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Recall that we are interested in the properties of \( \frac{\widehat{p}}{1 - \widehat{p}} \) as an estimate of \( \frac{p}{1 - p} \), where \( p \) is a binomial success probability. In our above notation, take \( g\left( p\right) = \frac{p}{1 - p} \) so \( {g}^{\prime }\left( p\right) = \frac{1}{{\left( 1 - p\right) }... | \[
\operatorname{Var}\left( \frac{\widehat{p}}{1 - \widehat{p}}\right) \approx {\left\lbrack {g}^{\prime }\left( p\right) \right\rbrack }^{2}\operatorname{Var}\left( \widehat{p}\right)
\]
\[
= {\left\lbrack \frac{1}{{\left( 1 - p\right) }^{2}}\right\rbrack }^{2}\frac{p\left( {1 - p}\right) }{n} = \frac{p}{n{\left( 1 -... | Yes |
Suppose \( X \) is a random variable with \( {\mathrm{E}}_{\mu }X = \mu \neq 0 \) . If we want to estimate a function \( g\left( \mu \right) \), a first-order approximation would give us\n\n\[ g\left( X\right) = g\left( \mu \right) + {g}^{\prime }\left( \mu \right) \left( {X - \mu }\right) \] | If we use \( g\left( X\right) \) as an estimator of \( g\left( \mu \right) \), we can say that approximately\n\n\[ {\mathrm{E}}_{\mu }g\left( X\right) \approx g\left( \mu \right) \]\n\n\[ {\operatorname{Var}}_{\mu }g\left( X\right) \approx {\left\lbrack {g}^{\prime }\left( \mu \right) \right\rbrack }^{2}{\operatorname{... | Yes |
Theorem 5.5.24 (Delta Method) Let \( {Y}_{n} \) be a sequence of random variables that satisfies \( \sqrt{n}\left( {{Y}_{n} - \theta }\right) \rightarrow \mathrm{n}\left( {0,{\sigma }^{2}}\right) \) in distribution. For a given function \( g \) and a specific value of \( \theta \), suppose that \( {g}^{\prime }\left( \... | Proof: The Taylor expansion of \( g\left( {Y}_{n}\right) \) around \( {Y}_{n} = \theta \) is\n\n\[ g\left( {Y}_{n}\right) = g\left( \theta \right) + {g}^{\prime }\left( \theta \right) \left( {{Y}_{n} - \theta }\right) + \text{ Remainder,}\]\n\nwhere the remainder \( \rightarrow 0 \) as \( {Y}_{n} \rightarrow \theta \) ... | No |
Suppose now that we have the mean of a random sample \( \bar{X} \) . For \( \mu \neq 0 \), we have\n\n\[ \sqrt{n}\left( {\frac{1}{\bar{X}} - \frac{1}{\mu }}\right) \rightarrow \mathrm{n}\left( {0,{\left( \frac{1}{\mu }\right) }^{4}{\operatorname{Var}}_{\mu }{X}_{1}}\right) \] \nin distribution. | If we do not know the variance of \( {X}_{1} \), to use the above approximation requires an estimate, say \( {S}^{2} \). Moreover, there is the question of what to do with the \( 1/\mu \) term, as we also do not know \( \mu \). We can estimate everything, which gives us the approximate variance\n\n\[ \widehat{\operator... | Yes |
Example 5.5.27 (Moments of a ratio estimator) Suppose \( X \) and \( Y \) are random variables with nonzero means \( {\mu }_{X} \) and \( {\mu }_{Y} \), respectively. The parametric function to be estimated is \( g\left( {{\mu }_{X},{\mu }_{Y}}\right) = {\mu }_{X}/{\mu }_{Y} \). | It is straightforward to calculate\n\n\[ \frac{\partial }{\partial {\mu }_{X}}g\left( {{\mu }_{X},{\mu }_{Y}}\right) = \frac{1}{{\mu }_{Y}} \]\n\nand\n\n\[ \frac{\partial }{\partial {\mu }_{Y}}g\left( {{\mu }_{X},{\mu }_{Y}}\right) = \frac{-{\mu }_{X}}{{\mu }_{Y}^{2}} \]\n\n\nThe first-order Taylor approximations (5.5.... | Yes |
Suppose that a particular electrical component is to be modeled with an exponential \( \\left( \\lambda \\right) \) lifetime. The manufacturer is interested in determining the probability that, out of \( c \) components, at least \( t \) of them will last \( h \) hours. | \[ {p}_{1} = P\\left( {\\text{ component lasts at least }h\\text{ hours }}\\right) \]\n\[ = P\\left( {X \\geq h \\mid \\lambda }\\right) \]\nand assuming that the components are independent, we can model the outcomes of the \( c \) components as Bernoulli trials, so\n\[ {p}_{2} = P\\text{(at least}t\\text{components la... | Yes |
The probability \( {p}_{2} \) can be calculated using the following steps. For \( j = 1,\ldots, n \) : a. Generate \( {X}_{1},\ldots ,{X}_{c} \) iid \( \sim \operatorname{exponential}\left( \lambda \right) \) . b. Set \( {Y}_{j} = 1 \) if at least \( t{X}_{i} \) s are \( \geq h \) ; otherwise, set \( {Y}_{j} = 0 \) . | Then, because \( {Y}_{j} \sim \operatorname{Bernoulli}\left( {p}_{2}\right) \) and \( \mathrm{E}{Y}_{j} = {p}_{2} \) , \[ \frac{1}{n}\mathop{\sum }\limits_{{j = 1}}^{n}{Y}_{j} \rightarrow {p}_{2}\text{ as }n \rightarrow \infty \] | No |
If \( Y \) is a continuous random variable with cdf \( {F}_{Y} \), then Theorem 2.1.10 implies that the random variable \( {F}_{Y}^{-1}\left( U\right) \) , where \( U \sim \operatorname{uniform}\left( {0,1}\right) \), has distribution \( {F}_{Y} \) . If \( Y \sim \operatorname{exponential}\left( \lambda \right) \), the... | \[ {F}_{Y}^{-1}\left( U\right) = - \lambda \log \left( {1 - U}\right) \] is an exponential \( \left( \lambda \right) \) random variable (see Exercise 5.49). | No |
Generate \( {U}_{1} \) and \( {U}_{2} \), two independent uniform \( \left( {0,1}\right) \) random variables, and set\n\n\[ R = \sqrt{-2\log {U}_{1}}\text{ and }\theta = {2\pi }{U}_{2}. \]\n\nThen\n\n\[ X = R\cos \theta \;\text{ and }\;Y = R\sin \theta \]\n\nare independent normal \( \left( {0,1}\right) \) random varia... | Thus, although we had no quick transformation for generating a single \( n\left( {0,1}\right) \) random variable, there is such a method for generating two variables. (See Exercise 5.50.) | No |
Example 5.6.5 (Binomial random variable generation) To generate a \( Y \sim \) \( \operatorname{binomial}\left( {4,\frac{5}{8}}\right) \), for example, generate \( U \sim \operatorname{uniform}\left( {0,1}\right) \) and set | \( \left( {5.6.8}\right) \)\n\n\[ \nY = \left\{ \begin{array}{ll} 0 & \text{if }0\; < U \leq {.020}, \\ 1 & \text{if }{.020} < U \leq {.152} \\ 2 & \text{if }{.152} < U \leq {.481} \\ 3 & \text{if }{.481} < U \leq {.847} \\ 4 & \text{if }{.847} < U \leq 1. \end{array}\right. \n\] | Yes |
Example 5.6.7 (Beta random variable generation-I) Suppose the goal is to generate \( Y \sim \operatorname{beta}\left( {a, b}\right) \). If both \( a \) and \( b \) are integers, then the direct transformation method (5.6.5) can be used. However, if \( a \) and \( b \) are not integers, then that method will not work. F... | \n\[ P\left( {V \leq y, U \leq \frac{1}{c}{f}_{Y}\left( V\right) }\right) = {\int }_{0}^{y}{\int }_{0}^{{f}_{Y}\left( v\right) /c}{dudv} \]\n\n\[ = \frac{1}{c}{\int }_{0}^{y}{f}_{Y}\left( v\right) {dv} \]\n\n\[ = \frac{1}{c}P\left( {Y \leq y}\right) \]\n\nSo we can calculate the beta probabilities from the uniform prob... | Yes |
Theorem 5.6.8 Let \( Y \sim {f}_{Y}\left( y\right) \) and \( V \sim {f}_{V}\left( V\right) \), where \( {f}_{Y} \) and \( {f}_{V} \) have common support with\n\n\[ \nM = \mathop{\sup }\limits_{y}{f}_{Y}\left( y\right) /{f}_{V}\left( y\right) < \infty .\n\]\n\nTo generate a random variable \( Y \sim {f}_{Y} \) :\n\n a. ... | Proof: The generated random variable \( Y \) has cdf\n\n\[ \nP\left( {Y \leq y}\right) = P\left( {V \leq y \mid \text{ stop }}\right)\n\]\n\n\[ \n= P\left( {V \leq y \mid U < \frac{1}{M}{f}_{Y}\left( V\right) /{f}_{V}\left( V\right) }\right)\n\]\n\n\[ \n= \frac{P\left( {V \leq y, U < \frac{1}{M}{f}_{Y}\left( V\right) /... | Yes |
Let \( {X}_{1},\ldots ,{X}_{n} \) be iid Bernoulli random variables with parameter \( \theta ,0 < \theta < 1. \) We will show that \( T\left( \mathbf{X}\right) = {X}_{1} + \cdots + {X}_{n} \) is a sufficient statistic for \( \theta \). | Note that \( T\left( \mathbf{X}\right) \) counts the number of \( {X}_{i} \) s that equal 1, so \( T\left( \mathbf{X}\right) \) has a binomial \( \left( {n,\theta }\right) \) distribution. The ratio of pmfs is thus\n\n\[ \frac{p\left( {\mathbf{x} \mid \theta }\right) }{q\left( {T\left( \mathbf{x}\right) \mid \theta }\r... | Yes |
