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Theorem 10.1.6 (Consistency of MLEs) Let \( {X}_{1},{X}_{2},\ldots \), be iid \( f\left( {x \mid \theta }\right) \), and let \( L(\theta |\mathbf{x}) = \mathop{\prod }\limits_{{i = 1}}^{n}f({x}_{i}|\theta )\;{be}\;{the}\;{likelihood}\;{function}.\;{Let}\;\hat{\theta }\;{denote}\;{the}\;{MLE}\;{of}\;\theta .\;{Let}\;\ta... | Proof: The proof proceeds by showing that \( \frac{1}{n}\log L\left( {\widehat{\theta } \mid \mathbf{x}}\right) \) converges almost surely to \( {\mathrm{E}}_{\theta }\left( {\log f\left( {X \mid \theta }\right) }\right) \) for every \( \theta \in \Theta \) . Under some conditions on \( f\left( {x \mid \theta }\right) ... | No |
Example 10.1.10 (Large-sample mixture variances) The hierarchical model\n\n\[ \n{Y}_{n} \mid {W}_{n} = {w}_{n} \sim \mathrm{n}\left( {0,{w}_{n} + \left( {1 - {w}_{n}}\right) {\sigma }_{n}^{2}}\right) ,\n\]\n\n\[ \n{W}_{n} \sim \operatorname{Bernoulli}\left( {p}_{n}\right)\n\]\n\ncan exhibit big discrepancies between th... | First, using Theorem 4.4.7 we have\n\n\[ \n\operatorname{Var}\left( {Y}_{n}\right) = {p}_{n} + \left( {1 - {p}_{n}}\right) {\sigma }_{n}^{2}\n\]\n\nIt then follows that the limiting variance of \( {Y}_{n} \) is finite only if \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {1 - {p}_{n}}\right) {\sigma }_{n}^{2... | Yes |
Theorem 10.1.12 (Asymptotic efficiency of MLEs) Let \( {X}_{1},{X}_{2},\ldots \), be iid \( f\left( {x \mid \theta }\right) \), let \( \widehat{\theta } \) denote the MLE of \( \theta \), and let \( \tau \left( \theta \right) \) be a continuous function of \( \theta \) . Under the regularity conditions in Miscellanea 1... | Proof: The proof of this theorem is interesting for its use of Taylor series and its exploiting of the fact that the MLE is defined as the zero of the likelihood function. We will outline the proof showing that \( \widehat{\theta } \) is asymptotically efficient; the extension to \( \tau \left( \widehat{\theta }\right)... | No |
Suppose that\n\n\[ \sqrt{n}\frac{{W}_{n} - \mu }{\sigma } \rightarrow Z\text{ in distribution,}\]\n\nwhere \( Z \sim \mathrm{n}\left( {0,1}\right) \) . | By applying Slutsky’s Theorem (Theorem 5.5.17) we conclude\n\n\[ {W}_{n} - \mu = \left( \frac{\sigma }{\sqrt{n}}\right) \left( {\sqrt{n}\frac{{W}_{n} - \mu }{\sigma }}\right) \rightarrow \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( \frac{\sigma }{\sqrt{n}}\right) Z = 0,\]\n\nso \( {W}_{n} - \mu \rightarrow 0 \... | Yes |
Suppose now that we want to estimate the variance of the Bernoulli distribution, \( p\left( {1 - p}\right) \). The MLE of this variance is given by \( \widehat{p}\left( {1 - \widehat{p}}\right) \), and an estimate of the variance of this estimator can be obtained by applying the approximation of (10.1.7). | We have\n\n\[ \widehat{\operatorname{Var}}\left( {\widehat{p}\left( {1 - \widehat{p}}\right) }\right) = \frac{{\left. {\left\lbrack \frac{\partial }{\partial p}\left( p\left( 1 - p\right) \right) \right\rbrack }^{2}\right| }_{p = \widehat{p}}}{-{\left. \frac{{\partial }^{2}}{\partial {p}^{2}}\log L\left( p \mid \mathbf... | Yes |
Suppose that \( {X}_{1},{X}_{2},\ldots \) are iid Poisson \( \left( \lambda \right) \), and we are interested in estimating the 0 probability. For example, the number of customers that come into a bank in a given time period is sometimes modeled as a Poisson random variable, and the 0 probability is the probability tha... | The \( {Y}_{i} \) s are Bernoulli \( \left( {e}^{-\lambda }\right) \), and hence it follows that\n\n\[ \mathrm{E}\left( \widehat{\tau }\right) = {e}^{-\lambda }\;\text{ and }\;\operatorname{Var}\left( \widehat{\tau }\right) = \frac{{e}^{-\lambda }\left( {1 - {e}^{-\lambda }}\right) }{n}. \]\n\nAlternatively, the MLE of... | Yes |
Example 10.2.1 (Robustness of the sample mean) Let \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \) . We know that \( \bar{X} \) has variance \( \operatorname{Var}\left( \bar{X}\right) = {\sigma }^{2}/n \), which is the Cramér-Rao Lower Bound. Hence, \( \bar{X} \) satisfie... | To investigate (2), the performance of \( \bar{X} \) under small deviations from the model, we first need to decide on what this means. A common interpretation is to use an \( \delta \) -contamination model; that is, for small \( \delta \), assume that we observe\n\n\[ \n{X}_{i} \sim \left\{ \begin{array}{ll} \mathrm{n... | No |
To find the limiting distribution of the median, we resort to an argument similar to that in the proof of Theorems 5.4.3 and 5.4.4, that is, an argument based on the binomial distribution.\n\nLet \( {X}_{1},\ldots ,{X}_{n} \) be a sample from a population with pdf \( f \) and cdf \( F \) (assumed to be differentiable),... | Some algebra then yields\n\n\[ P\left( {\sqrt{n}\left( {{M}_{n} - \mu }\right) \leq a}\right) = P\left( {\frac{\mathop{\sum }\limits_{i}{Y}_{i} - n{p}_{n}}{\sqrt{n{p}_{n}\left( {1 - {p}_{n}}\right) }} \geq \frac{\left( {n + 1}\right) /2 - n{p}_{n}}{\sqrt{n{p}_{n}\left( {1 - {p}_{n}}\right) }}}\right) .\n\nNow \( {p}_{n... | No |
Example 10.2.4 (AREs of the median to the mean) As there are simple expressions for the asymptotic variances of the mean and the median, the ARE is easily computed. The following table gives the AREs for three symmetric distributions. We find, as might be expected, that as the tails of the distribution get heavier, the... | \[ \begin{matrix} \text{ Median/mean asymptotic relative efficiencies } \\ \frac{\text{ Normal }}{.64}\;\frac{\text{ Logistic }}{.82}\;\frac{\text{ Double exponential }}{2} \end{matrix} \] | No |
Huber estimator asymptotic relative efficiencies, \( k = {1.5} \) | <table><thead><tr><th></th><th>Normal</th><th>Logistic</th><th>Double exponential</th></tr></thead><tr><td>vs. mean</td><td>.96</td><td>1.08</td><td>1.37</td></tr><tr><td>vs. median</td><td>1.51</td><td>1.31</td><td>.68</td></tr></table> | Yes |
Theorem 10.3.1 (Asymptotic distribution of the LRT—simple \( {H}_{0} \) ) For testing \( {H}_{0} : \theta = {\theta }_{0} \) versus \( {H}_{1} : \theta \neq {\theta }_{0} \), suppose \( {X}_{1},\ldots ,{X}_{n} \) are iid \( f\left( {x \mid \theta }\right) ,\widehat{\theta } \) is the MLE of \( \theta \), and \( f\left(... | Proof: First expand \( \log L\left( {\theta \mid \mathbf{x}}\right) = l\left( {\theta \mid \mathbf{x}}\right) \) in a Taylor series around \( \widehat{\theta } \), giving\n\n\[ l\left( {\theta \mid \mathbf{x}}\right) = l\left( {\widehat{\theta } \mid \mathbf{x}}\right) + {l}^{\prime }\left( {\widehat{\theta } \mid \mat... | Yes |
