Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Lemma 8.9. Let \( m \in {L}_{a}^{2}, a > n/2 \), and let \( \lambda > 0 \) . Define the operator \( {T}_{\lambda } \) by \( {\left( {T}_{\lambda }f\right) }^{ \frown }\left( \xi \right) = m\left( {\lambda \xi }\right) \widehat{f}\left( \xi \right) \) . Then\n\n\[{\int }_{{\mathbb{R}}^{n}}{\left| {T}_{\lambda }f\left( x... | Proof. If \( \widehat{K} = m \) then by our hypothesis, \( {\left( 1 + {\left| x\right| }^{2}\right) }^{a/2}K\left( x\right) = R\left( x\right) \in {L}^{2} \) , and the kernel of \( {T}_{\lambda } \) is \( {\lambda }^{-n}K\left( {{\lambda }^{-1}x}\right) \) . Hence,\n\n\[{\int }_{{\mathbb{R}}^{n}}{\left| {T}_{\lambda }... | Yes |
Theorem 8.10 (Hörmander). Let \( m \) be such that for some \( a > n/2 \) ,\n\n\[ \mathop{\sup }\limits_{j}{\begin{Vmatrix}m\left( {2}^{j} \cdot \right) \psi \end{Vmatrix}}_{{L}_{a}^{2}} < \infty \]\n\nThen the operator \( T \) associated with the multiplier \( m \) is bounded on \( {L}^{p}\left( {\mathbb{R}}^{n}\right... | Proof. We apply Theorem 8.6. First, define the family of operators \( \left\{ {S}_{j}\right\} \) by \( {\left( {S}_{j}f\right) }^{ \frown }\left( \xi \right) = \psi \left( {{2}^{-j}\xi }\right) \widehat{f}\left( \xi \right) \) . Then inequality (8.8) holds by our choice of \( \psi \) . Now let \( \widetilde{\psi } \) b... | Yes |
If for \( k = \left\lbrack {n/2}\right\rbrack + 1, m \in {C}^{k} \) away from the origin, and if for \( \left| \beta \right| \leq k \)\n\n\[ \mathop{\sup }\limits_{R}{R}^{\left| \beta \right| }{\left( \frac{1}{{R}^{n}}{\int }_{R < \left| \xi \right| < {2R}}{\left| {D}^{\beta }m\left( \xi \right) \right| }^{2}d\xi \righ... | Proof. If we make the change of variables \( \xi \mapsto {R\xi } \) in inequality (8.10) and use the fact that \( {D}^{\beta }m\left( {R \cdot }\right) \left( \xi \right) = {R}^{\left| \beta \right| }\left( {{D}^{\beta }m}\right) \left( {R\xi }\right) \), we get\n\n\[ \mathop{\sup }\limits_{R}{\left( {\int }_{1 < \left... | Yes |
Theorem 8.13. Let \( m \) be a bounded function which has uniformly bounded variation on each dyadic interval in \( \mathbb{R} \) . Then \( m \) is a multiplier on \( {L}^{p}\left( \mathbb{R}\right) \) , \( 1 < p < \infty \) . | Proof. The proof is similar to the proof of Corollary 3.8. Given a dyadic interval \( {I}_{j} \), let \( {T}_{j} \) be the operator associated with the multiplier \( m{\chi }_{{I}_{j}} \) . We will consider the case \( {I}_{j} = \left( {{2}^{j},{2}^{j + 1}}\right) \) ; the other case is handled in exactly the same way.... | Yes |
Theorem 8.14. Suppose \( m \) is a bounded function in the plane, twice differentiable in each quadrant of \( {\mathbb{R}}^{2} \) and such that\n\n\[ \mathop{\sup }\limits_{j}{\int }_{{I}_{j}}\left| {\frac{\partial m}{\partial {t}_{1}}\left( {{t}_{1},{t}_{2}}\right) }\right| d{t}_{1} < \infty \]\n\n\[ \mathop{\sup }\li... | Proof. As before, we restrict our attention to dyadic intervals in \( {\mathbb{R}}_{ + } \) . Let \( {I}_{i} = \left( {{2}^{i},{2}^{i + 1}}\right) ,{I}_{j} = \left( {{2}^{j},{2}^{j + 1}}\right) \), and fix \( \left( {{\xi }_{1},{\xi }_{2}}\right) \in {I}_{i} \times {I}_{j} \) . Then\n\n\[ m\left( {{\xi }_{1},{\xi }_{2}... | Yes |