Example 6.2.4 (Normal sufficient statistic) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \) , where \( {\sigma }^{2} \) is known. We wish to show that the sample mean, \( T\left( \mathbf{X}\right) = \bar{X} = \left( {{X}_{1} + \cdots + }\right. \) \( \left. {X}_{n}\right) /n ... | The joint pdf of the sample \( \mathbf{X} \) is\n\n\[ f\left( {\mathbf{x} \mid \mu }\right) = \mathop{\prod }\limits_{{i = 1}}^{n}{\left( 2\pi {\sigma }^{2}\right) }^{-1/2}\exp \left( {-{\left( {x}_{i} - \mu \right) }^{2}/\left( {2{\sigma }^{2}}\right) }\right) \]\n\n\[ = {\left( 2\pi {\sigma }^{2}\right) }^{-n/2}\exp ... | Yes |
Example 6.2.5 (Sufficient order statistics) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid from a pdf \( f \), where we are unable to specify any more information about the pdf (as is the case in nonparametric estimation). It then follows that the sample density is given by\n\n\[ \n f\left( \mathbf{x}\right) = \mathop{\prod ... | Of course, this is not much of a reduction, but we shouldn't expect more with so little information about the density \( f \) . | No |
Example 6.2.7 (Continuation of Example 6.2.4) For the normal model described earlier, we saw that the pdf could be factored as\n\n\[ f\left( {\mathbf{x} \mid \mu }\right) = {\left( 2\pi {\sigma }^{2}\right) }^{-n/2}\exp \left( {-\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - \bar{x}\right) }^{2}/\left( {2{\sigma ... | We can define\n\n\[ h\left( \mathbf{x}\right) = {\left( 2\pi {\sigma }^{2}\right) }^{-n/2}\exp \left( {-\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - \bar{x}\right) }^{2}/\left( {2{\sigma }^{2}}\right) }\right) ,\]\n\nwhich does not depend on the unknown parameter \( \mu \) . The factor in (6.2.4) that contains ... | Yes |
Example 6.2.8 (Uniform sufficient statistic) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid observations from the discrete uniform distribution on \( 1,\ldots ,\theta \) . That is, the unknown parameter, \( \theta \), is a positive integer and the pmf of \( {X}_{i} \) is\n\n\[ f\left( {x \mid \theta }\right) = \left\{ \begin... | The restriction \ | No |
Example 6.2.9 (Normal sufficient statistic, both parameters unknown)\n\nAgain assume that \( {X}_{1},\ldots ,{X}_{n} \) are iid \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \) but, unlike Example 6.2.4, assume that both \( \mu \) and \( {\sigma }^{2} \) are unknown so the parameter vector is \( \theta = \left( {\mu ... | \[ g\left( {\mathbf{t} \mid \theta }\right) = g\left( {{t}_{1},{t}_{2} \mid \mu ,{\sigma }^{2}}\right) \]\n\n\[ = {\left( 2\pi {\sigma }^{2}\right) }^{-n/2}\exp \left( {-\left( {n{\left( {t}_{1} - \mu \right) }^{2} + \left( {n - 1}\right) {t}_{2}}\right) /\left( {2{\sigma }^{2}}\right) }\right) . \]\n\nThen it can be s... | Yes |
Example 6.2.14 (Normal minimal sufficient statistic) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \), both \( \mu \) and \( {\sigma }^{2} \) unknown. Let \( \mathbf{x} \) and \( \mathbf{y} \) denote two sample points, and let \( \left( {\bar{x},{s}_{\mathbf{x}}^{2}}\right) \)... | \[ \frac{f\left( {\mathbf{x} \mid \mu ,{\sigma }^{2}}\right) }{f\left( {\mathbf{y} \mid \mu ,{\sigma }^{2}}\right) } = \frac{{\left( 2\pi {\sigma }^{2}\right) }^{-n/2}\exp \left( {-\left\lbrack {n{\left( \bar{x} - \mu \right) }^{2} + \left( {n - 1}\right) {s}_{\mathbf{x}}^{2}}\right\rbrack /\left( {2{\sigma }^{2}}\righ... | Yes |
Example 6.2.15 (Uniform minimal sufficient statistic) Suppose \( {X}_{1},\ldots ,{X}_{n} \) are iid uniform observations on the interval \( \left( {\theta ,\theta + 1}\right) , - \infty < \theta < \infty \) . Then the joint pdf of \( \mathbf{X} \) is\n\n\[ f\left( {\mathbf{x} \mid \theta }\right) = \left\{ \begin{array... | Thus, for two sample points \( \mathbf{x} \) and \( \mathbf{y} \), the numerator and denominator of the ratio \( f\left( {\mathbf{x} \mid \theta }\right) /f\left( {\mathbf{y} \mid \theta }\right) \) will be positive for the same values of \( \theta \) if and only if \( \mathop{\min }\limits_{i}{x}_{i} = \mathop{\min }\... | Yes |
Example 6.2.18 (Location family ancillary statistic) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid observations from a location parameter family with cdf \( F\left( {x - \theta }\right) , - \infty < \theta < \infty \) . We will show that the range, \( R = {X}_{\left( n\right) } - {X}_{\left( 1\right) } \), is an ancillary s... | We use Theorem 3.5.6 and work with \( {Z}_{1},\ldots ,{Z}_{n} \) iid observations from \( F\left( x\right) \) (corresponding to \( \theta = 0 \) ) with \( {X}_{1} = {Z}_{1} + \theta ,\ldots ,{X}_{n} = {Z}_{n} + \theta \) . Thus the cdf of the range statistic, \( R \), is\n\n\[ \n{F}_{R}\left( {r \mid \theta }\right) = ... | Yes |
Suppose that \( T \) has a binomial \( \left( {n, p}\right) \) distribution, \( 0 < p < 1 \) . Let \( g \) be a function such that \( {\mathrm{E}}_{p}g\left( T\right) = 0 \) . Then | \[ 0 = {\mathrm{E}}_{p}g\left( T\right) = \mathop{\sum }\limits_{{t = 0}}^{n}g\left( t\right) \left( \begin{matrix} n \\ t \end{matrix}\right) {p}^{t}{\left( 1 - p\right) }^{n - t} \] \[ = {\left( 1 - p\right) }^{n}\mathop{\sum }\limits_{{t = 0}}^{n}g\left( t\right) \left( \begin{matrix} n \\ t \end{matrix}\right) {\le... | Yes |
Example 6.2.23 (Uniform complete sufficient statistic) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid uniform \( \left( {0,\theta }\right) \) observations, \( 0 < \theta < \infty \) . Using an argument similar to that in Example 6.2.8, we can see that \( T\left( \mathbf{X}\right) = \mathop{\max }\limits_{i}{X}_{i} \) is a su... | Suppose \( g\left( t\right) \) is a function satisfying \( {\mathrm{E}}_{\theta }g\left( T\right) = 0 \) for all \( \theta \) . Since \( {\mathrm{E}}_{\theta }g\left( T\right) \) is constant as a function of \( \theta \), its derivative with respect to \( \theta \) is 0 . Thus we have that\n\n\[ 0 = \frac{d}{d\theta }{... | Yes |
Example 6.2.26 (Using Basu’s Theorem-I) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid exponential observations with parameter \( \theta \) . Consider computing the expected value of \[ g\left( \mathbf{X}\right) = \frac{{X}_{n}}{{X}_{1} + \cdots + {X}_{n}}. \] | We first note that the exponential distributions form a scale parameter family and thus, by Example 6.2.19, \( g\left( \mathbf{X}\right) \) is an ancillary statistic. The exponential distributions also form an exponential family with \( t\left( x\right) = x \) and so, by Theorem 6.2.25, \[ T\left( \mathbf{X}\right) = \... | Yes |
Let \( X \) have a negative binomial distribution with \( r = 3 \) and success probability \( p \) . If \( x = 2 \) is observed, then the likelihood function is the fifth-degree polynomial on \( 0 \leq p \leq 1 \) defined by | \[ L\left( {p \mid 2}\right) = {P}_{p}\left( {X = 2}\right) = \left( \begin{array}{l} 4 \\ 2 \end{array}\right) {p}^{3}{\left( 1 - p\right) }^{2}. \] | Yes |
Example 6.3.3 (Normal fiducial distribution) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \) , \( {\sigma }^{2} \) known. Using expression (6.2.4) for \( L\left( {\mu \mid \mathbf{x}}\right) \), we note first that (6.3.1) is satisfied if and only if \( \bar{x} = \bar{y} \), i... | The parameter \( \mu \) has a \( \mathrm{n}\left( {\bar{x},{\sigma }^{2}/n}\right) \) distribution. Hence, \( \left( {\mu - \bar{x}}\right) /\left( {\sigma /\sqrt{n}}\right) \) has a \( \mathrm{n}\left( {0,1}\right) \) distribution. Thus we have \[ {.95} = P\left( {-{1.96} < \frac{\mu - \bar{x}}{\sigma /\sqrt{n}} < {1.... | Yes |