Example 10.3.2 (Poisson LRT) For testing \( {H}_{0} : \lambda = {\lambda }_{0} \) versus \( {H}_{1} : \lambda \neq {\lambda }_{0} \) based on observing \( {X}_{1},\ldots ,{X}_{n} \) iid Poisson \( \left( \lambda \right) \), we have | \[ - 2\log \lambda \left( \mathbf{x}\right) = - 2\log \left( \frac{{e}^{-n{\lambda }_{0}}{\lambda }_{0}^{\sum {x}_{i}}}{{e}^{-n\widehat{\lambda }}{\widehat{\lambda }}^{\sum {x}_{i}}}\right) = {2n}\left\lbrack {\left( {{\lambda }_{0} - \widehat{\lambda }}\right) - \widehat{\lambda }\log \left( {{\lambda }_{0}/\widehat{\... | Yes |
Theorem 10.3.3 Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a pdf or pmf \( f\left( {x|\theta }\right) \) . Under the regularity conditions in Miscellanea 10.6.2, if \( \theta \in {\Theta }_{0} \) , then the distribution of the statistic \( - 2\log \lambda \left( \mathbf{X}\right) \) converges to a chi squ... | Rejection of \( {H}_{0} : \theta \in {\Theta }_{0} \) for small values of \( \lambda \left( \mathbf{X}\right) \) is equivalent to rejection for large values of \( - 2\log \lambda \left( \mathbf{X}\right) \) . Thus,\n\n\[ \n{H}_{0}\text{is rejected if and only if} - 2\log \lambda \left( \mathbf{X}\right) \geq {\chi }_{\... | Yes |
Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a Bernoulli \( \left( p\right) \) population. Consider testing \( {H}_{0} : p \leq {p}_{0} \) versus \( {H}_{1} \) : \( p > {p}_{0} \), where \( 0 < {p}_{0} < 1 \) is a specified value. The MLE of \( p \), based on a sample of size \( n \), is \( {\widehat{p}}_{... | \[ \frac{{\widehat{p}}_{n} - p}{\sqrt{\frac{{\widehat{p}}_{n}\left( {1 - {\widehat{p}}_{n}}\right) }{n}}} \rightarrow \mathrm{n}\left( {0,1}\right) \] The Wald test statistic \( {Z}_{n} \) is defined by replacing \( p \) by \( {p}_{0} \), and the large-sample Wald test rejects \( {H}_{0} \) if \( {Z}_{n} > {z}_{ \bulle... | No |
Example 10.3.7 (Tests based on the Huber estimator) If \( {X}_{1},\ldots ,{X}_{n} \) are iid from a pdf \( f\left( {x - \theta }\right) \), where \( f \) is symmetric around 0, then for the Huber M-estimator using the \( \rho \) function in (10.2.2) and the \( \psi \) function (10.2.7), we have an asymptotic variance | (10.3.7)\n\n\[ \frac{{\int }_{-k}^{k}{x}^{2}f\left( x\right) {dx} + {k}^{2}{P}_{0}\left( {\left| X\right| > k}\right) }{{\left\lbrack {P}_{0}\left( \left| X\right| \leq k\right) \right\rbrack }^{2}}. \]\n\nTherefore, based on the asymptotic normality of the M-estimator, we can (for example) test \( {H}_{0} : \theta = {... | Yes |
Example 10.4.3 (Binomial LRT interval) For \( Y = \mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i} \), where each \( {X}_{i} \) is an independent Bernoulli \( \left( p\right) \) random variable, we have the approximate \( 1 - \alpha \) confidence set | \[ \left\{ {p : - 2\log \left( \frac{{p}^{y}{\left( 1 - p\right) }^{n - y}}{{\widehat{p}}^{y}{\left( 1 - \widehat{p}\right) }^{n - y}}\right) \leq {\chi }_{1,\alpha }^{2}}\right\} . \] | Yes |
If \( {X}_{1},\ldots ,{X}_{n} \) are iid with mean \( \mu \) and variance \( {\sigma }^{2} \), then, from the Central Limit Theorem, | \[ \frac{\bar{X} - \mu }{\sigma /\sqrt{n}} \rightarrow \mathrm{n}\left( {0,1}\right) \] Moreover, from Slutsky’s Theorem, if \( {S}^{2} \rightarrow {\sigma }^{2} \) in probability, then \[ \frac{\bar{X} - \mu }{S/\sqrt{n}} \rightarrow \mathrm{n}\left( {0,1}\right) \] giving the approximate \( 1 - \alpha \) confidence i... | Yes |
Example 10.4.7 (Comparison of binomial intervals) For \( Y = \mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i},{X}_{1} \) , \( \ldots ,{X}_{n} \) iid from a Bernoulli \( \left( p\right) \) population, the Wald interval is | \[ \widehat{p} - {z}_{\bullet /2}\sqrt{\frac{\widehat{p}\left( {1 - \widehat{p}}\right) }{n}} \leq p \leq \widehat{p} + {z}_{\alpha /2}\sqrt{\frac{\widehat{p}\left( {1 - \widehat{p}}\right) }{n}}, \] | Yes |
Example 10.4.8 (Intervals based on the Huber estimator) In a development similar to Example 10.3.7, we can form asymptotic confidence intervals based on the Huber M-estimator. If \( {X}_{1},\ldots ,{X}_{n} \) are iid from a pdf \( f\left( {x - \theta }\right) \), where \( f \) is symmetric around 0, we have the approxi... | \[ {\widehat{\theta }}_{M} \pm {z}_{\alpha /2}\sqrt{\frac{\operatorname{Var}\left( {\widehat{\theta }}_{M}\right) }{n}} \] where \( \operatorname{Var}\left( {\widehat{\theta }}_{M}\right) \) is given by (10.3.7). Now we replace \( \operatorname{Var}\left( {\widehat{\theta }}_{M}\right) \) by the estimates (10.3.8) and ... | Yes |
Let \( {X}_{1},\ldots ,{X}_{n} \) be iid negative binomial \( \left( {r, p}\right) \) . We assume that \( r \) is known and we are interested in a confidence interval for \( p \) . Using the fact that \( Y = \sum {X}_{i} \sim \) negative binomial \( \left( {{nr}, p}\right) \), we can form intervals in a number of ways.... | In Exercise 2.38 it is established that, as \( p \rightarrow 0 \) ,\n\n\[ \n{2pY} \rightarrow {\chi }_{2nr}^{2}\;\text{in distribution.} \n\]\n\nSo, for small \( p,{2pY} \) is a pivot! Using this fact, we can construct a pivotal \( 1 - \alpha \) confidence interval, valid for small \( p \) :\n\n\[ \n\left\{ {p : \frac{... | No |
Theorem 11.2.5 Let \( \theta = \left( {{\theta }_{1},\ldots ,{\theta }_{k}}\right) \) be arbitrary parameters. Then\n\n\[ \n{\theta }_{1} = {\theta }_{2} = \cdots = {\theta }_{k} \Leftrightarrow \mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i}{\theta }_{i} = 0\;\text{ for all }\mathbf{a} \in \mathcal{A}, \n\]\n\nwhere \( \ma... | Proof: If \( {\theta }_{1} = \cdots = {\theta }_{k} = \theta \), then\n\n\[ \n\mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i}{\theta }_{i} = \mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i}\theta = \theta \mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i} = 0,\;\text{ (because a satisfies }\sum {a}_{i} = 0\text{ ) } \n\]\n\nproving one i... | Yes |
Example 11.2.6 (ANOVA contrasts) Special values of a will give particular tests or confidence intervals. For example, to compare treatments 1 and 2, take a \( = \) \( \left( {1, - 1,0,\ldots ,0}\right) \) . Then, using (11.2.6), to test \( {H}_{0} : {\theta }_{1} = {\theta }_{2} \) versus \( {H}_{1} : {\theta }_{1} \ne... | \[ \left| \frac{{\bar{Y}}_{1 \cdot } - {\bar{Y}}_{2 \cdot }}{\sqrt{{S}_{p}^{2}\left( {\frac{1}{{n}_{1}} + \frac{1}{{n}_{2}}}\right) }}\right| > {t}_{N - k,\alpha /2} \] | Yes |
Theorem 11.2.8 For \( {T}_{a} \) defined in expression (11.2.9), | Proof: To prove (11.2.12), use Lemma 11.2.7 and identify \( {v}_{i} \) with \( {\bar{U}}_{i} \) and \( {c}_{i} \) with \( {n}_{i} \) . The result is immediate. | No |
