \( {18}/{38} \) and \( X = - {10} \) with probability \( {20}/{38} \), or in a tabular form\n\n<table><thead><tr><th>\( x \)</th><th>10</th><th>-10</th><th>Total</th></tr></thead><tr><td>\( f\left( x\right) \)</td><td>18 38</td><td>20 38</td><td>1</td></tr></t... | Yes |
Example 3. Let \( X \) be a r.v. with p.d.f. \( f\left( x\right) = 3{x}^{2},0 < x < 1 \) . Then:\n\n(i) Calculate the quantities: \( {EX}, E{X}^{2} \), and \( \operatorname{Var}\left( X\right) \) .\n\n(ii) If the r.v. \( Y \) is defined by: \( Y = {3X} - 2 \), calculate the \( {EY} \) and the \( \operatorname{Var}\left... | (i) By (3), \( {EX} = {\int }_{0}^{1}x \times 3{x}^{2}\mathrm{\;d}x = {\left. \frac{3}{4}{x}^{4}\right| }_{0}^{1} = \frac{3}{4} = {0.75} \), whereas by (7), applied with \( k = 2, E{X}^{2} = {\int }_{0}^{1}{x}^{2} \times 3{x}^{2}\mathrm{\;d}x = \frac{3}{5} = {0.60} \), so that, by (9), \( \operatorname{Var}\left( X\rig... | Yes |
Proposition 1. If \( X \sim N\left( {\mu ,{\sigma }^{2}}\right) \), then \( Z = \frac{X - \mu }{\sigma } \) is \( \sim N\left( {0,1}\right) \) . | Proof.\n\[ \n{\varphi }_{Z}\left( t\right) = E\left( {e}^{itZ}\right) = E\left( {e}^{{it}\frac{X - \mu }{\sigma }}\right) \n\] \n\n\[ \n= \;{e}^{-{i\mu }\frac{t}{\sigma }}E\left( {e}^{i\frac{t}{\sigma }X}\right) \n\] \n\n\[ \n= {e}^{-{i\mu }\frac{t}{\sigma }}{\varphi }_{X}\left( \frac{t}{\sigma }\right) \n\] \n\n\[ \n=... | Yes |
Examine the median of the r.v. \( X \) distributed as follows:\n\n<table><thead><tr><th>\( x \)</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th><th>8</th><th>9</th><th>10</th></tr></thead><tr><td>\( f\left( x\right) \)</td><td>2/32</td><td>1/32</td><td>5/32</td><td>3/32</td><td>4/32</td><td>1/... | Discussion. We have \( P\left( {X \leq 6}\right) = {16}/{32} = {0.50} \geq {0.50} \) and \( P\left( {X \geq 6}\right) = {17}/{32} > \) \( {0.05} \geq {0.50} \), so that (47) is satisfied. Also,\n\n\[ P\left( {X \leq 7}\right) = {18}/{32} > {0.50} \geq {0.50}\;\text{ and }\;P\left( {X \geq 7}\right) = {16}/{32} = {0.50}... | Yes |
From a large collection of bolts which is known to contain \( 3\% \) defective bolts,1000 are chosen at random. If \( X \) is the number of the defective bolts among those chosen, what is the (approximate) probability that \( X \) does not exceed \( 5\% \) of 1000? | Discussion. With the selection of the \( i \) th bolt, associate the r.v. \( {X}_{i} \) to take the value 1, if the bolt is defective, and 0 otherwise. Then it may be assumed that the r.v.’s \( {X}_{i}, i = \) \( 1,\ldots ,{1000} \) are independently distributed as \( B\left( {1,{0.03}}\right) \) . Furthermore, it is c... | Yes |
Example 1. Each one of the r.v.’s \( X \) and \( Y \) takes on four values only, \( 0,1,2,3 \), with joint probabilities expressed best in a matrix form as in Table 4.1. | Discussion. The r.v.’s \( X \) and \( Y \) may represent, for instance, the number of customers waiting for service in two lines in a bank. Then, for example, for \( \left( {x, y}\right) \) with \( x = 2 \) and \( y = 1 \), we have \( {F}_{X, Y}\left( {x, y}\right) = {F}_{X, Y}\left( {2,1}\right) = \mathop{\sum }\limit... | Yes |