Theorem 6.3.6 (Birnbaum's Theorem) The Formal Likelihood Principle follows from the Formal Sufficiency Principle and the Conditionality Principle. The converse is also true. | Proof: We only outline the proof, leaving details to Exercise 6.33. Let \( {E}_{1},{E}_{2},{\mathbf{x}}_{1}^{ * } \) , and \( {\mathbf{x}}_{2}^{ * } \) be as defined in the Formal Likelihood Principle, and let \( {E}^{ * } \) be the mixed experiment from the Conditionality Principle. On the sample space of \( {E}^{ * }... | No |
Let \( X \) have a binomial distribution with sample size \( n \) known and success probability \( p \) unknown. Let \( T\left( x\right) \) be the estimate of \( p \) that is used when \( X = x \) is observed. Rather than using the number of successes, \( X \), to make an inference about \( p \), we could use the numbe... | (6.4.1)\n\n\[ T\left( x\right) = 1 - {T}^{ * }\left( {n - x}\right) = 1 - T\left( {n - x}\right) . \] | Yes |
Example 6.4.3 (Continuation of Example 6.4.1) For this problem, only two transformations are involved so we may set \( \mathcal{G} = \left\{ {{g}_{1},{g}_{2}}\right\} \), with \( {g}_{1}\left( x\right) = n - x \) and \( {g}_{2}\left( x\right) = x \) . Conditions (i) and (ii) are easily verified. The choice of \( {g}^{\... | For example,\n\n\[ \n{g}_{1}\left( {{g}_{1}\left( x\right) }\right) = {g}_{1}\left( {n - x}\right) = n - \left( {n - x}\right) = x. \n\] \n\nIn (ii), if \( {g}^{\prime } = g \), then \( {g}^{\prime \prime } = {g}_{2} \), while if \( {g}^{\prime } \neq g \), then \( {g}^{\prime \prime } = {g}_{1} \) satisfies the equali... | Yes |
Example 6.4.6 (Normal location invariance) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \) , both \( \mu \) and \( {\sigma }^{2} \) unknown. Consider the group of transformations defined by \( \mathcal{G} = \) \( \left\{ {{g}_{a}\left( \mathbf{x}\right) , - \infty < a < \inft... | For (i) note that\n\n\[ \n{g}_{-a}\left( {{g}_{a}\left( {{x}_{1},\ldots ,{x}_{n}}\right) }\right) = {g}_{-a}\left( {{x}_{1} + a,\ldots ,{x}_{n} + a}\right) \n\]\n\n\[ \n= \left( {{x}_{1} + a - a,\ldots ,{x}_{n} + a - a}\right) \n\]\n\n\[ \n= \left( {{x}_{1},\ldots ,{x}_{n}}\right) \text{.} \n\]\n\nSo if \( g = {g}_{a} ... | Yes |
Example 7.2.1 (Normal method of moments) Suppose \( {X}_{1},\ldots ,{X}_{n} \) are iid \( \mathbf{n}\left( {\theta ,{\sigma }^{2}}\right) \) . In the preceding notation, \( {\theta }_{1} = \theta \) and \( {\theta }_{2} = {\sigma }^{2} \) . We have \( {m}_{1} = \bar{X},{m}_{2} = \) \( \left( {1/n}\right) \sum {X}_{i}^{... | Solving for \( \theta \) and \( {\sigma }^{2} \) yields the method of moments estimators\n\n\[ \widetilde{\theta } = \bar{X}\;\text{ and }\;{\widetilde{\sigma }}^{2} = \frac{1}{n}\sum {X}_{i}^{2} - {\bar{X}}^{2} = \frac{1}{n}\sum {\left( {X}_{i} - \bar{X}\right) }^{2}. \] | Yes |
Example 7.2.2 (Binomial method of moments) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid binomial \( \left( {k, p}\right) \), that is,\n\n\[ P\left( {{X}_{i} = x \mid k, p}\right) = \left( \begin{array}{l} k \\ x \end{array}\right) {p}^{x}{\left( 1 - p\right) }^{k - x},\;x = 0,1,\ldots, k. \]\n\nHere we assume that both \( ... | Equating the first two sample moments to those of the population yields the system of equations\n\n\[ \bar{X} = {kp} \]\n\n\[ \frac{1}{n}\sum {X}_{i}^{2} = {kp}\left( {1 - p}\right) + {k}^{2}{p}^{2} \]\n\nwhich now must be solved for \( k \) and \( p \) . After a little algebra, we obtain the method of moments estimato... | Yes |
Example 7.2.3 (Satterthwaite approximation) If \( {Y}_{i}, i = 1,\ldots, k \), are independent \( {\chi }_{{r}_{i}}^{2} \) random variables, we have already seen (Lemma 5.3.2) that the distribution of \( \sum {Y}_{i} \) is also chi squared, with degrees of freedom equal to \( \sum {r}_{i} \) . Unfortunately, the distri... | Since \( \mathrm{E}\left( {{\chi }_{\nu }^{2}/\nu }\right) = 1 \), to match first moments we need\n\n\[ \mathrm{E}\left( {\mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i}{Y}_{i}}\right) = \mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i}\mathrm{E}{Y}_{i} = \mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i}{r}_{i} = 1 \]\n\nwhich gives us a... | Yes |
Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\theta ,1}\right) \), and let \( L\left( {\theta \mid \mathbf{x}}\right) \) denote the likelihood function. Then\n\n\[ L\left( {\theta \mid \mathbf{x}}\right) = \mathop{\prod }\limits_{{i = 1}}^{n}\frac{1}{{\left( 2\pi \right) }^{1/2}}{e}^{-\left( {1/2}\righ... | The equation \( \left( {d/{d\theta }}\right) L\left( {\theta \mid \mathbf{x}}\right) = 0 \) reduces to\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}\left( {{x}_{i} - \theta }\right) = 0 \]\n\nwhich has the solution \( \widehat{\theta } = \bar{x} \) . Hence, \( \bar{x} \) is a candidate for the MLE. To verify that \( \bar{x... | Yes |
Recall (Theorem 5.2.4) that for any number \( a \) ,\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - a\right) }^{2} \geq \mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - \bar{x}\right) }^{2} \]\n\nwith equality if and only if \( a = \bar{x} \) . | This implies that for any \( \theta \) ,\n\n\[ {e}^{-\left( {1/2}\right) \sum {\left( {x}_{i} - \theta \right) }^{2}} \leq {e}^{-\left( {1/2}\right) \sum {\left( {x}_{i} - \bar{x}\right) }^{2}} \]\n\nwith equality if and only if \( \theta = \bar{x} \) . Hence \( \bar{X} \) is the MLE. | Yes |
Let \( {X}_{1},\ldots ,{X}_{n} \) be iid Bernoulli \( \left( p\right) \) . Then the likelihood function is\n\n\[ L\left( {p \mid \mathbf{x}}\right) = \mathop{\prod }\limits_{{i = 1}}^{n}{p}^{{x}_{i}}{\left( 1 - p\right) }^{1 - {x}_{i}} = {p}^{y}{\left( 1 - p\right) }^{n - y}, \]\n\nwhere \( y = \sum {x}_{i} \) . | While this function is not all that hard to differentiate, it is much easier to differentiate the log likelihood\n\n\[ \log L\left( {p \mid \mathbf{x}}\right) = y\log p + \left( {n - y}\right) \log \left( {1 - p}\right) . \]\n\nIf \( 0 < y < n \), differentiating \( \log L\left( {p \mid \mathbf{x}}\right) \) and settin... | Yes |
Example 7.2.8 (Restricted range MLE) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\theta ,1}\right) \), where it is known that \( \theta \) must be nonnegative. With no restrictions on \( \theta \), we saw that the MLE of \( \theta \) is \( \bar{X} \) ; however, if \( \bar{X} \) is negative, it will be... | If \( \bar{x} \) is negative, it is easy to check (see Exercise 7.4) that the likelihood function \( L\left( {\theta \mid \mathbf{x}}\right) \) is decreasing in \( \theta \) for \( \theta \geq 0 \) and is maximized at \( \widehat{\theta } = 0 \) . Hence, in this case, the MLE of \( \theta \) is\n\n\[ \widehat{\theta } ... | No |
Example 7.2.9 (Binomial MLE, unknown number of trials) Let \( {X}_{1},\ldots \) , \( {X}_{n} \) be a random sample from a binomial \( \left( {k, p}\right) \) population, where \( p \) is known and \( k \) is unknown. For example, we flip a coin we know to be fair and observe \( {x}_{i} \) heads but we do not know how m... | Maximizing \( L\left( {k \mid \mathbf{x}, p}\right) \) by differentiation is difficult because of the factorials and because \( k \) must be an integer. Thus we try a different approach.\n\nOf course, \( L\left( {k \mid \mathbf{x}, p}\right) = 0 \) if \( k < \mathop{\max }\limits_{i}{x}_{i} \) . Thus the MLE is an inte... | Yes |