Theorem 11.2.10 Under the ANOVA assumptions, if \( \mathbf{M} = \sqrt{\left( {k - 1}\right) {F}_{k - 1, N - k,\alpha }}, \) then the probability is \( 1 - \alpha \) that\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i}{\bar{Y}}_{i \cdot } - \mathbf{M}\sqrt{{S}_{p}^{2}\mathop{\sum }\limits_{{i = 1}}^{k}\frac{{a}_{i}^{2}... | Proof: The simultaneous probability statement requires \( \mathbf{M} \) to satisfy\n\n\[ P\left( {\left| {\mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i}{\bar{Y}}_{i \cdot } - \mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i}{\theta }_{i}}\right| \leq \mathbf{M}\sqrt{{S}_{p}^{2}\mathop{\sum }\limits_{{i = 1}}^{k}\frac{{a}_{i}^{2}}... | Yes |
Theorem 11.2.11 For any numbers \( {y}_{ij}, i = 1,\ldots, k \), and \( j = 1,\ldots ,{n}_{i} \) ,\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{k}\mathop{\sum }\limits_{{j = 1}}^{{n}_{i}}{\left( {y}_{ij} - \overline{\bar{y}}\right) }^{2} = \mathop{\sum }\limits_{{i = 1}}^{k}{n}_{i}{\left( {\bar{y}}_{i \cdot } - \overline{\ba... | Proof: The proof is quite simple and relies only on the fact that, when we are dealing with means, the cross-term often disappears. Write\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{k}\mathop{\sum }\limits_{{j = 1}}^{{n}_{i}}{\left( {y}_{ij} - \overline{\bar{y}}\right) }^{2} = \mathop{\sum }\limits_{{i = 1}}^{k}\mathop{\sum... | No |
Lemma 11.3.2 Let \( {Y}_{1},\ldots ,{Y}_{n} \) be uncorrelated random variables with Var \( {Y}_{i} = {\sigma }^{2} \) for all \( i = 1,\ldots, n \) . Let \( {c}_{1},\ldots ,{c}_{n} \) and \( {d}_{1},\ldots ,{d}_{n} \) be two sets of constants. Then\n\n\[ \operatorname{Cov}\left( {\mathop{\sum }\limits_{{i = 1}}^{n}{c}... | Proof: This type of result has been encountered before. It is similar to Lemma 5.3.3 and Exercise 11.11. However, here we do not need either normality or independence of \( {Y}_{1},\ldots ,{Y}_{n} \) . | No |
The regression ANOVA for the grape crop yield data follows. | <table><thead><tr><th>Source of variation</th><th>Degrees of freedom</th><th>Sum of squares</th><th>Mean square</th><th>\\( F \\) statistic</th></tr></thead><tr><td>Regression</td><td>1</td><td>6.66</td><td>6.66</td><td>50.23</td></tr><tr><td>Residual</td><td>10</td><td>1.33</td><td>.133</td><td></td></tr><tr><td>Total... | No |
Example 12.3.1 (Challenger data) A by now infamous data set is that of space shuttle O-ring failures, which have been linked to temperature. The data in Table 12.3.1 give the temperatures at takeoff and whether or not an O-ring failed. | Solving (12.3.6) and (12.3.7) using \( F\left( {\alpha + \beta {x}_{i}}\right) = {e}^{\alpha + \beta {x}_{i}}/\left( {1 + {e}^{\alpha + \beta {x}_{i}}}\right) \) yields MLEs \( \widehat{\alpha } = {15.043} \) and \( \widehat{\beta } = - {.232} \) . Figure 12.3.1 shows the fitted curve along with the data. | Yes |
The estimated information matrix of the estimates from the Challenger data is given by\n\n\[\n\begin{array}{l} I\left( {\widehat{\alpha },\widehat{\beta }}\right) = \left( \begin{matrix} \mathop{\sum }\limits_{{j = 1}}^{J}{\widehat{F}}_{j}\left( {1 - {\widehat{F}}_{j}}\right) & \mathop{\sum }\limits_{{j = 1}}^{J}{x}_{j... | The likelihood asymptotics tell use that, for example, \( \widehat{\beta } \pm {z}_{\alpha /2}\operatorname{se}\left( \widehat{\beta }\right) \) is, for large samples, an approximate \( {100}\left( {1 - \alpha }\right) \% \) confidence interval for \( \beta \) . So for the Challenger data we have a \( {95}\% \) confide... | Yes |
Example 12.4.2 (Catastrophic observations) McPherson (1990) describes an experiment in which the levels of carbon dioxide \( \left( {\mathrm{{CO}}}_{2}\right) \) and oxygen \( \left( {\mathrm{O}}_{2}\right) \) were measured in the pouches of 24 potoroos (a marsupial). Interest is in the regression of \( {\mathrm{{CO}}}... | \[ \text{least squares:}y = {18.67} - {.89x}\text{,} \] \[ \text{least absolute deviation:}y = {18.59} - {.89x}\text{.} \] However, an aberrant observation can upset least squares. When entering the data the \( {\mathrm{O}}_{2} \) value of 18 on Animal 15 was incorrectly entered as 10 (we really did this). For this new... | Yes |
Example 12.4.3 (Asymptotic normality of the LAD estimator) We adapt the argument leading to (10.2.6) to derive the asymptotic distribution of the LAD estimator. Also, to simplify things, we consider only the model\n\n\\[ \n{Y}_{i} = \\beta {x}_{i} + {\\varepsilon }_{i} \n\\]\n\nthat is, we set \\( \\alpha = 0 \\) . (Th... | First look at the numerator. As \\( {E}_{\\beta }\\psi \\left( {{Y}_{i} - {\\widehat{\\beta }}_{L}{x}_{i}}\\right) = 0 \\) and \\( \\operatorname{Var}\\psi \\left( {{Y}_{i} - {\\widehat{\\beta }}_{L}{x}_{i}}\\right) = {x}_{i}^{2} \\), it follows that\n\n(12.4.3)\n\n\\[ \n\\frac{-1}{\\sqrt{n}}\\mathop{\\sum }\\limits_{{... | Yes |
Theorem 1. Suppose r.v.’s \( {X}_{1},\ldots ,{X}_{n} \) are independent and \( {X}_{k} \sim N\left( {{a}_{k},{\sigma }_{k}^{2}}\right) \) , \( k = 1,2,\ldots, n \) . Let \( {c}_{1},{c}_{2},\ldots ,{c}_{n} \) be constants and let \( T = \mathop{\sum }\limits_{{k = 1}}^{n}{c}_{k}{X}_{k} \), then \( T \sim N\left( {\mu ,{... | Proof. Idea: characteristic function (Compute the mean and variance first)\n\nSince \( {X}_{k} \sim N\left( {{a}_{k},{\sigma }_{k}^{2}}\right) \), its characteristic function (c.f.) is\n\n\[ \n{\varphi }_{k}\left( t\right) = E\left( {e}^{{it}{X}_{k}}\right) = {e}^{i{a}_{k}t - \frac{1}{2}{t}^{2}{\sigma }_{k}^{2}}. \n\]\... | Yes |
Corollary 1. In Theorem 1, if \( {a}_{1} = \cdots = {a}_{n} = a,{\sigma }_{1}^{2} = \cdots = {\sigma }_{n}^{2} \), then | \[ T \sim N\left( {a\mathop{\sum }\limits_{{k = 1}}^{n}{c}_{k},{\sigma }^{2}\mathop{\sum }\limits_{{k = 1}}^{n}{c}_{k}^{2}}\right) . \] | Yes |
Theorem 2. Suppose r.v.’s \( {X}_{1},\cdots ,{X}_{n} \) i.i.d. \( \sim N\left( {a,{\sigma }^{2}}\right) \) . 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 \( n \times n \) co... | Proof. (1) By equation (1), \( {Y}_{i} = \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{ik}{X}_{k} \), then by Corollary 1,\n\n\[ {Y}_{i} \sim N\left( {a\mathop{\sum }\limits_{{k = 1}}^{n}{a}_{ik},{\sigma }^{2}\mathop{\sum }\limits_{{k = 1}}^{n}{a}_{ik}^{2}}\right) .\n\nTherefore, \( E\left( {Y}_{i}\right) = a\mathop{\sum }\l... | Yes |