Example 2. Let the r.v.’s \( X \) and \( Y \) have the joint p.d.f. \( {f}_{X, Y}\left( {x, y}\right) = {\lambda }_{1}{\lambda }_{2}{\mathrm{e}}^{-{\lambda }_{1}x - {\lambda }_{2}y} \) , \( x, y > 0,{\lambda }_{1},{\lambda }_{2} > 0 \) . For example, \( X \) and \( Y \) may represent the lifetimes of two components in ... | Discussion. The corresponding joint d.f. is\n\n\[ \n{F}_{X, Y}\left( {x, y}\right) = {\int }_{0}^{y}{\int }_{0}^{x}{\lambda }_{1}{\lambda }_{2} \times {\mathrm{e}}^{-{\lambda }_{1}s - {\lambda }_{2}t}\mathrm{\;d}s\mathrm{\;d}t = {\int }_{0}^{y}{\lambda }_{2}{\mathrm{e}}^{-{\lambda }_{2}t}\left( {{\int }_{0}^{x}{\lambda... | Yes |
Example 5. Refer to Example 1 and derive the marginal and conditional p.d.f.'s involved. | Discussion. From Table 4.1, we have: \( {f}_{X}\left( 0\right) = {0.25},{f}_{X}\left( 1\right) = {0.53},{f}_{X}\left( 2\right) = {0.18} \) , and \( {f}_{X}\left( 3\right) = {0.04} \) ; also, \( {f}_{Y}\left( 0\right) = {0.26},{f}_{Y}\left( 1\right) = {0.54},{f}_{Y}\left( 2\right) = {0.15} \), and \( {f}_{Y}\left( 3\rig... | Yes |
In reference to Example 1, calculate: \( E\left( {X \mid Y = 0}\right) \) and \( E\left( {Y \mid X = 2}\right) \) . | In Example 5, we have calculated the conditional p.d.f.’s \( {f}_{X \mid Y}\left( {\cdot \mid 0}\right) \) and \( {f}_{Y \mid X}\left( {\cdot \mid 2}\right) \) . Therefore:\n\n\[ E\left( {X \mid Y = 0}\right) = 0 \times \frac{5}{26} + 1 \times \frac{21}{26} + 2 \times 0 + 3 \times 0 = \frac{21}{26} \simeq {0.808},\;\te... | Yes |
Theorem 3. For two r.v.’s \( X \) and \( Y \) with finite first and second moments, and (positive) standard deviations \( {\sigma }_{X} \) and \( {\sigma }_{Y} \), it holds:\n\n\[ \operatorname{Var}\left( {X + Y}\right) = {\sigma }_{X}^{2} + {\sigma }_{Y}^{2} + 2\operatorname{Cov}\left( {X, Y}\right) = {\sigma }_{X}^{2... | Proof. Since (40) follows immediately from (39), and \( \operatorname{Cov}\left( {X, Y}\right) = {\sigma }_{X}{\sigma }_{Y} \times \rho \left( {X, Y}\right) \) , it suffices to establish only the first equality in (39). Indeed,\n\n\[ \operatorname{Var}\left( {X + Y}\right) = E{\left\lbrack \left( X + Y\right) - E\left(... | Yes |
If the r.v.’s \( {X}_{1} \) and \( {X}_{2} \) have the Bivariate Normal distribution with parameters \( {\mu }_{1},{\mu }_{2},{\sigma }_{1}^{2},{\sigma }_{2}^{2} \), and \( \rho \): (i) Calculate the quantities: \( E\left( {{c}_{1}{X}_{1} + {c}_{2}{X}_{2}}\right) ,\operatorname{Var}\left( {{c}_{1}{X}_{1} + {c}_{2}{X}_{... | (i) \( E\left( {{c}_{1}{X}_{1} + {c}_{2}{X}_{2}}\right) = {c}_{1}E{X}_{1} + {c}_{2}E{X}_{2} = {c}_{1}{\mu }_{1} + {c}_{2}{\mu }_{2} \), since \( {X}_{i} \sim N\left( {{\mu }_{i},{\sigma }_{i}^{2}}\right) \), so that \( E{X}_{i} = {\mu }_{i}, i = 1,2 \). Also,\n\n\[ \operatorname{Var}\left( {{c}_{1}{X}_{1} + {c}_{2}{X}_... | Yes |
Example 1. Examine the r.v.’s \( X \) and \( Y \) from an independence viewpoint, if their joint p.d.f. is given by: \( {f}_{X, Y}\left( {x, y}\right) = {4xy},0 < x < 1,0 < y < 1 \) (and 0 otherwise). | Discussion. We will use part (ii) of Theorem 1 for which the marginal p.d.f.'s are needed. To this end, we have:\n\n\[ \n{f}_{X}\left( x\right) = {4x}{\int }_{0}^{1}y\mathrm{\;d}y = {2x},\;0 < x < 1; \n\]\n\n\[ \n{f}_{Y}\left( y\right) = {4y}{\int }_{0}^{1}x\mathrm{\;d}x = {2y},\;0 < y < 1. \n\]\n\nHence, for all \( 0 ... | Yes |