Theorem 7.2.10 (Invariance property of MLEs) If \( \widehat{\theta } \) is the MLE of \( \theta \), then for any function \( \tau \left( \theta \right) \), the MLE of \( \tau \left( \theta \right) \) is \( \tau \left( \widehat{\theta }\right) \) . | Proof: Let \( \widehat{\eta } \) denote the value that maximizes \( {L}^{ * }\left( {\eta \mid \mathbf{x}}\right) \) . We must show that \( {L}^{ * }\left( {\widehat{\eta } \mid \mathbf{x}}\right) = \) \( {L}^{ * }\left\lbrack {\tau \left( \widehat{\theta }\right) \mid \mathbf{x}}\right\rbrack \) . Now, as stated above... | Yes |
Example 7.2.11 (Normal MLEs, \( \mu \) and \( \sigma \) unknown) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\theta ,{\sigma }^{2}}\right) \), with both \( \theta \) and \( {\sigma }^{2} \) unknown. Then | \[ L\left( {\theta ,{\sigma }^{2} \mid \mathbf{x}}\right) = \frac{1}{{\left( 2\pi {\sigma }^{2}\right) }^{n/2}}{e}^{-\left( {1/2}\right) {\sum }_{i = 1}^{n}{\left( {x}_{i} - \theta \right) }^{2}/{\sigma }^{2}} \] and \[ \log L\left( {\theta ,{\sigma }^{2} \mid \mathbf{x}}\right) = - \frac{n}{2}\log {2\pi } - \frac{n}{2... | Yes |
We observe \( {X}_{1},\ldots ,{X}_{n} \) and \( {Y}_{1},\ldots \) , \( {Y}_{n} \), all mutually independent, where \( {Y}_{i} \sim \operatorname{Poisson}\left( {\beta {\tau }_{i}}\right) \) and \( {X}_{i} \sim \operatorname{Poisson}\left( {\tau }_{i}\right) \) . This would model, for instance, the incidence of a diseas... | The joint pmf is therefore\n\n\[ f\left( {\left( {{x}_{1},{y}_{1}}\right) ,\left( {{x}_{2},{y}_{2}}\right) ,\ldots ,\left( {{x}_{n},{y}_{n}}\right) \mid \beta ,{\tau }_{1},{\tau }_{2},\ldots ,{\tau }_{n}}\right) \]\n\n(7.2.11)\n\n\[ = \mathop{\prod }\limits_{{i = 1}}^{n}\frac{{e}^{-\beta {\tau }_{i}}{\left( \beta {\tau... | No |
The incomplete-data likelihood is obtained from (7.2.11) by summing over \( {x}_{1} \) . This gives\n\n\[ L\left( {\beta ,{\tau }_{1},{\tau }_{2},\ldots ,{\tau }_{n} \mid {y}_{1},\left( {{x}_{2},{y}_{2}}\right) ,\ldots ,\left( {{x}_{n},{y}_{n}}\right) }\right) \]\n\n\( \left( {7.2.15}\right) \)\n\n\[ = \left\lbrack {\m... | The EM algorithm allows us to maximize \( L\left( {\theta \mid \mathbf{y}}\right) \) by working with only \( L\left( {\theta \mid \mathbf{y},\mathbf{x}}\right) \) and the conditional pdf or pmf of \( \mathbf{X} \) given \( \mathbf{y} \) and \( \theta \), defined by\n\n(7.2.17)\n\n\[ L\left( {\theta \mid \mathbf{y},\mat... | Yes |
Let \( \left( {\mathbf{x},\mathbf{y}}\right) = \left( {\left( {{x}_{1},{y}_{1}}\right) \text{,}}\right. \) \( \left. {\left( {{x}_{2},{y}_{2}}\right) ,\ldots ,\left( {{x}_{n},{y}_{n}}\right) }\right) \) denote the complete data and \( \left( {{\mathbf{x}}_{\left( -1\right) },\mathbf{y}}\right) = \left( {{y}_{1},\left( ... | \[ \mathrm{E}\left\lbrack {\log L\left( {\beta ,{\tau }_{1},{\tau }_{2},\ldots ,{\tau }_{n} \mid \left( {\mathbf{x},\mathbf{y}}\right) }\right) \mid {\tau }^{\left( r\right) },\left( {{\mathbf{x}}_{\left( -1\right) },\mathbf{y}}\right) }\right\rbrack \]\n\[ = \mathop{\sum }\limits_{{{x}_{1} = 0}}^{\infty }\log \left( {... | Yes |
Example 7.3.3 (Normal MSE) Let \( {X}_{1},\ldots ,{X}_{n} \) be iidn \( \left( {\mu ,{\sigma }^{2}}\right) \). The statistics \( \bar{X} \) and \( {S}^{2} \) are both unbiased estimators since\n\n\[ \mathrm{E}\bar{X} = \mu ,\;\mathrm{E}{S}^{2} = {\sigma }^{2},\;\text{ for all }\mu \text{ and }{\sigma }^{2}. \]\n\n(This... | The MSE of \( \bar{X} \) remains \( {\sigma }^{2}/n \) even if the normality assumption is dropped. However, the above expression for the MSE of \( {S}^{2} \) does not remain the same if the normality assumption is relaxed (see Exercise 5.8). | No |
Let \( {X}_{1},\ldots ,{X}_{n} \) be iid Bernoulli \( \left( p\right) \) . The MSE of \( \widehat{p} \), the MLE, as an estimator of \( p \), is | \[ {\mathrm{E}}_{p}{\left( \widehat{p} - p\right) }^{2} = {\operatorname{Var}}_{p}\bar{X} = \frac{p\left( {1 - p}\right) }{n}. \] | Yes |
Example 7.3.6 (MSE of equivariant estimators) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( f(x - \) \( \theta ) \) . For an estimator \( W\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) to satisfy \( W\left( {{g}_{a}\left( \mathbf{x}\right) }\right) = {\bar{g}}_{a}\left( {W\left( \mathbf{x}\right) }\right) \), we must have | (7.3.2)\n\n\[ W\left( {{x}_{1},\ldots ,{x}_{n}}\right) + a = W\left( {{x}_{1} + a,\ldots ,{x}_{n} + a}\right) ,\]\n\nwhich specifies the equivariant estimators with.respect to the group of transformations defined by \( \mathcal{G} = \left\{ {{g}_{a}\left( \mathbf{x}\right) : - \infty < a < \infty }\right\} \), where \(... | No |
Example 7.3.8 (Poisson unbiased estimation) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid Poisson \( \left( \lambda \right) \), and let \( \bar{X} \) and \( {S}^{2} \) be the sample mean and variance, respectively. Recall that for the Poisson pmf both the mean and variance are equal to \( \lambda \) . Therefore, applying Th... | \[ {\mathrm{E}}_{\lambda }\bar{X} = \lambda ,\;\text{ for all }\lambda \] and \[ {\mathrm{E}}_{\lambda }{S}^{2} = \lambda ,\;\text{ for all }\lambda \] so both \( \bar{X} \) and \( {S}^{2} \) are unbiased estimators of \( \lambda \) . | Yes |
Corollary 7.3.10 (Cramér-Rao Inequality, iid case) If the assumptions of Theorem 7.3.9 are satisfied and, additionally, if \( {X}_{1},\ldots ,{X}_{n} \) are iid with \( \operatorname{pdff}\left( {x \mid \theta }\right) \), then\n\n\[ \n{\operatorname{Var}}_{\theta }W\left( \mathbf{X}\right) \geq \frac{{\left( \frac{d}{... | Proof: We only need to show that\n\n\[ \n{\mathrm{E}}_{\theta }\left( {\left( \frac{\partial }{\partial \theta }\log f\left( \mathbf{X} \mid \theta \right) \right) }^{2}\right) = n{\mathrm{E}}_{\theta }\left( {\left( \frac{\partial }{\partial \theta }\log f\left( X \mid \theta \right) \right) }^{2}\right) .\n\]\n\nSinc... | Yes |
Example 7.3.12 (Conclusion of Example 7.3.8) Here \( \tau \left( \lambda \right) = \lambda \), so \( {\tau }^{\prime }\left( \lambda \right) = 1 \) . | Also, since we have an exponential family, using Lemma 7.3.11 gives us\n\n\[ \n{\mathrm{E}}_{\lambda }\left( {\left( \frac{\partial }{\partial \lambda }\log \mathop{\prod }\limits_{{i = 1}}^{n}f\left( {X}_{i} \mid \lambda \right) \right) }^{2}\right) = - n{\mathrm{E}}_{\lambda }\left( {\frac{{\partial }^{2}}{\partial {... | Yes |
Example 7.3.14 (Normal variance bound) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \) , and consider estimation of \( {\sigma }^{2} \), where \( \mu \) is unknown. The normal pdf satisfies the assumptions of the Cramér-Rao Theorem and Lemma 7.3.11, so we have | \[ \frac{{\partial }^{2}}{\partial {\left( {\sigma }^{2}\right) }^{2}}\log \left( {\frac{1}{{\left( 2\pi {\sigma }^{2}\right) }^{1/2}}{e}^{-\left( {1/2}\right) {\left( x - \mu \right) }^{2}/{\sigma }^{2}}}\right) = \frac{1}{2{\sigma }^{4}} - \frac{{\left( x - \mu \right) }^{2}}{{\sigma }^{6}} \] and \[ - \mathrm{E}\lef... | Yes |
Here we have\n\n\[ L\left( {\mu ,{\sigma }^{2} \mid \mathbf{x}}\right) = \frac{1}{{\left( 2\pi {\sigma }^{2}\right) }^{n/2}}{e}^{-\left( {1/2}\right) {\sum }_{i = 1}^{n}{\left( {x}_{i} - \mu \right) }^{2}/{\sigma }^{2}}, \]\n\nand hence\n\n\[ \frac{\partial }{\partial {\sigma }^{2}}\log L\left( {\mu ,{\sigma }^{2} \mid... | Thus, taking \( \mathbf{a}\left( {\sigma }^{2}\right) = n/\left( {2{\sigma }^{4}}\right) \) shows that the best unbiased estimator of \( {\sigma }^{2} \) is \( \mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - \mu \right) }^{2}/n \), which is calculable only if \( \mu \) is known. If \( \mu \) is unknown, the bound ... | Yes |