Theorem 8.15. The Bochner-Riesz multipliers \( {T}^{a} \) satisfy the following:\n\n(1) If \( a > \frac{n - 1}{2} \) then \( {T}^{a} \) is bounded on \( {L}^{p}\left( {\mathbb{R}}^{n}\right) ,1 \leq p \leq \infty \) .\n\n(2) If \( 0 < a \leq \frac{n - 1}{2} \) then \( {T}^{a} \) is bounded on \( {L}^{p}\left( {\mathbb{... | For the values of \( p \) not considered in Theorem 8.15, see Section 8.3 below. The value \( \left( {n - 1}\right) /2 \) is called the critical index.\n\nThe singularity of the Bochner-Riesz multipliers is on the circle \( \left| \xi \right| = 1 \) ; therefore, to prove Theorem 8.15 we are going to decompose the multi... | No |
Lemma 8.16. Given \( 0 < \delta < 1 \), let \( \phi \) be a function on \( \mathbb{R} \) which is supported on \( 1 - {4\delta } < t < 1 - \delta \) and is such that \( 0 \leq \phi \leq 1 \) and \( \left| {{D}^{\beta }\phi }\right| \leq C{\delta }^{-\left| \beta \right| } \) for any \( \beta \) . Then for any \( \epsil... | Proof. Fix \( K \) so that \( \widehat{K}\left( \xi \right) = \phi \left( \left| \xi \right| \right) \) and let \( a \) be a positive even integer. Then by inequality (1.20) and the Plancherel theorem,\n\n\[{\begin{Vmatrix}\left( 1 + {\left| x\right| }^{a}\right) K\end{Vmatrix}}_{2} \leq C{\begin{Vmatrix}\left( I + {\l... | Yes |
Lemma 8.17. If \( m \) is a function with compact support which is a multiplier on \( {L}^{p} \) for some \( p \), then \( \widehat{m} \in {L}^{p} \) . | Proof. Let \( f \in \mathcal{S} \) be such that \( \widehat{f} = 1 \) on the support of \( m \) . Then \( f \in {L}^{p} \) and so \( {T}_{m}f \in {L}^{p} \), but \( {\left( {T}_{m}f\right) }^{ \frown } = m\widehat{f} = m \) . | No |
Lemma 8.18. The Fourier transform of \( {\left( 1 - {\left| \xi \right| }^{2}\right) }_{ + }^{a} \) is | \[ {K}^{a}\left( x\right) = {\pi }^{-a}\Gamma \left( {a + 1}\right) {\left| x\right| }^{-\frac{n}{2} - a}{J}_{\frac{n}{2} + a}\left( {{2\pi }\left| x\right| }\right) ,\] where \( {J}_{\mu } \) is the Bessel function \[ {J}_{\mu }\left( t\right) = \frac{{\left( \frac{t}{2}\right) }^{\mu }}{\Gamma \left( {\mu + \frac{1}{... | Yes |
Lemma 8.20. If \( \Omega \in {L}^{q}\left( {S}^{n - 1}\right), q > 1 \), and\n\n\[ K\left( x\right) = \frac{\Omega \left( {x}^{\prime }\right) }{{\left| x\right| }^{n}}{\chi }_{\{ 1 < \left| x\right| \leq 2\} }, \]\n\nthen for \( a < 1/{q}^{\prime } \)\n\n\[ \left| {\widehat{K}\left( \xi \right) }\right| \leq C{\left| ... | Proof. If we rewrite the Fourier transform in polar coordinates we get\n\n\[ \widehat{K}\left( \xi \right) = {\int }_{{S}^{n - 1}}\Omega \left( u\right) {\int }_{1}^{2}{e}^{-{2\pi ir}\left( {u \cdot \xi }\right) }\frac{dr}{r}{d\sigma }\left( u\right) = {\int }_{{S}^{n - 1}}\Omega \left( u\right) I\left( {u \cdot \xi }\... | Yes |