Theorem 3. Suppose r.v.’s \( {X}_{1},\cdots ,{X}_{n}\;i.i.d. \sim N\left( {a,{\sigma }^{2}}\right) . \) Let \( \bar{X} = \frac{1}{n}\mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i} \) and \( {S}^{2} = \frac{1}{n - 1}\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {X}_{i} - \bar{X}\right) }^{2} \) be the sample mean and sample var... | ## Proof of Theorem 3 Proof. (1) holds by Corollary 2. (2) [Writing \( {S}^{2} \) as sum of squares of \( n - 1 \) independent Normal r.v.’s by appropriate linear transformation] Let \[ \mathbf{A} = \left( \begin{matrix} \frac{1}{\sqrt{n}} & \frac{1}{\sqrt{n}} & \cdots & \frac{1}{\sqrt{n}} \\ {a}_{21} & {a}_{22} & \cdo... | Yes |
Let \( {X}_{1},\ldots ,{X}_{n} \) be i.i.d. r.v.’s distributed as \( N\left( {\mu ,{\sigma }^{2}}\right) \). Then the joint p.d.f. of the order statistics \( {Y}_{1},\ldots ,{Y}_{n} \) is given by | \[ g\left( {{y}_{1},\ldots ,{y}_{n}}\right) = n!{\left( \frac{1}{\sqrt{2\pi }\sigma }\right) }^{n}\exp \left\lbrack {-\frac{1}{2{\sigma }^{2}}\mathop{\sum }\limits_{{j = 1}}^{n}{\left( {y}_{j} - \mu \right) }^{2}}\right\rbrack ,\] if \( - \infty < {y}_{1} < \cdots < {y}_{n} < \infty \), and zero otherwise. | Yes |
Let \( \mathbf{X} = \left( {{X}_{1},\cdots ,{X}_{n}}\right) \) is a random sample from \( N\left( {\mu ,{\sigma }^{2}}\right) \), 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 \( \theta = \left( {\mu ,{\sigma }^{2}}\right) , \) th... | Yes |
Example 7. Let \( \mathbf{X} = \left( {{X}_{1},\cdots ,{X}_{n}}\right) \) is a random sample from Gamma distribution \( \Gamma \left( {\gamma ,\lambda }\right) \), then the sample distribution family belongs to exponential family. | Discussion. The joint p.d.f of \( \mathbf{X} \) is\n\n\[ f\left( {\mathbf{x};\gamma ,\lambda }\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\left\lbrack {\frac{{\lambda }^{\gamma }}{\Gamma \left( \gamma \right) }{x}_{i}^{\gamma - 1}\exp \left\{ {-\lambda {x}_{i}}\right\} {I}_{0,\infty }\left( {x}_{i}\right) }\right\rbr... | Yes |
Example 8. Binomial distribution family \( \{ B\left( {n,\theta }\right) ,0 < \theta < 1\} \) 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 9. Poisson distribution family \( \{ P\left( \theta \right) ,\theta > 0\} \) belongs to exponential family. | Discussion. Let \( X \sim P\left( \theta \right) \), then its p.d.f. is\n\n\[ f\left( {x;\theta }\right) = P\left( {X = x}\right) = \frac{{e}^{-\theta }{\theta }^{x}}{x!} = {e}^{-\theta }\exp \{ x\log \theta \} \frac{1}{x!} \]\n\n\[ = C\left( \theta \right) \exp \left\{ {{Q}_{1}\left( \theta \right) {T}_{1}\left( x\rig... | Yes |
Theorem 7. Let \( \mathbf{X} = \left( {{X}_{1},\cdots ,{X}_{n}}\right) \) be a random sample from an exponential family, i.e., the joint p.d.f. of \( \mathbf{X} \) is \[ f\left( {\mathbf{x};\theta }\right) = C\left( \theta \right) \exp \left\{ {\mathop{\sum }\limits_{{i = 1}}^{k}{\theta }_{i}{T}_{i}\left( \mathbf{x}\ri... | - Proof: 陈希孺. 数理统计引论. 北京: 科学出版社, 1981, 1998. (Theorem 1.6.1 on Page 80) | No |
Theorem 1. Let \( {X}_{1},\cdots ,{X}_{n} \) be a random sample from the population \( X \) with mean \( E\left( X\right) = \mu \) and variance \( \operatorname{Var}\left( X\right) = {\sigma }^{2} \) . Then, the sample mean \( \bar{X} = \frac{1}{n}\mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i} \) and the sample variance \(... | ## Proof of Theorem 1\n\nProof. (1)\n\n\[ E\left( \bar{X}\right) = E\left( {\frac{1}{n}\mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i}}\right) = \frac{1}{n}\mathop{\sum }\limits_{{i = 1}}^{n}E\left( {X}_{i}\right) = \mu \]\n\n(2)\n\n\[ E\left( {S}^{2}\right) = \frac{1}{n - 1}\left\lbrack {\mathop{\sum }\limits_{{i = 1}}^{n}... | 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)... | By Theorem 3 in topic \( 2,\;\left( {n - 1}\right) {S}^{2}/{\sigma }^{2} \sim {\chi }_{n - 1}^{2}\;\mathrm{{and}}\;\mathrm{{thus}}\;{Var}\lbrack \left( {n - 1}\right) {S}^{2}/{\sigma }^{2}\rbrack = \n\n\( 2\left( {n - 1}\right) = {\left( n - 1}\right) }^{2}\operatorname{Var}\left( {S}^{2}\right) /{\sigma }^{4}. \) | 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 \( g\left( \theta \right) = \theta \left( {1 - \theta }\right) \) . | Yes |
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 \( {\sigma }^{2} \). | Discussion. The conditions referred to above are satisfied here, and, specifically, \( {E}_{\theta }{X}_{1} = \mu ,{E}_{\theta }{X}_{1}^{2} = {\sigma }^{2} + {\mu }^{2},\theta = \left( {\mu ,{\sigma }^{2}}\right) \) . Here we need the first two sample moments, \( \bar{X} \) and \( \frac{1}{n}\mathop{\sum }\limits_{{i =... | Yes |
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. | Recall that \( {E}_{\theta }{X}_{1} = \frac{\alpha + \beta }{2} \) and \( {\sigma }_{\theta }^{2}\left( {X}_{1}\right) = \frac{{\left( \alpha - \beta \right) }^{2}}{12},\theta = \left( {\alpha ,\beta }\right) \), so that:\n\n\[ \bar{X} = \frac{\alpha + \beta }{2},\;\frac{1}{n}\mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i}^... | Yes |
Let \( \\left( {{X}_{i},{Y}_{i}}\\right), i = 1,\\cdots, n \) be a random sample from a two-dimensional population. Find the moment estimates of population covariance \( {\\sigma }_{12} = E\\left\\lbrack {(X - E\\left( X\\right) (Y - E\\left( Y\\right) }\\right. \) ] and population correlation \( \\rho = {\\sigma }_{12... | The moment estimate of \( {\\sigma }_{12} \) is the sample covariance\n\n\[ \n{m}_{12} = \\frac{1}{n}\\mathop{\\sum }\\limits_{{i = 1}}^{n}\\left( {{X}_{i} - \\bar{X}}\\right) \\left( {{Y}_{i} - \\bar{Y}}\\right) \n\]\n\nwhere \( \\bar{X} = \\mathop{\\sum }\\limits_{{i = 1}}^{n}{X}_{i}/n,\\bar{Y} = \\mathop{\\sum }\\li... | Yes |
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... | Observe that \( {0.6} = 6/{10} = \) \( t/n \) . Actually, the value \( t/n \) maximizes the likelihood function among all values of \( \theta \) with \( 0 < \theta < 1 \) . Then \( \widehat{\theta } = t/n \) will be the maximum likelihood estimate (MLE) | 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... | Since \( f\left( {{x}_{i};\theta }\right) = {\theta }^{{x}_{i}}{\left( 1 - \theta \right) }^{1 - {x}_{i}},{x}_{i} = 0 \) or \( 1, i = 1,\ldots, n \), the likelihood function is\n\n\[ L\left( {\theta \mid \mathbf{x}}\right) = \mathop{\prod }\limits_{{i = 1}}^{n}f\left( {{x}_{i};\theta }\right) = {\theta }^{t}{\left( 1 -... | Yes |
Example 6. Determine the MLE \( \widehat{\theta } = \widehat{\theta }\left( \mathbf{x}\right) \) of \( \theta \in \left( {0,\infty }\right) \) in the \( P\left( \theta \right) \) distribution in terms of the random sample \( {X}_{1},\ldots ,{X}_{n} \) with observed values \( {x}_{1},\ldots ,{x}_{n} \) . | Discussion. Here \( f\left( {{x}_{i};\theta }\right) = \frac{{\mathrm{e}}^{-\theta }{\theta }^{{x}_{i}}}{{x}_{i}!},{x}_{i} = 0,1,\ldots, i = 1,\ldots, n \), so that\n\n\[ \log L\left( {\theta \mid \mathbf{x}}\right) = \log \left( {\mathop{\prod }\limits_{{i = 1}}^{n}\frac{{\mathrm{e}}^{-\theta }{\theta }^{{x}_{i}}}{{x}... | Yes |