Example 7.3.18 (Conditioning on an insufficient statistic) Let \( {X}_{1},{X}_{2} \) be iid \( \mathrm{n}\left( {\theta ,1}\right) \) . The statistic \( \bar{X} = \frac{1}{2}\left( {{X}_{1} + {X}_{2}}\right) \) has \[ {\mathrm{E}}_{\theta }\bar{X} = \theta \;\text{ and }\;{\operatorname{Var}}_{\theta }\bar{X} = \frac{1... | \[ \phi \left( {X}_{1}\right) = {\mathrm{E}}_{\theta }\left( {\bar{X} \mid {X}_{1}}\right) \] \[ = \frac{1}{2}{\mathrm{E}}_{\theta }\left( {{X}_{1} \mid {X}_{1}}\right) + \frac{1}{2}{\mathrm{E}}_{\theta }\left( {{X}_{2} \mid {X}_{1}}\right) \] \[ = \frac{1}{2}{X}_{1} + \frac{1}{2}\theta \] since \( {\mathrm{E}}_{\theta... | Yes |
Theorem 7.3.19 If \( W \) is a best unbiased estimator of \( \tau \left( \theta \right) \), then \( W \) is unique. | Proof: Suppose \( {W}^{\prime } \) is another best unbiased estimator, and consider the estimator \( {W}^{ * } = \frac{1}{2}\left( {W + {W}^{\prime }}\right) \) . Note that \( {\mathrm{E}}_{\theta }{W}^{ * } = \tau \left( \theta \right) \) and\n\n\[ \n{\operatorname{Var}}_{\theta }{W}^{ * } = {\operatorname{Var}}_{\the... | No |
Theorem 7.3.20 If \( {\mathrm{E}}_{\theta }W = \tau \left( \theta \right), W \) is the best unbiased estimator of \( \tau \left( \theta \right) \) if and only if \( W \) is uncorrelated with all unbiased estimators of 0. | Proof: If \( W \) is best unbiased, the above argument shows that \( W \) must satisfy \( {\operatorname{Cov}}_{\theta }\left( {W, U}\right) = 0 \) for all \( \theta \), for any \( U \) satisfying \( {\mathrm{E}}_{\theta }U = 0 \) . Hence the necessity is established.\n\nSuppose now that we have an unbiased estimator \... | Yes |
Let \( X \) be an observation from a uniform \( \left( {\theta ,\theta + 1}\right) \) distribution. Then | \[ {\mathrm{E}}_{\theta }X = {\int }_{\theta }^{\theta + 1}{xdx} = \theta + \frac{1}{2} \] and so \( X - \frac{1}{2} \) is an unbiased estimator of \( \theta \), and it is easy to check that \( {\operatorname{Var}}_{\theta }X = \frac{1}{12} \) . | Yes |
For \( {X}_{1},\ldots ,{X}_{n} \) iid uniform \( \left( {0,\theta }\right) \), we saw that \( \frac{n + 1}{n}Y \) is an unbiased estimator of \( \theta \), where \( Y = \max \left\{ {{X}_{1},\ldots }\right. \) , \( \left. {X}_{n}\right\} \) . The conditions of the Cramér-Rao Theorem are not satisfied, and we have not y... | In Example 6.2.23, however, it was shown that \( Y \) is a complete sufficient statistic. This means that the family of pdfs of \( Y \) is complete, and there are no unbiased estimators of zero that are based on \( Y \) . (By sufficiency, in the form of the Rao-Blackwell Theorem, we need consider only unbiased estimato... | Yes |
Example 7.3.26 (Risk of normal variance) Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \) population. Consider estimating \( {\sigma }^{2} \) using squared error loss. We will consider estimators of the form \( {\delta }_{b}\left( \mathbf{X}\right) = b{S}^{2... | \[ R\left( {\left( {\mu ,{\sigma }^{2}}\right) ,{\delta }_{b}}\right) = \operatorname{Var}b{S}^{2} + {\left( \mathrm{E}b{S}^{2} - {\sigma }^{2}\right) }^{2} \] \[ = {b}^{2}\operatorname{Var}{S}^{2} + {\left( b\mathrm{E}{S}^{2} - {\sigma }^{2}\right) }^{2} \] \[ = \frac{{b}^{2}2{\sigma }^{4}}{n - 1} + {\left( b - 1\righ... | Yes |
Example 7.3.29 (Normal Bayes estimates) Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a \( \mathrm{n}\left( {\theta ,{\sigma }^{2}}\right) \) population and let \( \pi \left( \theta \right) \) be \( \mathrm{n}\left( {\mu ,{\tau }^{2}}\right) \) . The values \( {\sigma }^{2},\mu \), and \( {\tau }^{2} \) are... | \[ \mathrm{E}\left( {\theta \mid \bar{x}}\right) = \frac{{\tau }^{2}}{{\tau }^{2} + \left( {{\sigma }^{2}/n}\right) }\bar{x} + \frac{{\sigma }^{2}/n}{{\tau }^{2} + \left( {{\sigma }^{2}/n}\right) }\mu \] and \[ \operatorname{Var}\left( {\theta \mid \bar{x}}\right) = \frac{{\tau }^{2}{\sigma }^{2}/n}{{\tau }^{2} + \left... | Yes |
Theorem 7.5.1 (Lehmann-Scheffé) Unbiased estimators based on complete sufficient statistics are unique. | Proof: Suppose \( T \) is a complete sufficient statistic, and \( \phi \left( T\right) \) is an estimator with \( {E}_{\theta }\phi \left( T\right) = \tau \left( \theta \right) \) . From Theorem 7.3.23 we know that \( \phi \left( T\right) \) is the best unbiased estimator of \( \tau \left( \theta \right) \), and from T... | Yes |
Example 8.2.2 (Normal LRT) Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a \( \mathrm{n}\left( {\theta ,1}\right) \) population. Consider testing \( {H}_{0} : \theta = {\theta }_{0} \) versus \( {H}_{1} : \theta \neq {\theta }_{0} \) . Here \( {\theta }_{0} \) is a number fixed by the experimenter prior to ... | (8.2.1)\n\[ \lambda \left( \mathbf{x}\right) = \frac{{\left( 2\pi \right) }^{-n/2}\exp \left\lbrack {-\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - {\theta }_{0}\right) }^{2}/2}\right\rbrack }{{\left( 2\pi \right) }^{-n/2}\exp \left\lbrack {-\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {x}_{i} - \bar{x}\right) }^{... | Yes |
Example 8.2.3 (Exponential LRT) Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from an exponential population with pdf\n\n\[ f\left( {x \mid \theta }\right) = \left\{ \begin{array}{ll} {e}^{-\left( {x - \theta }\right) } & x \geq \theta \\ 0 & x < \theta , \end{array}\right. \]\n\nwhere \( - \infty < \theta < \in... | Clearly \( L\left( {\theta \mid \mathbf{x}}\right) \) is an increasing function of \( \theta \) on \( - \infty < \theta \leq {x}_{\left( 1\right) } \) . Thus, the denominator of \( \lambda \left( \mathbf{x}\right) \), the unrestricted maximum of \( L\left( {\theta \mid \mathbf{x}}\right) \), is\n\n\[ L\left( {{x}_{\lef... | Yes |
Theorem 8.2.4 If \( T\left( \mathbf{X}\right) \) is a sufficient statistic for \( \theta \) and \( {\lambda }^{ * }\left( t\right) \) and \( \lambda \left( \mathbf{x}\right) \) are the LRT statistics based on \( T \) and \( \mathbf{X} \), respectively, then \( {\lambda }^{ * }\left( {T\left( \mathbf{x}\right) }\right) ... | Proof: From the Factorization Theorem (Theorem 6.2.6), the pdf or pmf of \( \mathbf{X} \) can be written as \( f\left( {\mathbf{x} \mid \theta }\right) = g\left( {T\left( \mathbf{x}\right) \mid \theta }\right) h\left( \mathbf{x}\right) \), where \( g\left( {t \mid \theta }\right) \) is the pdf or pmf of \( T \) and \( ... | Yes |
Example 8.2.7 (Normal Bayesian test) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\theta ,{\sigma }^{2}}\right) \) and let the prior distribution on \( \theta \) be \( \mathrm{n}\left( {\mu ,{\tau }^{2}}\right) \), where \( {\sigma }^{2},\mu \), and \( {\tau }^{2} \) are known. Consider testing \( {H}_... | From Example 7.2.16, the posterior \( \pi \left( {\theta \mid \bar{x}}\right) \) is normal with mean \( \left( {n{\tau }^{2}\bar{x} + {\sigma }^{2}\mu }\right) /\left( {n{\tau }^{2} + {\sigma }^{2}}\right) \) and variance \( {\sigma }^{2}{\tau }^{2}/\left( {n{\tau }^{2} + {\sigma }^{2}}\right) \). If we decide to accep... | Yes |
Example 8.2.8 (Normal union-intersection test) Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \) population. Consider testing \( {H}_{0} : \mu = {\mu }_{0} \) versus \( {H}_{1} : \mu \neq \) \( {\mu }_{0} \), where \( {\mu }_{0} \) is a specified number. We c... | The LRT of \( {H}_{0\mathrm{\;L}} : \mu \leq {\mu }_{0} \) versus \( {H}_{1\mathrm{\;L}} : \mu > {\mu }_{0} \) is\n\n\[ \n\text{reject}{H}_{\mathrm{{OL}}} : \mu \leq {\mu }_{0}\text{in favor of}{H}_{1\mathrm{\;L}} : \mu > {\mu }_{0}\text{if}\frac{\widetilde{X} - {\mu }_{0}}{S/\sqrt{n}} \geq {t}_{\mathrm{L}} \n\]\n\n(se... | No |
Example 8.3.3 (Normal power function) Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a \( \mathrm{n}\left( {\theta ,{\sigma }^{2}}\right) \) population, \( {\sigma }^{2} \) known. An LRT of \( {H}_{0} : \theta \leq {\theta }_{0} \) versus \( {H}_{1} : \theta > {\theta }_{0} \) is a test that rejects \( {H}_{... | \[ \beta \left( \theta \right) = {P}_{\theta }\left( {\frac{\widetilde{X} - {\theta }_{0}}{\sigma /\sqrt{n}} > c}\right) \] \[ = {P}_{\theta }\left( {\frac{\bar{X} - \theta }{\sigma /\sqrt{n}} > c + \frac{{\theta }_{0} - \theta }{\sigma /\sqrt{n}}}\right) \] \[ = P\left( {Z > c + \frac{{\theta }_{0} - \theta }{\sigma /... | Yes |