Example 1.1. Vandermonde Matrix. Fix a sequence of numbers \( \left\{ {{x}_{1},{x}_{2}}\right. \) , \( \left. {\ldots ,{x}_{m}}\right\} \) . If \( p \) and \( q \) are polynomials of degree \( < n \) and \( \alpha \) is a scalar, then \( p + q \) and \( {\alpha p} \) are also polynomials of degree \( < n \) . Moreover,... | In this example, it is clear that the matrix-vector product \( {Ac} \) need not be thought of as \( m \) distinct scalar summations, each giving a different linear combination of the entries of \( c \), as (1.1) might suggest. Instead, \( A \) can be viewed as a matrix of columns, each giving sampled values of a monomi... | Yes |
A simple example of a matrix-matrix product is the outer product. This is the product of an \( m \) -dimensional column vector \( u \) with an \( n \) -dimensional row vector \( v \) ; the result is an \( m \times n \) matrix of rank 1. The outer product can be written | \[ \left\lbrack \begin{array}{llll} & & & \\ u & & & \\ & & & \\ & & & \\ & & & \end{array}\right\rbrack = \left\lbrack \begin{matrix} & & & \\ & & & \\ {v}_{1}u & {v}_{2}u & \cdots & {v}_{n} \\ & & & \\ & & & \end{matrix}\right\rbrack = \left\lbrack \begin{matrix} & & & \\ & & & \\ {v}_{1}u & {v}_{2}u & \cdots & {v}_{... | Yes |
Example 1.3. As a second illustration, consider \( B = {AR} \), where \( R \) is the upper-triangular \( n \times n \) matrix with entries \( {r}_{ij} = 1 \) for \( i \leq j \) and \( {r}_{ij} = 0 \) for \( i > j \) . This product can be written\n\n\[ \n\left\\lbrack \begin{matrix} & & & & & \\ & & & & & \\ {b}_{1} & \... | The column formula (1.6) now gives\n\n\[ \n{b}_{j} = A{r}_{j} = \mathop{\\sum }\\limits_{{k = 1}}^{j}{a}_{k} \n\]\n\n(1.7)\n\nThat is, the \( j \) th column of \( B \) is the sum of the first \( j \) columns of \( A \) . The matrix \( R \) is a discrete analogue of an indefinite integral operator. | Yes |
Theorem 1.1. \( \operatorname{range}\left( A\right) \) is the space spanned by the columns of \( A \) . | Proof. By (1.2), any \( {Ax} \) is a linear combination of the columns of \( A \) . Conversely, any vector \( y \) in the space spanned by the columns of \( A \) can be written as a linear combination of the columns, \( y = \mathop{\sum }\limits_{{j = 1}}^{n}{x}_{j}{a}_{j} \) . Forming a vector \( x \) out of the coeff... | Yes |
Theorem 1.2. A matrix \( A \in {\mathbb{C}}^{m \times n} \) with \( m \geq n \) has full rank if and only if it maps no two distinct vectors to the same vector. | Proof. \( \Rightarrow \) ) If \( A \) is of full rank, its columns are linearly independent, so they form a basis for range \( \left( A\right) \) . This means that every \( b \in \operatorname{range}\left( A\right) \) has a unique linear expansion in terms of the columns of \( A \), and therefore, by (1.2), every \( b ... | Yes |