Example 7. Determine the MLE \( \widehat{\theta } = \widehat{\theta }\left( \mathbf{x}\right) \) of \( \theta \in \left( {0,\infty }\right) \) in the Negative Exponential distribution \( f\left( {x;\theta }\right) = \theta {\mathrm{e}}^{-{\theta x}}, x > 0 \), on the basis of the random sample \( {X}_{1},\ldots ,{X}_{n... | Discussion. Since \( f\left( {{x}_{i};\theta }\right) = \theta {\mathrm{e}}^{-\theta {x}_{i}},{x}_{i} > 0,\;i = 1,\ldots, n \), we have\n\n\( \log L\left( {\theta \mid \mathbf{x}}\right) = \log \left( {{\theta }^{n}{\mathrm{e}}^{-{n\theta }\bar{x}}}\right) = n\log \theta - n\bar{x}\theta \), so that\n\n\( \frac{\partia... | Yes |
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. | Discussion. With \( {x}_{1},\ldots ,{x}_{n} \) being the observed values of \( {X}_{1},\ldots ,{X}_{n} \), we have:\n\n(i) Let \( \mu \) be unknown. Then\n\n\[ \log L\left( {\mu \mid \mathbf{x}}\right) = \log \left\{ {\mathop{\prod }\limits_{{i = 1}}^{n}\frac{1}{\sqrt{2\pi }\sigma }\exp \left\lbrack {-\frac{1}{2{\sigma... | Yes |
Example 9. Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from the Uniform \( U\left( {\alpha ,\beta }\right) (\alpha < \) \( \beta \) ) distribution, where only one of \( \alpha \) and \( \beta \) is unknown. Determine the MLE of the (unknown) parameter. | ## Discussion.\n\n(i) Let \( \alpha \) be unknown. Since\n\n\[ f\left( {{x}_{i};\alpha }\right) = \frac{1}{\beta - \alpha }{I}_{\left\lbrack \alpha ,\beta \right\rbrack }\left( {x}_{i}\right) ,\;i = 1,\ldots, n\text{, it follows that} \]\n\n\[ L\left( {\alpha \mid \mathbf{x}}\right) = \frac{1}{{\left( \beta - \alpha \r... | Yes |
Example 10. Refer to Example 8 and suppose that both \( \mu \) and \( {\sigma }^{2} \) are unknown. Determine their MLE's. | Discussion. Here\n\n\[\n\log L\left( {\mu ,{\sigma }^{2} \mid \mathbf{x}}\right) = - \frac{n}{2}\log \left( {2\pi }\right) - \frac{n}{2}\log {\sigma }^{2} - \frac{1}{2{\sigma }^{2}}\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - \mu \right) }^{2},\n\]\n\nand then the two likelihood equations produce the unique sol... | 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... | Discussion. The number of independent parameters is \( r - 1 \), since, for example, \( {p}_{r} = 1 - {p}_{1} - \cdots - {p}_{r - 1} \) . Looking at the \( \log L\left( {{p}_{1},\ldots ,{p}_{r} \mid \mathbf{x}}\right) \) and taking partial derivatives with respect to \( {p}_{i}, i = 1,\ldots, r - 1 \) (and remembering ... | No |
Refer to Example 9, assume that both \( \alpha \) and \( \beta \) are unknown, and determine their MLE's. | Since always \( \alpha \leq {x}_{\left( 1\right) } \leq {x}_{\left( n\right) } \leq \beta \), the right-hand side of (3) is maximized if \( \left. {{I}_{\lbrack \alpha ,\infty )}\left( {x}_{\left( 1\right) }\right) = 1\text{ and }{I}_{( - \infty ,\beta \rbrack }\left( {x}_{\left( n\right) }\right) = 1\text{, which happ... | Yes |
Theorem 3 (Invariance property of MLEs). If \( \widehat{\theta } \) is the MLE of \( \theta \), then for any function \( g\left( \theta \right) \), the MLE of \( g\left( \theta \right) \) is \( g\left( \widehat{\theta }\right) \) . | Proof. If \( g\left( \cdot \right) \) is a one-to-one function defined on \( \Theta \) onto \( {\Theta }^{ * } \), then the equation \( {\theta }^{ * } = g\left( \theta \right) \) can be solved for \( \theta \). Let \( \theta = {g}^{-1}\left( {\theta }^{ * }\right) \). Then,\n\n\[ L\left( {\theta ;\mathbf{x}}\right) = ... | Yes |
Theorem 4. Let \( \\mathbf{X} = \\left( {{X}_{1},\\cdots ,{X}_{n}}\\right) \) be a random sample from the distribution \( \\{ f\\left( {x;\\theta }\\right) ,\\theta \\in \\Theta \\} \) . Let \( T = T\\left( {{X}_{1},\\cdots ,{X}_{n}}\\right) \) be a sufficient statistic of \( \\theta \) . If the MLE of \( \\theta \) ex... | Proof. By Factorization Theorem, the p.d.f. of \( \\mathbf{X} \) can be written\n\n\[ L\\left( {\\theta ;\\mathbf{x}}\\right) = \\mathop{\\prod }\\limits_{{i = 1}}^{n}f\\left( {{x}_{i};\\theta }\\right) = g\\left( {T\\left( \\mathbf{x}\\right) ,\\theta }\\right) h\\left( \\mathbf{x}\\right) . \]\n\nLet \( \\widehat{\\t... | Yes |
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 |
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, the conditional distribution of \( \mathbf{X} \) given \( T \) does not depend on \( \theta \) . Hence, the condition expectation \( E\left( {\widehat{g}\left( \mathbf{X}\right) \mid T}\right) \) does not depend on \( \theta \) . Therefore, \( h\left( T\right) \) is a statistic which... | Yes |
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}... | Discussion. By Lemma 1, the following estimate is unbiased and has smaller variance.\n\n\[ h\\left( t\\right) = E\\left( {{X}_{1} \\mid T = t}\\right) = 1 \\cdot P\\left( {{X}_{1} = 1 \\mid T = t}\\right) + 0 \\cdot P\\left( {{X}_{1} = 0 \\mid T = t}\\right) \]\n\n\[ = \\frac{P\\left( {{X}_{1} = 1, T = t}\\right) }{P\\... | Yes |
Theorem 7. Let \( \widehat{g}\left( \mathbf{X}\right) \) be an unbiased estimate of \( g\left( \theta \right) \) and suppose that \( {\operatorname{Var}}_{\theta }\left( {\widehat{g}\left( \mathbf{X}\right) }\right) < \infty ,\forall \theta \in \Theta \) . If for any statistic \( l\left( \mathbf{X}\right) \) satisfying... | Proof. For any unbiased estimate of \( g\left( \theta \right) ,{\widehat{g}}_{1}\left( \mathbf{X}\right) \), let \( l\left( \mathbf{X}\right) = {\widehat{g}}_{1}\left( \mathbf{X}\right) - \widehat{g}\left( \mathbf{X}\right) \), then \( l\left( \mathbf{X}\right) \) is an unbiased estimate of 0 . Hence, \( {\operatorname... | Yes |
Corollary 1. Let \( T = T\left( \mathbf{X}\right) \) be a sufficient statistic of \( \theta \) and \( h\left( {T\left( \mathbf{X}\right) }\right) \) be an unbiased estimate of \( g\left( \theta \right) \) and suppose that \( {\operatorname{Var}}_{\theta }\left( {h\left( T\right) }\right) < \infty ,\forall \theta \in \T... | \[ {\operatorname{Cov}}_{\theta }\left( {h\left( T\right) ,\delta \left( T\right) }\right) = {E}_{\theta }\left\lbrack {h\left( T\right) \cdot \delta \left( T\right) }\right\rbrack = 0,\forall \theta \in \Theta ,\] then \( h\left( T\right) \) is a UMVUE of \( g\left( \theta \right) \) . | No |