Theorem 8.3.12 (Neyman-Pearson Lemma) Consider testing \( {H}_{0} : \theta = {\theta }_{0} \) versus \( {H}_{1} : \theta = {\theta }_{1} \), where the pdf or pmf corresponding to \( {\theta }_{i} \) is \( f\left( {\mathbf{x} \mid {\theta }_{i}}\right), i = 0,1 \), using a test with rejection region \( R \) that satisfi... | Proof: We will prove the theorem for the case that \( f\left( {\mathbf{x} \mid {\theta }_{0}}\right) \) and \( f\left( {\mathbf{x} \mid {\theta }_{1}}\right) \) are pdfs of continuous random variables. The proof for discrete random variables can be accomplished by replacing integrals with sums. (See Exercise 8.21.)\n\n... | No |
Corollary 8.3.13 Consider the hypothesis problem posed in Theorem 8.3.12. Suppose \( T\left( \mathbf{X}\right) \) is a sufficient statistic for \( \theta \) and \( g\left( {t \mid {\theta }_{i}}\right) \) is the pdf or pmf of \( T \) corresponding to \( {\theta }_{i}, i = 0,1 \) . Then any test based on \( T \) with re... | Proof: In terms of the original sample \( \mathbf{X} \), the test based on \( T \) has the rejection region \( R = \{ \mathbf{x} : T\left( \mathbf{x}\right) \in S\} \) . By the Factorization Theorem, the pdf or pmf of \( \mathbf{X} \) can be written as \( f\left( {\mathbf{x} \mid {\theta }_{i}}\right) = g\left( {T\left... | Yes |
Example 8.3.15 (UMP normal test) Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a \( \mathrm{n}\left( {\theta ,{\sigma }^{2}}\right) \) population, \( {\sigma }^{2} \) known. The sample mean \( \bar{X} \) is a sufficient statistic for \( \theta \) . Consider testing \( {H}_{0} : \theta = {\theta }_{0} \) ver... | \[ \bar{x} < \frac{\left( {2{\sigma }^{2}\log k}\right) /n - {\theta }_{0}^{2} + {\theta }_{1}^{2}}{2\left( {{\theta }_{1} - {\theta }_{0}}\right) }.\]\nThe fact that \( {\theta }_{1} - {\theta }_{0} < 0 \) was used to obtain this inequality. The right-hand side increases from \( - \infty \) to \( \infty \) as \( k \) ... | Yes |
Consider testing \( {H}_{0}^{\prime } : \theta \geq \) \( {\theta }_{0} \) versus \( {H}_{1}^{\prime } : \theta < {\theta }_{0} \) using the test that rejects \( {H}_{0}^{\prime } \) if\n\n\[ \bar{X} < - \frac{\sigma {z}_{\alpha }}{\sqrt{n}} + {\theta }_{0} \] | As \( \bar{X} \) is sufficient and its distribution has an MLR (see Exercise 8.25), it follows from Theorem 8.3.17 that the test is a UMP level \( \alpha \) test in this problem.\n\nAs the power function of this test,\n\n\[ \beta \left( \theta \right) = {P}_{\theta }\left( {\bar{X} < - \frac{\sigma {z}_{\alpha }}{\sqrt... | No |
Example 8.3.19 (Nonexistence of UMP test) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\theta ,{\sigma }^{2}}\right) \) , \( {\sigma }^{2} \) known. Consider testing \( {H}_{0} : \theta = {\theta }_{0} \) versus \( {H}_{1} : \theta \neq {\theta }_{0} \) . For a specified value of \( \alpha \) , a level... | Now consider Test 2, which rejects \( {H}_{0} \) if \( \bar{X} > \sigma {z}_{\alpha }/\sqrt{n} + {\theta }_{0} \) . Test 2 is also a level \( \alpha \) test. Let \( {\beta }_{i}\left( \theta \right) \) denote the power function of Test \( i \) . For any \( {\theta }_{2} > {\theta }_{0} \) ,\n\n\[ \n{\beta }_{2}\left( {... | Yes |
Example 8.3.20 (Unbiased test) When no UMP level \( \alpha \) test exists within the class of all tests, we might try to find a UMP level \( \alpha \) test within the class of unbiased tests. The power function, \( {\beta }_{3}\left( \theta \right) \), of Test 3, which rejects \( {H}_{0} : \theta = {\theta }_{0} \) in ... | \[ \bar{X} > \sigma {z}_{\alpha /2}/\sqrt{n} + {\theta }_{0}\;\text{ or }\;\bar{X} < - \sigma {z}_{\alpha /2}/\sqrt{n} + {\theta }_{0} \] as well as \( {\beta }_{1}\left( \theta \right) \) and \( {\beta }_{2}\left( \theta \right) \) from Example 8.3.19, is shown in Figure 8.3.3. Test 3 is actually a UMP unbiased level ... | Yes |
Theorem 8.3.21 Consider testing \( {H}_{0} : \theta \in {\Theta }_{0} \) versus \( {H}_{1} : \theta \in {\Theta }_{0}^{\mathrm{c}} \), where \( {\Theta }_{0} = \) \( \mathop{\bigcap }\limits_{{\gamma \in \Gamma }}{\Theta }_{\gamma } \) and \( {\lambda }_{\gamma }\left( \mathbf{x}\right) \) is defined in the previous pa... | Proof: Since \( {\Theta }_{0} = \mathop{\bigcap }\limits_{{\gamma \in \Gamma }}{\Theta }_{\gamma } \subset {\Theta }_{\gamma } \) for any \( \gamma \), from Definition 8.2.1 we see that for any \( \mathbf{x} \) ,\n\n\[ {\lambda }_{\gamma }\left( \mathbf{x}\right) \geq \lambda \left( \mathbf{x}\right) \;\text{ for each ... | Yes |
Theorem 8.3.23 Let {\\alpha }_{\\gamma } be the size of the test of {H}_{0\\gamma } with rejection region {R}_{\\gamma }. Then the IUT with rejection region R = \\mathop{\\bigcap }\\limits_{{\\gamma \\in \\Gamma }}{R}_{\\gamma } is a level \\alpha = \\mathop{\\sup }\\limits_{{\\gamma \\in \\Gamma }}{\\alpha }_{\\gamma ... | Proof: Let \\theta \\in {\\Theta }_{0} . Then \\theta \\in {\\Theta }_{\\gamma } for some \\gamma and\n\n\\[ \n{P}_{\\theta }\\left( {\\mathbf{X} \\in R}\\right) \\leq {P}_{\\theta }\\left( {\\mathbf{X} \\in {R}_{\\gamma }}\\right) \\leq {\\alpha }_{\\gamma } \\leq \\alpha .\n\\]\n\nSince \\theta \\in {\\Theta }_{0} wa... | Yes |
Theorem 8.3.24 Consider testing \( {H}_{0} : \theta \in \mathop{\bigcup }\limits_{{j = 1}}^{k}{\Theta }_{j} \), where \( k \) is a finite positive integer. For each \( j = 1,\ldots, k \), let \( {R}_{j} \) be the rejection region of a level \( \alpha \) test of \( {H}_{0j} \) . Suppose that for some \( i = 1,\ldots, k ... | Proof: By Theorem 8.3.23, \( R \) is a level \( \alpha \) test, that is,\n\n(8.3.7)\n\n\[ \mathop{\sup }\limits_{{\theta \in {\Theta }_{0}}}{P}_{\theta }\left( {\mathbf{X} \in R}\right) \leq \alpha \]\n\nBut, because all the parameter points \( {\theta }_{l} \) satisfy \( {\theta }_{l} \in {\Theta }_{i} \subset {\Theta... | Yes |
Theorem 8.3.27 Let \( W\left( \mathbf{X}\right) \) be a test statistic such that large values of \( W \) give evidence that \( {H}_{1} \) is true. For each sample point \( \mathbf{x} \), define\n\n\[ p\left( \mathbf{x}\right) = \mathop{\sup }\limits_{{\theta \in {\Theta }_{0}}}{P}_{\theta }\left( {W\left( \mathbf{X}\ri... | Proof: Fix \( \theta \in {\Theta }_{0} \) . Let \( {F}_{\theta }\left( w\right) \) denote the cdf of \( - W\left( \mathbf{X}\right) \) . Define\n\n\[ {p}_{\theta }\left( \mathbf{x}\right) = {P}_{\theta }\left( {W\left( \mathbf{X}\right) \geq W\left( \mathbf{x}\right) }\right) = {P}_{\theta }\left( {-W\left( \mathbf{X}\... | Yes |
Again consider the normal model of Example 8.3.28, but consider testing \( {H}_{0} : \mu \leq {\mu }_{0} \) versus \( {H}_{1} : \mu > {\mu }_{0} \) . By Exercise 8.37, the LRT rejects \( {H}_{0} \) for large values of \( W\left( \mathbf{X}\right) = \left( {\bar{X} - {\mu }_{0}}\right) /\left( {S/\sqrt{n}}\right) \) . T... | Consider any \( \mu \leq {\mu }_{0} \) and any \( \sigma \) :\n\n\[ \n{P}_{\mu ,\sigma }\left( {W\left( \mathbf{X}\right) \geq W\left( \mathbf{x}\right) }\right) = {P}_{\mu ,\sigma }\left( {\frac{\bar{X} - {\mu }_{0}}{S/\sqrt{n}} \geq W\left( \mathbf{x}\right) }\right) \n\]\n\n\[ \n= {P}_{\mu ,\sigma }\left( {\frac{\ba... | Yes |