Theorem 2.1. The vectors in an orthogonal set \( S \) are linearly independent. | Proof. If the vectors in \( S \) are not independent, then some \( {v}_{k} \in S \) can be expressed as a linear combination of other members \( {v}_{1},\ldots ,{v}_{n} \in S \) ,\n\n\[ \n{v}_{k} = \mathop{\sum }\limits_{\substack{{i = 1} \\ {i \neq k} }}^{n}{c}_{i}{v}_{i} \n\]\n\nSince \( {v}_{k} \neq 0,{v}_{k}^{ * }{... | Yes |
The \( p \) -Norm of a Diagonal Matrix. Let \( D \) be the diagonal matrix\n\n\[ D = \left\lbrack \begin{array}{llll} {d}_{1} & & & \\ & {d}_{2} & & \\ & & \ddots & \\ & & & {d}_{m} \end{array}\right\rbrack . \]\n\nThen, as in the second row of Figure 3.1, the image of the 2-norm unit sphere under \( D \) is an \( m \)... | This result for the 2-norm generalizes to any \( p \) : if \( D \) is diagonal, then \( {\begin{Vmatrix}D\end{Vmatrix}}_{p} = \mathop{\max }\limits_{{1 \leq i \leq m}}\left| {d}_{i}\right| . | Yes |
Example 3.5. The 2-Norm of a Row Vector. Consider a matrix \( A \) containing a single row. This matrix can be written as \( A = {a}^{ * } \), where \( a \) is a column vector. The Cauchy-Schwarz inequality allows us to obtain the induced matrix 2-norm. For any \( x \), we have \( \parallel {Ax}{\parallel }_{2} = \left... | This bound is tight: observe that \( \parallel {Aa}{\parallel }_{2} = \parallel a{\parallel }_{2}^{2} \) . Therefore, we have \[ \parallel A{\parallel }_{2} = \mathop{\sup }\limits_{{x \neq 0}}\left\{ {\parallel {Ax}{\parallel }_{2}/\parallel x{\parallel }_{2}}\right\} = \parallel a{\parallel }_{2}. \] | Yes |
The 2-Norm of an Outer Product. More generally, consider the rank-one outer product \( A = u{v}^{ * } \), where \( u \) is an \( m \) -vector and \( v \) is an \( n \) -vector. For any \( n \) -vector \( x \), we can bound \( \parallel {Ax}{\parallel }_{2} \) as follows: | \[ \parallel {Ax}{\parallel }_{2} = \parallel u{v}^{ * }x{\parallel }_{2} = \parallel u{\parallel }_{2}\left| {{v}^{ * }x}\right| \leq \parallel u{\parallel }_{2}\parallel v{\parallel }_{2}\parallel x{\parallel }_{2}. \] (3.13) Therefore \( \parallel A{\parallel }_{2} \leq \parallel u{\parallel }_{2}\parallel v{\parall... | Yes |
Theorem 3.1. For any \( A \in {\mathbb{C}}^{m \times n} \) and unitary \( Q \in {\mathbb{C}}^{m \times m} \), we have\n\n\[ \parallel {QA}{\parallel }_{2} = \parallel A{\parallel }_{2},\;\text{ 且 }\parallel {QA}{\parallel }_{F} = \parallel A{\parallel }_{F}. \]\n | Proof. Since \( \parallel {Qx}{\parallel }_{2} = \parallel x{\parallel }_{2} \) for every \( x \), by (2.10), the invariance in the 2-norm follows from (3.6). For the Frobenius norm we may use (3.18). | No |
Theorem 5.1. The rank of \( A \) is \( r \), the number of nonzero singular values. | Proof. The rank of a diagonal matrix is equal to the number of its nonzero entries, and in the decomposition \( A = {U\sum }{V}^{ * }, U \) and \( V \) are of full rank. Therefore \( \operatorname{rank}\left( A\right) = \operatorname{rank}\left( \sum \right) = r \) . | Yes |
Theorem 5.3. \( \parallel A{\parallel }_{2} = {\sigma }_{1} \) and \( \parallel A{\parallel }_{F} = \sqrt{{\sigma }_{1}^{2} + {\sigma }_{2}^{2} + \cdots + {\sigma }_{r}^{2}} \) . | Proof. The first result was already established in the proof of Theorem 4.1: since \( A = {U\sum }{V}^{ * } \) with unitary \( U \) and \( V,\parallel A{\parallel }_{2} = \parallel \sum {\parallel }_{2} = \max \left\{ \left| {\sigma }_{j}\right| \right\} = {\sigma }_{1} \) , by Theorem 3.1. For the second, note that by... | Yes |