Example 10 (cont’). Let \( \mathbf{X} = \left( {{X}_{1},\cdots ,{X}_{n}}\right) \) be a random sample from \( N\left( {\mu ,{\sigma }^{2}}\right) \) . Determine the UMVUE of \( \mu \) and \( {\sigma }^{2} \) . Denote \( \theta = \left( {\mu ,{\sigma }^{2}}\right) \) . | Discussion. We have shown that \( T = \left( {{T}_{1},{T}_{2}}\right) \) is sufficient for \( \left( {\mu ,{\sigma }^{2}}\right) \), where \( {T}_{1} = \bar{X},{T}_{2} = \mathop{\sum }\limits_{{i = 1}}^{n}{\left( {X}_{i} - \bar{X}\right) }^{2}.{T}_{1} \) and \( {T}_{2} \) are independent and \( {T}_{1} \sim N\left( {\m... | Yes |
Theorem 8 (Lehmann-Scheffé Theorem). Let \( \\mathbf{X} = \\left( {{X}_{1},\\cdots ,{X}_{n}}\\right) \) be a random sample from a population with p.d.f. \( f\\left( {x;\\theta }\\right) ,\\theta \\in \\Theta \) . Let \( T\\left( \\mathbf{X}\\right) \) be a sufficient and complete statistic of \( g\\left( \\theta \\righ... | Proof. (1) Uniqueness of unbiased estimator based on \( T \) . Let \( {\\widehat{g}}_{1}\\left( {T\\left( \\mathbf{X}\\right) }\\right) \) be any unbiased estimator of \( g\\left( \\theta \\right) \) . Let \( \\delta \\left( T\\right) = \\widehat{g}\\left( {T\\left( \\mathbf{X}\\right) }\\right) - {\\widehat{g}}_{1}\\l... | Yes |
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... | (1) \( \\bar{X} = T/n \) is an unbiased estimator of \( \\theta \), thus it is the UMVUE of \( \\theta \). (2) Let \( \\delta \\left( T\\right) \) be an unbiased estimator of \( {\\theta }^{r} \), then, \[ \\mathop{\\sum }\\limits_{{t = 0}}^{\\infty }\\delta \\left( t\\right) \\frac{{\\left( n\\theta \\right) }^{t}}{t!... | Yes |
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 (1) \( \mu \) and \( {\sigma }^{2},\left( 2\right) {\sigma }^{r}, r > 0... | Discussion. \( \left( {{T}_{1},{T}_{2}}\right) = \left( {\mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i},\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {X}_{i} - \bar{X}\right) }^{2}}\right) \) is sufficient and complete statistics for \( \left( {\mu ,{\sigma }^{2}}\right) \) . (1) \( \bar{X} = {T}_{1}/n \) is the UMVUE of \( \... | Yes |
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. The p.d.f. of \( {X}_{1},\cdots ,{X}_{n} \) is \( f\left( {\mathbf{x};\theta }\right) = \mathop{\prod }\limits_{{i = 1}}^{n}f\left( {{x}_{i};\theta }\right) \) . Let\n\n\[ \nS\left( {\mathbf{x};\theta }\right) = \frac{\partial \log f\left( {\mathbf{x};\theta }\right) }{\partial \theta } = \mathop{\sum }\limits_{... | Yes |
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. The p.d.f. of \( B\\left( {1,\\theta }\\right) \) is \( f\\left( {x;\\theta }\\right) = {\\theta }^{x}{\\left( 1 - \\theta \\right) }^{1 - x}, x = 0,1,0 < \\theta < 1 \) . Since Bernoulli distribution belongs to exponential family, the regularity conditions in Theorem 9 hold. The Fisher information is\n\n\[ \nI\... | Yes |
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. The p.d.f. of \( P\\left( \\theta \\right) \) is \( f\\left( {x;\\theta }\\right) = {e}^{-\\theta }{\\theta }^{x}/x!, x = 0,1,\\cdots ,\\theta > 0 \) . Since Poisson distribution belongs to exponential family, the regularity conditions in Theorem 9 hold. The Fisher information is\n\n\[ \nI\\left( \\theta \\right... | Yes |
Example 21. 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. The p.d.f. of \( N\\left( {\\mu ,{\\sigma }^{2}}\\right) \) is\n\n\[ f\\left( {x;\\mu ,{\\sigma }^{2}}\\right) = \\frac{1}{\\sqrt{2\\pi }\\sigma }\\exp \\left\\{ {-\\frac{{\\left( x - \\mu \\right) }^{2}}{2{\\sigma }^{2}}}\\right\\} .\n\]\n\nSince normal distribution belongs to exponential family, the regularity... | Yes |
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... | The p.d.f. of \( N\left( {\mu ,{\sigma }^{2}}\right) \) is\n\n\[ f\left( {x;\theta }\right) = \frac{1}{\sqrt{{2\pi }{\theta }_{2}}}\exp \left\{ {-\frac{{\left( x - {\theta }_{1}\right) }^{2}}{2{\theta }_{2}}}\right\} .\n\]\n\nIt is easy to obtain\n\n\[ \frac{\partial \log f\left( {x;\theta }\right) }{\partial {\theta }... | Yes |
Lemma 1. Let \( {\widehat{\theta }}_{L}\left( \mathbf{X}\right) \) be a lower confidence limit for \( \theta \) with confidence level \( 1 - {\alpha }_{1} \) and let \( {\widehat{\theta }}_{U}\left( \mathbf{X}\right) \) be an upper confidence limit for \( \theta \) with confidence level \( 1 - {\alpha }_{2} \), and sup... |  | Yes |
Example 1. Let \( {X}_{1},\ldots ,{X}_{n} \) be a random interval from the \( N\left( {\mu ,{\sigma }^{2}}\right) \) distribution, where only one of \( \mu \) or \( {\sigma }^{2} \) is unknown. Construct a confidence interval for it with confidence coefficient \( 1 - \alpha \) . | ## Discussion.\n\n- Let \( \mu \) be unknown. The natural point estimate of \( \mu \) is the sample mean\n\n\( \bar{X} = \frac{1}{n}\mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i} \) whose distribution is \( \bar{X} \sim N\left( {\mu ,{\sigma }^{2}/n}\right) \) . Standardization,\n\n\[ Z = \frac{\sqrt{n}\left( {\bar{X} - \m... | Yes |
On the basis of the random sample \( {X}_{1},\ldots ,{X}_{n} \) from the \( U\left( {0,\theta }\right) \left( {\theta > 0}\right) \) distribution, construct a confidence interval for \( \theta \) with confidence coefficient \( 1 - \alpha \) . | Discussion. It has been seen (just apply Example 6 (ii) in Chapter 9 with \( \alpha = 0 \) and \( \beta = \theta \) ) that \( X = {X}_{\left( n\right) } \) is a sufficient statistic for \( \theta \) . Also, the p.d.f. of \( X \) is given by (see Example 14 in Chapter 9) \( {f}_{X}\left( {x;\theta }\right) = \frac{n}{{\... | Yes |
Example 3. Let \( \\mathbf{X} = \\left( {{X}_{1},\\cdots ,{X}_{n}}\\right) \) be a random sample from Exponential distribution \( \\operatorname{Exp}\\left( \\lambda \\right) \) with p.d.f. \( f\\left( {x;\\lambda }\\right) = \\lambda {e}^{-{\\lambda x}}{I}_{\\left( 0,\\infty \\right) }\\left( x\\right) ,\\lambda > 0 \... | Discussion. Since \( \\bar{X} = \\mathop{\\sum }\\limits_{{i = 1}}^{n}{X}_{i}/n \) is an unbiased estimate of \( 1/\\lambda \) (it is UMVUE), we will construct a confidence interval for \( \\lambda \) based on \( \\bar{X} \) . Let \( T = {2\\lambda n}\\bar{X} \), it can be shown that\n\n\[ \n{2\\lambda n}\\bar{X} = {2\... | Yes |
Example 4. Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from the \( N\left( {\mu ,{\sigma }^{2}}\right) \) distribution, where both \( \mu \) and \( {\sigma }^{2} \) are unknown. Construct confidence intervals for \( \mu \) and \( {\sigma }^{2} \) , each with confidence coefficient \( 1 - \alpha \) . | Discussion. We have that:\n\n\[ \frac{\sqrt{n}\left( {\bar{X} - \mu }\right) }{\sigma } \sim N\left( {0,1}\right) \;\text{ and }\;\frac{\left( {n - 1}\right) {S}^{2}}{{\sigma }^{2}} = \mathop{\sum }\limits_{{i = 1}}^{n}{\left( \frac{{X}_{i} - \bar{X}}{\sigma }\right) }^{2} \sim {\chi }_{n - 1}^{2}, \]\n\nwhere \( {S}^{... | Yes |