Example 8.3.31 (Risk of UMP test) Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \) population, \( {\sigma }^{2} \) known. The UMP level \( \alpha \) test of \( {H}_{0} : \theta \geq {\theta }_{0} \) versus \( {H}_{1} : \theta < {\theta }_{0} \) is the test t... | \[ \beta \left( \theta \right) = {P}_{\theta }\left( {Z < - {z}_{\alpha } - \frac{\theta - {\theta }_{0}}{\sigma /\sqrt{n}}}\right) \] where \( Z \) has a \( \mathrm{n}\left( {0,1}\right) \) distribution. For \( \alpha = {.10} \), the risk function (8.3.12) for \( {c}_{\mathrm{I}} = 8 \) and \( {c}_{11} = 3 \) is shown... | Yes |
When we estimate \( \mu \) by \( \bar{X} \) , the probability that we are exactly correct, that is, \( P\left( {\bar{X} = \mu }\right) \), is 0 . However, with an interval estimator, we have a positive probability of being correct. The probability that \( \mu \) is covered by the interval \( \left\lbrack {\bar{X} - 1,\... | \[ P\left( {\mu \in \left\lbrack {\bar{X} - 1,\bar{X} + 1}\right\rbrack }\right) = P\left( {\bar{X} - 1 \leq \mu \leq \bar{X} + 1}\right) \] \[ = P\left( {-1 \leq \bar{X} - \mu \leq 1}\right) \] \[ = P\left( {-2 \leq \frac{\bar{X} - \mu }{\sqrt{1/4}} \leq 2}\right) \] \[ = P\left( {-2 \leq Z \leq 2}\right) \;\left( {\f... | Yes |
Example 9.1.6 (Scale uniform interval estimator) Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a uniform \( \left( {0,\theta }\right) \) population and let \( Y = \max \left\{ {{X}_{1},\ldots ,{X}_{n}}\right\} \) . We are interested in an interval estimator of \( \theta \) . We consider two candidate estima... | \[ {P}_{\theta }\left( {\theta \in \left\lbrack {{aY},{bY}}\right\rbrack }\right) = {P}_{\theta }\left( {{aY} \leq \theta \leq {bY}}\right) \] \[ = {P}_{\theta }\left( {\frac{1}{b} \leq \frac{Y}{\theta } \leq \frac{1}{a}}\right) \] \[ = {P}_{\theta }\left( {\frac{1}{b} \leq T \leq \frac{1}{a}}\right) \] \( \left( {T = ... | Yes |
Example 9.2.1 (Inverting a normal test) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \) and consider testing \( {H}_{0} : \mu = {\mu }_{0} \) versus \( {H}_{1} : \mu \neq {\mu }_{0} \) . For a fixed \( \alpha \) level, a reasonable test (in fact, the most powerful unbiased te... | \[ \bar{x} - {z}_{\alpha /2}\frac{\sigma }{\sqrt{n}} \leq {\mu }_{0} \leq \bar{x} + {z}_{\alpha /2}\frac{\sigma }{\sqrt{n}}. \] Since the test has size \( \alpha \), this means that \( P\left( {{H}_{0}\text{is rejected} \mid \mu = {\mu }_{0}}\right) = \alpha \) or, stated in another way, \( P\left( {{H}_{0}\text{is acc... | Yes |
For each \( {\theta }_{0} \in \Theta \), let \( A\left( {\theta }_{0}\right) \) be the acceptance region of a level \( \alpha \) test of \( {H}_{0} : \theta = {\theta }_{0} \) . For each \( \mathbf{x} \in \mathcal{X} \), define a set \( C\left( \mathbf{x}\right) \) in the parameter space by\n\n\[ \nC\left( \mathbf{x}\r... | Proof: For the first part, since \( A\left( {\theta }_{0}\right) \) is the acceptance region of a level \( \alpha \) test,\n\n\[ \n{P}_{{\theta }_{0}}\left( {\mathbf{X} \notin A\left( {\theta }_{0}\right) }\right) \leq \alpha \;\text{ and hence }\;{P}_{{\theta }_{0}}\left( {\mathbf{X} \in A\left( {\theta }_{0}\right) }... | Yes |
Example 9.2.8 (Gamma pivot) Suppose that \( {X}_{1},\ldots ,{X}_{n} \) are iid exponential \( \left( \lambda \right) \) . Then \( T = \sum {X}_{i} \) is a sufficient statistic for \( \lambda \) and \( T \sim \operatorname{gamma}\left( {n,\lambda }\right) \) . In the gamma pdf \( t \) and \( \lambda \) appear together a... | \[ Q\left( {T,\lambda }\right) \sim \operatorname{gamma}\left( {n,\lambda \left( {2/\lambda }\right) }\right) = \operatorname{gamma}\left( {n,2}\right) \] which does not depend on \( \lambda \) . The quantity \( Q\left( {T,\lambda }\right) = {2T}/\lambda \) is a pivot with a \( \operatorname{gamma}\left( {n,2}\right) \... | Yes |
In Example 9.2.3 we obtained a confidence interval for the mean, \( \lambda \), of the exponential \( \left( \lambda \right) \) pdf by inverting a level \( \alpha \) LRT of \( {H}_{0} : \lambda = {\lambda }_{0} \) versus \( {H}_{1} : \lambda \neq {\lambda }_{0} \). Now we also see that if we have a sample \( {X}_{1},\l... | If we choose constants \( a \) and \( \bar{b} \) to satisfy \( P\left( {a \leq {\chi }_{2n}^{2} \leq b}\right) = 1 - \alpha \), then\n\n\[ \n{P}_{\lambda }\left( {a \leq \frac{2T}{\lambda } \leq b}\right) = {P}_{\lambda }\left( {a \leq Q\left( {T,\lambda }\right) \leq b}\right) = P\left( {a \leq {\chi }_{2n}^{2} \leq b... | Yes |
Example 9.2.10 (Normal pivotal interval) It follows from Theorem 5.3.1 that if \( {X}_{1},\ldots ,{X}_{n} \) are iid \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \), then \( \left( {\bar{X} - \mu }\right) /\left( {\sigma /\sqrt{n}}\right) \) is a pivot. If \( {\sigma }^{2} \) is known, we can use this pivot to calcu... | \[ P\left( {-a \leq \frac{\bar{X} - \mu }{\sigma /\sqrt{n}} \leq a}\right) = P\left( {-a \leq Z \leq a}\right) ,\left( {Z\text{ is standard normal }}\right) \] and (by now) familiar algebraic manipulations give us the confidence interval \[ \left\{ {\mu : \bar{x} - a\frac{\sigma }{\sqrt{n}} \leq \mu \leq \bar{x} + a\fr... | Yes |
Example 9.2.11 (Shortest length binomial set) Sterne (1954) proposed the following method for constructing binomial confidence sets, a method that produces a set with the shortest length. Given \( \alpha \), for each value of \( p \) find the size \( \alpha \) acceptance region composed of the most probable \( x \) val... | To see the unexpected problems with this seemingly reasonable construction, consider a small example. Let \( X \sim \operatorname{binomial}\left( {3, p}\right) \) and use confidence coefficient \( 1 - \alpha = \) .442. Table 9.2.2 gives the acceptance regions obtained by the Sterne construction and the confidence sets ... | No |
Theorem 9.2.12 (Pivoting a continuous cdf) Let \( T \) be a statistic with continuous cdf \( {F}_{T}\left( {t \mid \theta }\right) \) . Let \( {\alpha }_{1} + {\alpha }_{2} = \alpha \) with \( 0 < \alpha < 1 \) be fixed values. Suppose that for each \( t \in \mathcal{T} \), the functions \( {\theta }_{\mathrm{L}}\left(... | Proof: We will prove only part (i). The proof of part (ii) is similar and is left as Exercise 9.19.\n\nAssume that we have constructed the \( 1 - \alpha \) acceptance region\n\n\[ \n\left\{ {t : {\alpha }_{1} \leq {F}_{T}\left( {t \mid {\theta }_{0}}\right) \leq 1 - {\alpha }_{2}}\right\}\n\]\n\nSince \( {F}_{T}\left( ... | No |
If \( {X}_{1},\ldots ,{X}_{n} \) are iid with pdf \( f\left( {x \mid \mu }\right) = {e}^{-\left( {x - \mu }\right) }{I}_{\lbrack \mu ,\infty )}\left( x\right) \), then \( Y = \min \left\{ {X}_{1}\right. \) , \( \left. {\ldots ,{X}_{n}}\right\} \) is sufficient for \( \mu \) with pdf \[ {f}_{Y}\left( {y \mid \mu }\right... | Fix \( \alpha \) and define \( {\mu }_{\mathrm{L}}\left( y\right) \) and \( {\mu }_{\mathrm{U}}\left( y\right) \) to satisfy \[ {\int }_{{\mu }_{\mathrm{U}}\left( y\right) }^{y}n{e}^{-n\left( {u - {\mu }_{\mathrm{U}}\left( y\right) }\right) }{du} = \frac{\alpha }{2},\;{\int }_{y}^{\infty }n{e}^{-n\left( {u - {\mu }_{\m... | Yes |
Theorem 9.2.14 (Pivoting a discrete cdf) Let \( T \) be a discrete statistic with cdf \( {F}_{T}\left( {t \mid \theta }\right) = P\left( {T \leq t \mid \theta }\right) \) . Let \( {\alpha }_{1} + {\alpha }_{2} = \alpha \) with \( 0 < \alpha < 1 \) be fixed values. Suppose that for each \( t \in \mathcal{T},{\theta }_{\... | Proof: We will only sketch the proof of part (i). The details, as well as the proof of part (ii), are left to Exercise 9.20.\n\nFirst recall Exercise 2.10, where it was shown that \( {F}_{T}\left( {T \mid \theta }\right) \) is stochastically greater than a uniform random variable, that is, \( {P}_{\theta }\left( {{F}_{... | No |