Theorem 5.4. The nonzero singular values of \( A \) are the square roots of the nonzero eigenvalues of \( {A}^{ * }A \) or \( A{A}^{ * } \) . (These matrices have the same nonzero eigenvalues.) | Proof. From the calculation\n\n\[ \n{A}^{ * }A = {\left( U\sum {V}^{ * }\right) }^{ * }\left( {{U\sum }{V}^{ * }}\right) = V{\sum }^{ * }{U}^{ * }{U\sum }{V}^{ * } = V\left( {{\sum }^{ * }\sum }\right) {V}^{ * }, \n\] \n\nwe see that \( {A}^{ * }A \) is similar to \( {\sum }^{ * }\sum \) and hence has the same \( n \) ... | Yes |
Theorem 5.5. If \( A = {A}^{ * } \), then the singular values of \( A \) are the absolute values of the eigenvalues of \( A \) . | Proof. As is well known (see Exercise 2.3), a hermitian matrix has a complete set of orthogonal eigenvectors, and all of the eigenvalues are real. An equivalent statement is that (5.1) holds with \( X \) equal to some unitary matrix \( Q \) and \( \Lambda \) a real diagonal matrix. But then we can write\n\n\[ A = {Q\La... | Yes |
Theorem 5.6. For \( A \in {\mathbb{C}}^{m \times m},\left| {\det \left( A\right) }\right| = \mathop{\prod }\limits_{{i = 1}}^{m}{\sigma }_{i} \) . | Proof. The determinant of a product of square matrices is the product of the determinants of the factors. Furthermore, the determinant of a unitary matrix is always 1 in absolute value; this follows from the formula \( {U}^{ * }U = I \) and the property \( \det \left( {U}^{ * }\right) = {\left( \det \left( U\right) \ri... | Yes |
Theorem 5.7. \( A \) is the sum of \( r \) rank-one matrices:\n\n\[ A = \mathop{\sum }\limits_{{j = 1}}^{r}{\sigma }_{j}{u}_{j}{v}_{j}^{ * } \] | Proof. If we write \( \sum \) as a sum of \( r \) matrices \( {\sum }_{j} \), where \( {\sum }_{j} = \operatorname{diag}\left( {0,\ldots ,0,{\sigma }_{j},0}\right. \) , \( \ldots ,0) \), then (5.3) follows from (4.3). | No |
For any \( \nu \) with \( 0 \leq \nu \leq r \), define\n\n\[ \n{A}_{\nu } = \mathop{\sum }\limits_{{j = 1}}^{\nu }{\sigma }_{j}{u}_{j}{v}_{j}^{ * }\n\]\n\n(5.4)\n\nif \( \nu = p = \min \{ m, n\} \), define \( {\sigma }_{\nu + 1} = 0 \) . Then\n\n\[ \n{\begin{Vmatrix}A - {A}_{\nu }\end{Vmatrix}}_{2} = \mathop{\inf }\lim... | Proof. Suppose there is some \( B \) with \( \operatorname{rank}\left( B\right) \leq \nu \) such that \( \parallel A - B{\parallel }_{2} < \) \( {\begin{Vmatrix}A - {A}_{\nu }\end{Vmatrix}}_{2} = {\sigma }_{\nu + 1} \) . Then there is an \( \left( {n - \nu }\right) \) -dimensional subspace \( W \subseteq {\mathbb{C}}^{... | Yes |
Theorem 6.1. A projector \( P \) is orthogonal if and only if \( P = {P}^{ * } \) . | Proof. If \( P = {P}^{ * } \), then the inner product between a vector \( {Px} \in {S}_{1} \) and a vector \( \left( {I - P}\right) y \in {S}_{2} \) is zero:\n\n\[ \n{x}^{ * }{P}^{ * }\left( {I - P}\right) y = {x}^{ * }\left( {P - {P}^{2}}\right) y = 0.\n\]\n\nThus the projector is orthogonal, providing the proof in th... | No |