Example 6. On the basis of the random sample \( {X}_{1},\ldots ,{X}_{n} \) from the \( N\left( {\mu ,{\sigma }^{2}}\right) \) distribution, construct a confidence region for the pair \( \left( {\mu ,{\sigma }^{2}}\right) \) with confidence coefficient \( 1 - \alpha \) . | Discussion. In solving this problem, we draw heavily on what we have done in the previous example. Let \( \bar{X} \) be the sample mean and define \( {S}^{2} \) by \( {S}^{2} \equiv \frac{1}{n - 1}\mathop{\sum }\limits_{{i = 1}}^{n} \) \( {\left( {X}_{i} - \bar{X}\right) }^{2} \) . Then 因此所求的区域形状可能不标准\n\n\[ \frac{\sqrt... | Yes |
On the basis of the random sample \( {X}_{1},\ldots ,{X}_{n} \) from the \( B\left( {1,\theta }\right) \) distribution, construct a confidence interval for \( \theta \) with confidence coefficient approximately \( 1 - \alpha \) . | Recall that \( {E}_{\theta }{X}_{1} = \theta \) and \( {\sigma }_{\theta }^{2}\left( {X}_{1}\right) = \theta \left( {1 - \theta }\right) \), so that, by the CLT,\n\n\[ \frac{\sqrt{n}\left( {{\bar{X}}_{n} - \theta }\right) }{\sqrt{\theta \left( {1 - \theta }\right) }} \simeq N\left( {0,1}\right) \]\n\n(17)\n\nLet \( T =... | Yes |
Construct a confidence interval for \( \theta \) with confidence coefficient approximately \( 1 - \alpha \) on the basis of the random sample \( {X}_{1},\ldots ,{X}_{n} \) from the \( P\left( \theta \right) \) distribution. | Here \( {E}_{\theta }{X}_{1} = {\sigma }_{\theta }^{2}\left( {X}_{1}\right) = \theta \), so that, working as in the previous example, and employing Theorem 6(iii) in Chapter 7, we have\n\n反解出 \( \theta \) 比较复杂, \( \frac{\sqrt{n}\left( {{\bar{X}}_{n} - \theta }\right) }{\sqrt{\theta }} \simeq N\left( {0,1}\right) ,\; \)... | No |
Example 9. Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from the Negative Exponential distribution in the following parameterization: \( f\left( {x;\theta }\right) = \frac{1}{\theta }{e}^{-x/\theta }, x > 0 \) . Construct a confidence interval for \( \theta \) with confidence coefficient approximately \( 1 - \a... | Discussion. In the adopted parameterization above, \( {E}_{\theta }{X}_{1} = \theta \) and \( {\sigma }_{\theta }^{2}\left( {X}_{1}\right) = {\theta }^{2} \) . Then working as in the previous example, we have that\n\n\[ \frac{\sqrt{n}\left( {{\bar{X}}_{n} - \theta }\right) }{\theta } \simeq N\left( {0,1}\right) ,\;\tex... | Yes |
For the simple case of \( \widehat{\theta } = {\bar{X}}_{n} \) . Calculate its Jackknife estimates of bias and variance. | It is easy to show that\n\n\[ \n{\widehat{\theta }}_{\left( i\right) } = \frac{1}{n - 1}\mathop{\sum }\limits_{{j \neq i}}{X}_{j} = \frac{n{\bar{X}}_{n} - {X}_{i}}{n - 1} \n\]\n\nand, therefore, \( {\widehat{\theta }}_{\left( \cdot \right) } = {\bar{X}}_{n},{\widehat{\operatorname{Bias}}}_{\left( \theta \right) } = 0 \... | Yes |
Consider the case of \( \widehat{\theta } = {\bar{X}}_{n}^{2} \) as an estimate of \( {\mu }^{2} \) . Calculate its Jackknife estimates of bias and variance. | Note that \( \operatorname{Bias}\left( \widehat{\theta }\right) = {\alpha }_{2}/n \) . Let \( {\bar{X}}_{n, i}^{2} = {\widehat{\theta }}_{\left( i\right) } \) . The Jackknife bias estimator is\n\n\[ \n{\widehat{\operatorname{Bias}}}_{\left( \theta \right) } = \frac{n - 1}{n}\mathop{\sum }\limits_{{i = 1}}^{n}\left( {{\... | Yes |
Theorem 1. 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 ... | ## Confidence Interval: Linear Regression Model\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... | Yes |
For the simple case of \( \widehat{\theta } = {\bar{X}}_{n} \) . Calculate its Jackknife estimates of bias and variance. | It is easy to show that\n\n\[ \n{\widehat{\theta }}_{\left( i\right) } = \frac{1}{n - 1}\mathop{\sum }\limits_{{j \neq i}}{X}_{j} = \frac{n{\bar{X}}_{n} - {X}_{i}}{n - 1} \n\] \n\nand, therefore, \( {\widehat{\theta }}_{\left( \cdot \right) } = {\bar{X}}_{n},{\widehat{\operatorname{Bias}}}_{\left( \theta \right) } = 0 ... | Yes |
Consider the case of \( \widehat{\theta } = {\bar{X}}_{n}^{2} \) as an estimate of \( {\mu }^{2} \) . Calculate its Jackknife estimates of bias and variance. | Note that \( \operatorname{Bias}\left( \widehat{\theta }\right) = {\alpha }^{2}/n \) . Let \( {\bar{X}}_{n, i}^{2} = {\widehat{\theta }}_{\left( i\right) } \) . The Jackknife bias estimator is\n\n\[ \n{\widehat{\operatorname{Bias}}}_{\left( \theta \right) } = \frac{n - 1}{n}\mathop{\sum }\limits_{{i = 1}}^{n}\left( {{\... | Yes |
Example 1. The rate of scrap of a product made in a manufactory is \( p \) in the long run. A store will purchase products from this manufactory if the rate of scrap is lower than 0.01. Draw a random sample of size 100 and found that 3 of them are scraps. Should the store purchase these products? | Discussion. There are two possibilities\n\n\[ \n{H}_{0} : 0 < p \leq {0.01} \leftrightarrow {H}_{1} : {0.01} < p < 1. \]\n\nThe store should determine which of \( {H}_{0},{H}_{1} \) is more possible. | No |
Whether or not the rate of unemployment is acceptable, e.g., lower that \( {\theta }_{0} = {6.25}\% \) ? | Let \( \theta \) be the true rate of unemployment. The agency needs to determine whether \( \theta > {\theta }_{0} \) or \( \theta \leq {\theta }_{0} \) . That is, to decide whether the following hypothesis \( {H}_{0} \) is acceptable or not.\n\n\[ \n{H}_{0} : \theta > {\theta }_{0} \leftrightarrow {H}_{1} : \theta \le... | No |
Example 4. Suppose that the mean \( \theta \) of a r.v. \( X \) represents the dosage of a drug which is used for the treatment of a certain disease. For this medication to be both safe and effective, \( \theta \) must satisfy the requirements \( {\theta }_{1} < \theta < {\theta }_{2} \), for two specified values \( {\... | Discussion. Of course, we have to assume a certain distribution for the r.v. \( X \), which for good reasons is taken to be \( N\left( {\theta ,{\sigma }^{2}}\right) ,\sigma \) known. | Yes |
Example 5. It is claimed that a new treatment is more effective than the standard treatment for prolonging the lives of terminal cancer patients. The standard treatment has been used for a long time, and from records in medical journals the mean survival period is known to have a certain numerical value (in years). The... | Discussion. Suppose that the survival time for a terminal cancer patient treated with the standard treatment is a r.v. \( X \sim N\left( {{\theta }_{1},{\sigma }_{1}^{2}}\right) \) . Likewise, let the r.v. \( Y \) stand for the survival time for such a patient subject to the new treatment, and let \( Y \sim N\left( {{\... | Yes |