Example 9.2.15 (Poisson interval estimator) Let \( {X}_{1},\ldots ,{X}_{n} \) be a random sample from a Poisson population with parameter \( \lambda \) and define \( Y = \sum {X}_{i}.Y \) is sufficient for \( \lambda \) and \( Y \sim \operatorname{Poisson}\left( {n\lambda }\right) \) . Applying the above method with \(... | Recall the identity, from Example 3.3.1, linking the Poisson and gamma families. Applying that identity to the sums in (9.2.16), we can write (remembering that \( {y}_{0} \) is the observed value of \( Y \) )\n\n\[ \frac{\alpha }{2} = \mathop{\sum }\limits_{{k = 0}}^{{y}_{0}}{e}^{-{n\lambda }}\frac{{\left( n\lambda \ri... | Yes |
We now construct a credible set for the problem of Example 9.2.15. Let \( {X}_{1},\ldots ,{X}_{n} \) be iid Poisson \( \left( \lambda \right) \) and assume that \( \lambda \) has a gamma prior pdf, \( \lambda \sim \operatorname{gamma}\left( {a, b}\right) \). The posterior pdf of \( \lambda \) (see Exercise 7.24) is \[ ... | We can form a credible set for \( \lambda \) in many different ways, as any set \( A \) satisfying (9.2.18) will do. One simple way is to split the \( \alpha \) equally between the upper and lower endpoints. From (9.2.19) it follows that \( \frac{2\left( {{nb} + 1}\right) }{b}\lambda \sim {\chi }_{2\left( {a + \sum {x}... | Yes |
The \( {90}\% \) confidence and credible sets of Example 9.2.16 maintain their respective probability guarantees, but how do they fare under the other criteria? First, lets look at the credible probability of the confidence set (9.2.17), which is given by | \[ P\left\{ {\frac{1}{2n}{\chi }_{{2\sum x},1 - \alpha /2}^{2} \leq \lambda \leq \frac{1}{2n}{\chi }_{2\left( {{\sum x} + 1}\right) ,\alpha /2}^{2}}\right\} \] where \( \lambda \) has the distribution (9.2.19). Figure 9.2.4 shows the credible probability of the set (9.2.20), which is constant at \( 1 - \alpha \), along... | No |
Example 9.2.18 (Coverage of a normal credible set) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\theta ,{\sigma }^{2}}\right) \), and let \( \theta \) have the prior pdf \( \mathrm{n}\left( {\mu ,{\tau }^{2}}\right) \), where \( \mu ,\sigma \), and \( \tau \) are all known. In Example 7.2.16 we saw tha... | We now calculate the coverage probability of the Bayesian region (9.2.23). Under the classical model \( \bar{X} \) is the random variable, \( \theta \) is fixed, and \( \bar{X} \sim \mathrm{n}\left( {\theta ,{\sigma }^{2}/n}\right) \). For ease of notation define \( \gamma = {\sigma }^{2}/\left( {n{\tau }^{2}}\right) \... | Yes |
Example 9.3.1 (Optimizing length) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \), where \( \sigma \) is known. From the method of Section 9.2.2 and the fact that\n\n\[ Z = \frac{\bar{X} - \mu }{\sigma /\sqrt{n}} \]\n\nis a pivot with a standard normal distribution, any \( a ... | In Example 9.2.1 we took \( a = - {z}_{\alpha /2} \) and \( b = {z}_{\alpha /2} \), but no mention was made of optimality. If we take \( 1 - \alpha = {.90} \), then any of the following pairs of numbers give \( {90}\% \) intervals:\n\nThree \( {90}\% \) normal confidence intervals\n\n<table><thead><tr><th>\( a \)</th><... | Yes |
For normal intervals based on the pivot \( \frac{\bar{X} - \mu }{S/\sqrt{n}} \) we know that the shortest length \( 1 - \alpha \) confidence interval of the form\n\n\[ \bar{x} - b\frac{s}{\sqrt{n}} \leq \mu \leq \bar{x} - a\frac{s}{\sqrt{n}} \]\n\nhas \( a = - {t}_{n - 1,\alpha /2} \) and \( b = {t}_{n - 1,\alpha /2} \... | It is easy to see that if we had considered the criterion of expected length and wanted to find a \( 1 - \alpha \) interval to minimize\n\n\[ {\mathrm{E}}_{\sigma }\left( {\operatorname{Length}\left( S\right) }\right) = \left( {b - a}\right) \frac{{\mathrm{E}}_{\sigma }S}{\sqrt{n}} = \left( {b - a}\right) c\left( n\rig... | No |
Example 9.3.4 (Shortest pivotal interval) Suppose \( X \sim \operatorname{gamma}\left( {k,\beta }\right) \). The quantity \( Y = X/\beta \) is a pivot, with \( Y \sim \operatorname{gamma}\left( {k,1}\right) \), so we can get a confidence interval by finding constants \( a \) and \( b \) to satisfy\n\n(9.3.1)\n\n\[ P\le... | Although Theorem 9.3.2 is not directly applicable here, a modified argument can solve this problem. Condition (a) in Theorem 9.3.2 defines \( b \) as a function of \( a \), say \( b\left( a\right) \). We must solve the following constrained minimization problem:\n\n\[ \text{Minimize, with respect to a:}\frac{1}{a} - \f... | No |
Example 9.3.6 (UMA confidence bound) Let \( {X}_{1},\ldots ,{X}_{n} \) be iid \( \mathrm{n}\left( {\mu ,{\sigma }^{2}}\right) \), where \( {\sigma }^{2} \) is known. The interval\n\n\[ C\left( \bar{x}\right) = \left\{ {\mu : \mu \geq \bar{x} - {z}_{\alpha }\frac{\sigma }{\sqrt{n}}}\right\} \]\n\nis a \( 1 - \alpha \) U... | The more common two-sided interval,\n\n\[ C\left( \bar{x}\right) = \left\{ {\mu : \bar{x} - {z}_{\alpha /2}\frac{\sigma }{\sqrt{n}} \leq \mu \leq \bar{x} + {z}_{\alpha /2}\frac{\sigma }{\sqrt{n}}}\right\} ,\]\n\nis not UMA, since it is obtained by inverting the two-sided acceptance region from the test of \( {H}_{0} : ... | Yes |
The two-sided normal interval\n\n\[ C\left( \bar{x}\right) = \left\{ {\mu : \bar{x} - {z}_{\alpha /2}\frac{\sigma }{\sqrt{n}} \leq \mu \leq \bar{x} + {z}_{\alpha /2}\frac{\sigma }{\sqrt{n}}}\right\} \]\n\nis an unbiased interval. | It can be obtained by inverting the unbiased test of \( {H}_{0} : \mu = {\mu }_{0} \) versus \( {H}_{1} : \mu \neq {\mu }_{0} \) given in Example 8.3.20. Similarly, the interval (9.2.14) based on the \( t \) distribution is also an unbiased interval, since it also can be obtained by inverting a unbiased test (see Exerc... | No |
Theorem 9.3.9 (Pratt) Let \( X \) be a real-valued random variable with \( X \sim f\left( {x \mid \theta }\right) \) , where \( \theta \) is a real-valued parameter. Let \( C\left( x\right) = \left\lbrack {L\left( x\right), U\left( x\right) }\right\rbrack \) be a confidence interval for \( \theta \) . If \( L\left( x\r... | Proof: From the definition of expected value we can write\n\n\[ \n{\mathrm{E}}_{{\theta }^{ * }}\left( {\operatorname{Length}\left\lbrack {C\left( X\right) }\right\rbrack }\right) = {\int }_{\mathcal{X}}\operatorname{Length}\left\lbrack {C\left( x\right) }\right\rbrack f\left( {x \mid {\theta }^{ * }}\right) {dx} \n\]\... | Yes |
Example 9.3.11 (Poisson HPD region) In Example 9.2.16 we derived a \( 1 - \alpha \) credible set for a Poisson parameter. We now construct an HPD region. By Corollary 9.3.10, this region is given by \( \{ \lambda : \pi \left( {\lambda \mid \sum x}\right) \geq k\} \), where \( k \) is chosen so that | \[ 1 - \alpha = {\int }_{\{ \lambda : \pi \left( {\lambda \mid {\sum x}}\right) \geq k\} }\pi \left( {\lambda \mid \sum x}\right) {d\lambda }. \] Recall that the posterior pdf of \( \lambda \) is gamma \( \left( {a+\sum x,{\left\lbrack n + \left( 1/b\right) \right\rbrack }^{-1}}\right) \), so we need to find \( {\lambd... | Yes |
Example 10.1.2 (Consistency of \( \bar{X} \) ) Let \( {X}_{1},{X}_{2},\ldots \) be iid \( \mathrm{n}\left( {\theta ,1}\right) \), and consider the sequence\n\n\[ \n{\bar{X}}_{n} = \frac{1}{n}\mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i} \n\]\n\nRecall that \( {\bar{X}}_{n} \sim \mathrm{n}\left( {\theta ,1/n}\right) \), so... | \[ \n= {\int }_{-\epsilon }^{\epsilon }{\left( \frac{n}{2\pi }\right) }^{\frac{1}{2}}{e}^{-\left( {n/2}\right) {y}^{2}}{dy}\;\text{ (substitute }y = {\bar{x}}_{n} - \theta \text{ ) } \n\]\n\n\[ \n= {\int }_{-\epsilon \sqrt{n}}^{\epsilon \sqrt{n}}{\left( \frac{1}{2\pi }\right) }^{\frac{1}{2}}{e}^{-\left( {1/2}\right) {t... | Yes |
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