Theorem 7.1. Every \( A \in {\mathbb{C}}^{m \times n}\left( {m \geq n}\right) \) has a full QR factorization, hence also a reduced QR factorization. | Proof. Suppose first that \( A \) has full rank and that we want just a reduced QR factorization. In this case, a proof of existence is provided by the Gram-Schmidt algorithm itself. By construction, this process generates orthonormal columns of \( \widehat{Q} \) and entries of \( \widehat{R} \) such that (7.4) holds. ... | Yes |
Theorem 7.2. Each \( A \in {\mathbb{C}}^{m \times n}\left( {m \geq n}\right) \) of full rank has a unique reduced \( {QR} \) factorization \( A = \widehat{Q}\widehat{R} \) with \( {r}_{jj} > 0. \) | Proof. Again, the proof is provided by the Gram-Schmidt iteration. From (7.4), the orthonormality of the columns of \( \widehat{Q} \), and the upper-triangularity of \( \widetilde{R} \), it follows that any reduced QR factorization of \( A \) must satisfy (7.6)-(7.8). By the assumption of full rank, the denominators (7... | Yes |
Theorem 8.1. Algorithms 7.1 and 8.1 require \( \sim {2m}{n}^{2} \) flops to compute a \( {QR} \) factorization of an \( m \times n \) matrix. | Theorem 8.1 can be established as follows. To be definite, consider the modified Gram-Schmidt algorithm, Algorithm 8.1. When \( m \) and \( n \) are large, the work is dominated by the operations in the innermost loop:\n\n\[ {r}_{ij} = {q}_{i}^{ * }{v}_{j} \]\n\n\[ {v}_{j} = {v}_{j} - {r}_{ij}{q}_{i}. \]\n\nThe first l... | Yes |
Suppose we are given \( m \) distinct points \( {x}_{1},\ldots ,{x}_{m} \in \mathbb{C} \) and data \( {y}_{1},\ldots ,{y}_{m} \in \mathbb{C} \) at these points. Then there exists a unique polynomial interpolant to these data in these points, that is, a polynomial of degree at most \( m - 1 \) , \[ p\left( x\right) = {c... | The relationship of the data \( \left\{ {x}_{i}\right\} ,\left\{ {y}_{i}\right\} \) to the coefficients \( \left\{ {c}_{i}\right\} \) can be expressed by the square Vandermonde system seen already in Example 1.1: \[ \left\lbrack \begin{matrix} 1 & {x}_{1} & {x}_{1}^{2} & & {x}_{1}^{m - 1} \\ 1 & {x}_{2} & {x}_{2}^{2} &... | No |
Theorem 11.1. Let \( A \in {\mathbb{C}}^{m \times n}\left( {m \geq n}\right) \) and \( b \in {\mathbb{C}}^{m} \) be given. A vector \( x \in {\mathbb{C}}^{n} \) minimizes the residual norm \( \parallel r{\parallel }_{2} = \parallel b - {Ax}{\parallel }_{2} \), thereby solving the least squares problem (11.2), if and on... | Proof. The equivalence of (11.8) and (11.10) follows from the properties of orthogonal projectors discussed in Lecture 6, and the equivalence of (11.8) and (11.9) follows from the definition of \( r \) . To show that \( y = {Pb} \) is the unique point in range \( \left( A\right) \) that minimizes \( \parallel b - y{\pa... | Yes |
Consider the trivial problem of obtaining the scalar \( x/2 \) from \( x \in \mathbb{C} \) | The Jacobian of the function \( f : x \mapsto x/2 \) is just the derivative \( J = {f}^{\prime } = 1/2 \), so by (12.6),\n\n\[ \n\kappa = \frac{\parallel J\parallel }{\parallel f\left( x\right) \parallel /\parallel x\parallel } = \frac{1/2}{\left( {x/2}\right) /x} = 1.\n\]\n\nThis problem is well-conditioned by any sta... | Yes |