A government agency wishes to assess the prevailing rate of unemployment in a particular city. It is felt that this assessment can be done quickly and effectively by sampling a small fraction \( n \), say, of the labor force in the city. The obvious question to be considered here is: Whether or not the rate of unemploy... | Let \( \theta \) be the true rate of unemployment. The agency needs to determine whether \( \theta > {\theta }_{0} \) or \( \theta \leq {\theta }_{0} \) . That is, to decide whether the following hypothesis \( {H}_{0} \) is acceptable or not.\n\n\[ \n{H}_{0} : \theta > {\theta }_{0} \leftrightarrow {H}_{1} : \theta \le... | No |
It is claimed that a new treatment is more effective than the standard treatment for prolonging the lives of terminal cancer patients. The standard treatment has been used for a long time, and from records in medical journals the mean survival period is known to have a certain numerical value (in years). The new treatm... | Discussion. Suppose that the survival time for a terminal cancer patient treated with the standard treatment is a r.v. \( X \sim N\left( {{\theta }_{1},{\sigma }_{1}^{2}}\right) \) . Likewise, let the r.v. \( Y \) stand for the survival time for such a patient subject to the new treatment, and let \( Y \sim N\left( {{\... | Yes |
Example 1. The rate of scrap of a product made in a manufactory is \( p \) in the long run. A store will purchase products from this manufactory if the rate of scrap is lower than 0.01. Draw a random sample of size 100 and found that 3 of them are scraps. Should the store purchase these products? | Discussion. There are two possibilities\n\n\[ \n{H}_{0} : 0 < p \leq {0.01} \leftrightarrow {H}_{1} : {0.01} < p < 1. \]\n\nThe store should determine which of \( {H}_{0},{H}_{1} \) is more possible. | No |
A government agency wishes to assess the prevailing rate of unemployment in a particular city. It is felt that this assessment can be done quickly and effectively by sampling a small fraction \( n \), say, of the labor force in the city. The obvious question to be considered here is: Whether or not the rate of unemploy... | Let \( \theta \) be the true rate of unemployment. The agency needs to determine whether \( \theta > {\theta }_{0} \) or \( \theta \leq {\theta }_{0} \) . That is, to decide whether the following hypothesis \( {H}_{0} \) is acceptable or not.\n\n\[ \n{H}_{0} : \theta > {\theta }_{0} \leftrightarrow {H}_{1} : \theta \le... | Yes |
Example 5. It is claimed that a new treatment is more effective than the standard treatment for prolonging the lives of terminal cancer patients. The standard treatment has been used for a long time, and from records in medical journals the mean survival period is known to have a certain numerical value (in years). The... | Discussion. Suppose that the survival time for a terminal cancer patient treated with the standard treatment is a r.v. \( X \sim N\left( {{\theta }_{1},{\sigma }_{1}^{2}}\right) \) . Likewise, let the r.v. \( Y \) stand for the survival time for such a patient subject to the new treatment, and let \( Y \sim N\left( {{\... | Yes |
为研究正常成年男女血液红细胞平均数的差别, 检验某地正常成年男子156人，女子74人，计算男女红细胞的平均数和样本标准差分别为\n\n男: \( \bar{X} = {465.13} \) 万 \( /{\mathrm{{mm}}}^{3},{S}_{X} = {54.80} \) 万 \( /{\mathrm{{mm}}}^{3} \)\n\n女: \( \bar{Y} = {422.16}万/{\mathrm{{mm}}}^{3},{S}_{X} = {49.20}万/{\mathrm{{mm}}}^{3} \)\n\n假定正常男女红细胞数分别服从正态分布, 且方差相同。检验正常成年人红细胞数是否与性别有关（显著水平 \( \a... | ## Numerical Example\n\nDiscussion. Let \( {X}_{1},\cdots ,{X}_{m} \) i.i.d. \( \sim N\left( {{\mu }_{1},{\sigma }_{1}^{2}}\right) ,{Y}_{1},\cdots ,{Y}_{n} \) i.i.d. \( \sim N\left( {{\mu }_{2},{\sigma }_{2}^{2}}\right) \),\n\nsuppose that \( {X}_{i} \) ’s and \( {Y}_{i} \) ’s are independent. The test problem is\n\n\[... | Yes |
假定正常男女红细胞数分别服从正态分布, 方差未必相同。 检验正常成年人红细胞数是否与性别有关（显著水平 \( \alpha = \) 0.01) | Let \( {X}_{1},\cdots ,{X}_{m} \) i.i.d. \( \sim N\left( {{\mu }_{1},{\sigma }_{1}^{2}}\right) ,{Y}_{1},\cdots ,{Y}_{n} \) i.i.d. \( \sim N\left( {{\mu }_{2},{\sigma }_{2}^{2}}\right) \) , suppose that \( {X}_{i} \) ’s and \( {Y}_{i} \) ’s are independent. The test problem is\n\n\[ \n{H}_{0} : {\mu }_{1} - {\mu }_{2} =... | Yes |
Determine the LR test for testing the hypothesis \( {H}_{0} : \theta = 0 \) (against the alternative \( {H}_{A} : \theta \neq 0 \) ) at level of significance \( \alpha \) on the basis of one observation from the p.d.f. \( f\left( {x;\theta }\right) = \frac{1}{\pi } \times \frac{1}{1 + {\left( x - \theta \right) }^{2}},... | Next, clearly, \( L\left( {\theta \mid x}\right) \left( { = f\left( {x;\theta }\right) }\right) \) is maximized for \( \theta = x \), so that \( \lambda = \frac{1}{\pi } \times \frac{1}{1 + {x}^{2}}/\frac{1}{\pi } \) \( = \frac{1}{1 + {x}^{2}} \), and \( \lambda < {\lambda }_{0} \), if and only if \( {x}^{2} > \frac{1}... | Yes |
Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample of size \( n \) from the Negative Exponential p.d.f. \( f\left( {x;\theta }\right) = \theta {e}^{-{\theta x}}, x > 0\left( {\theta > 0}\right) \) . Derive the LR test for testing the hypothesis \( {H}_{0} : \theta = {\theta }_{0} \) (against the alternative \( {H}_{A... | Here\n\n\[ L\left( {\theta \mid \mathbf{x}}\right) = {\theta }^{n}{e}^{-{\theta t}},\;\text{ where }\;\mathbf{x} = \left( {{x}_{1},\ldots ,{x}_{n}}\right) \text{ and }t = \mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i}. \]\n\nWe also know that the MLE of \( \theta \) is \( \widehat{\theta } = 1/\bar{x} = n/t \) . Therefore,... | Yes |
Theorem 1. (Neyman-Pearson Fundamental Lemma) Let \( \mathbf{X} = \left( {{X}_{1},\cdots ,{X}_{n}}\right) \) be a random sample with p.d.f. \( f\left( {\mathbf{x};\theta }\right) \), we are interested in testing the simple hypothesis:\n\n\[ \n{H}_{0} : \theta = {\theta }_{0} \leftrightarrow {H}_{1} : \theta = {\theta }... | Proof. (1) Existence. Let \( G\left( \cdot \right) \) be the d.f. of the r.v. \( f\left( {\mathbf{X};{\theta }_{1}}\right) /f\left( {\mathbf{X};{\theta }_{0}}\right) \), i.e.,\n\n\[ \nG\left( y\right) = P\left( {\frac{f\left( {\mathbf{X};{\theta }_{1}}\right) }{f\left( {\mathbf{X};{\theta }_{0}}\right) } \leq y}\right)... | Yes |
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