Consider the problem of computing \( \sqrt{x} \) for \( x > 0 \) . | The Jacobian of \( f : x \mapsto \sqrt{x} \) is the derivative \( J = {f}^{\prime } = 1/\left( {2\sqrt{x}}\right) \), so we have\n\n\[ \kappa = \frac{\parallel J\parallel }{\parallel f\left( x\right) \parallel /\parallel x\parallel } = \frac{1/\left( {2\sqrt{x}}\right) }{\sqrt{x}/x} = \frac{1}{2}. \]\n\nAgain, this is ... | Yes |
Consider the problem of obtaining the scalar \( f\left( x\right) = {x}_{1} - {x}_{2} \) from the vector \( x = {\left( {x}_{1},{x}_{2}\right) }^{ * } \in {\mathbb{C}}^{2} \) . For simplicity, we use the \( \infty \) -norm on the data space \( {\mathbb{C}}^{2} \) . The Jacobian of \( f \) is | \[ J = \left\lbrack \begin{array}{ll} \frac{\partial f}{\partial {x}_{1}} & \frac{\partial f}{\partial {x}_{2}} \end{array}\right\rbrack = \left\lbrack \begin{array}{ll} 1 & - 1 \end{array}\right\rbrack \] with \( \parallel J{\parallel }_{\infty } = 2 \) . The condition number is thus \[ \kappa = \frac{\parallel J{\par... | Yes |
Consider the computation of \( f\left( x\right) = \tan x \) for \( x \) near \( {10}^{100} \) . In this problem, minuscule relative perturbations in \( x \) can result in arbitrarily large changes in \( \tan x \) . The result: \( \tan \left( {10}^{100}\right) \) is effectively uncomputable on most computers. | The same minuscule perturbations result in arbitrary changes in the derivative of \( \tan x \), so there is little point in trying to calculate the Jacobian other than to observe that it is not small. | Yes |
Let \( A \in {\mathbb{C}}^{m \times m} \) be nonsingular and consider the equation \( {Ax} = b \) . The problem of computing \( b \), given \( x \), has condition number | \[ \kappa = \parallel A\parallel \frac{\parallel x\parallel }{\parallel b\parallel } \leq \parallel A\parallel \begin{Vmatrix}{A}^{-1}\end{Vmatrix} \] \( \left( {12.13}\right) \) with respect to perturbations of \( x \) . The problem of computing \( x \), given \( b \), has condition number \[ \kappa = \begin{Vmatrix}{... | Yes |
Theorem 14.1. For problems \( f \) and algorithms \( \widetilde{f} \) defined on finite-dimensional spaces \( X \) and \( Y \), the properties of accuracy, stability, and backward stability all hold or fail to hold independently of the choice of norms in \( X \) and \( Y \) . | Proof. It is well known (and easily proved) that in a finite-dimensional vector space, all norms are equivalent in the sense that if \( \| \cdot \| \) and \( \| \cdot {\| }^{\prime } \) are two norms on the same space, then there exist positive constants \( {C}_{1} \) and \( {C}_{2} \) such that \( {C}_{1}\parallel x\p... | Yes |
Suppose we are given vectors \( x, y \in {\mathbb{C}}^{m} \) and wish to compute the inner product \( \alpha = {x}^{ * }y \) . The obvious algorithm is to compute the pairwise products \( {\bar{x}}_{i}{y}_{i} \) with \( \otimes \) and add them with \( \oplus \) to obtain a computed result \( \widetilde{\alpha } \) . | It can be shown that this algorithm is backward stable; this is done implicitly in Lecture 17. | No |
On the other hand, suppose we wish to compute the rank-one outer product \( A = x{y}^{ * } \) for vectors \( x \in {\mathbb{C}}^{m}, y \in {\mathbb{C}}^{n} \) . The obvious algorithm is to compute the \( {mn} \) products \( {x}_{i}{\bar{y}}_{j} \) with \( \otimes \) and collect them into a matrix \( \widetilde{A} \) . | This algorithm is stable, but it is not backward stable. The explanation is that the matrix \( \widetilde{A} \) will be most unlikely to have rank exactly 1, and thus it cannot generally be written in the form \( \left( {x + {\delta x}}\right) {\left( y + \delta y\right) }^{ * } \) . As a rule, for problems where the d... | No |
Example 15.3. Suppose we use \( \oplus \) to compute \( x + 1 \), given \( x \in \mathbb{C} : \widetilde{f}\left( x\right) = \) \( {fl}\left( x\right) \bigoplus 1 \) . This algorithm is stable but not backward stable. | The reason is that for \( x \approx 0 \), the addition \( \oplus \) will introduce absolute errors of size \( O\left( {\epsilon }_{\text{machine }}\right) \) . Relative to the size of \( x \), these are unbounded, so they cannot be interpreted as caused by small relative perturbations in the data. This example indicate... | Yes |
What is it reasonable to expect of a computer program or calculator that computes \( \sin x \) or \( \cos x \) ? | Again the answer is stability, not backward stability. For \( \cos x \), this follows from the fact that \( \cos 0 \neq 0 \), as in the previous example. For both \( \sin x \) and \( \cos x \), backward stability is also ruled out by the fact that the function has derivative equal to zero at certain points. For example... | Yes |
Theorem 15.1. Suppose a backward stable algorithm is applied to solve a problem \( f : X \rightarrow Y \) with condition number \( \kappa \) on a computer satisfying the axioms (13.5) and (13.7). Then the relative errors satisfy\n\n\[ \frac{\parallel \widetilde{f}\left( x\right) - f\left( x\right) \parallel }{\parallel... | Proof. By the definition (14.5) of backward stability, we have \( \widetilde{f}\left( x\right) = f\left( \widetilde{x}\right) \) for some \( \widetilde{x} \in X \) satisfying\n\n\[ \frac{\parallel \widetilde{x} - x\parallel }{\parallel x\parallel } = O\left( {\epsilon }_{\text{machine }}\right) \]\n\nBy the definition ... | Yes |
Theorem 16.1. Let the \( {QR} \) factorization \( A = {QR} \) of a matrix \( A \in {\mathbb{C}}^{m \times n} \) be \( \textit{computed by Householder triangularization (Algorithm 10.1) on a computer} \) satisfying the axioms (13.5) and (13.7), and let the computed factors \( \widetilde{Q} \) and \( \widetilde{R} \) be ... | As always in this book, the expression \( O\left( {\epsilon }_{\text{machine }}\right) \) in (16.3) has the precise meaning discussed in Lecture 14. The bound holds as \( {\epsilon }_{\text{machine }} \rightarrow 0 \), uniformly for all matrices \( A \) of any fixed dimensions \( m \) and \( n \), but not uniformly wit... | Yes |
Theorem 16.3. The solution \( \widetilde{x} \) computed by Algorithm 16.1 satisfies\n\n\[ \frac{\parallel \widetilde{x} - x\parallel }{\parallel x\parallel } = O\left( {\kappa \left( A\right) {\epsilon }_{\text{machine }}}\right) \] | (16.7) | No |
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