Corollary 8.21. If \( \Omega \in {L}^{q}\left( {S}^{n - 1}\right), q > 1 \), and \( \int \Omega \left( u\right) {d\sigma }\left( u\right) = 0 \), then the singular integral\n\n\[ \n{Tf}\left( x\right) = \text{ p. v. }{\int }_{{\mathbb{R}}^{n}}\frac{\Omega \left( {y}^{\prime }\right) }{{\left| y\right| }^{n}}f\left( {x ... | If we define \( K \) as in Lemma 8.20 then\n\n\[ \n{Tf} = \mathop{\sum }\limits_{{j = - \infty }}^{\infty }{K}_{j} * f \n\]\n\nand the desired result follows immediately from Theorem 8.19. | Yes |
Corollary 8.22. If \( K \) and \( {K}_{j} \) are defined as in Theorem 8.19 then the square function\n\n\[ g\left( f\right) = {\left( \mathop{\sum }\limits_{{j = - \infty }}^{\infty }{\left| {K}_{j} * f\right| }^{2}\right) }^{1/2} \]\n\nis bounded on \( {L}^{p},1 < p < \infty \) . | Proof. First note that given a sequence \( \epsilon = \left\{ {\epsilon }_{j}\right\} \) such that for each \( j \) , \( {\epsilon }_{j} = \pm 1 \), if we define the operator\n\n\[ {T}_{\epsilon }f = \mathop{\sum }\limits_{j}{\epsilon }_{j}{K}_{j} * f \]\n\nthen it follows from the proof of Theorem 8.19 that \( {\begin... | Yes |
Theorem 8.25. Let \( {\left\{ {\sigma }_{j}\right\} }_{j \in \mathbb{Z}} \) be a sequence of finite Borel measures with \( \begin{Vmatrix}{\sigma }_{j}\end{Vmatrix} \leq C \) and such that for some \( a > 0 \)\n\n\[ \left| {{\widehat{\sigma }}_{j}\left( \xi \right) }\right| \leq C\min \left( {{\left| {2}^{j}{\xi }_{1}\... | Proof. Define the operator \( {S}_{j} \) by \( {\left( {S}_{j}\right) }^{ \frown }\left( \xi \right) = {\chi }_{{\Delta }_{j}}\left( {\xi }_{1}\right) \widehat{f}\left( \xi \right) \), where\n\n\[ {\Delta }_{j} = \left( {-{2}^{j + 1}, - {2}^{j}}\right\rbrack \cup \left\lbrack {{2}^{j},{2}^{j + 1}}\right) .\n\]\n\nThen\... | Yes |
Theorem 8.26. Let \( \left\{ {\mu }_{j}\right\} \) be a sequence of positive Borel measures on \( {\mathbb{R}}^{2} \) with \( \begin{Vmatrix}{\mu }_{j}\end{Vmatrix} \leq C \) and such that for some \( a > 0 \)\n\n\[ \left| {{\widehat{\mu }}_{j}\left( \xi \right) }\right| \leq C{\left| {2}^{j}{\xi }_{1}\right| }^{-a} \]... | Proof. Fix \( \phi \in \mathcal{S}\left( \mathbb{R}\right) \) such that \( \widehat{\phi }\left( 0\right) = 1 \), and for each \( j \) define the measure \( {\widetilde{\sigma }}_{j} \) by\n\n\[ \widehat{{\widehat{\sigma }}_{j}}\left( \xi \right) = {\widehat{\mu }}_{j}\left( \xi \right) - {\widehat{\mu }}_{j}\left( {0,... | Yes |
Lemma 8.27 (Van der Corput's Lemma). Let\n\n\[ \nI\left( {a, b}\right) = {\int }_{a}^{b}{e}^{{ih}\left( t\right) }{dt} \]\n\nThen\n\n(1) if \( \left| {{h}^{\prime }\left( t\right) }\right| \geq \lambda > 0 \) and \( {h}^{\prime } \) is monotonic,\n\n\[ \n\left| {I\left( {a, b}\right) }\right| \leq C{\lambda }^{-1} \]\n... | Proof. (1) If we integrate by parts we get\n\n\[ \nI\left( {a, b}\right) = {\int }_{a}^{b}i{h}^{\prime }\left( t\right) {e}^{{ih}\left( t\right) } \cdot \frac{dt}{i{h}^{\prime }\left( t\right) } = \frac{{e}^{{ih}\left( b\right) }}{i{h}^{\prime }\left( b\right) } - \frac{{e}^{{ih}\left( a\right) }}{i{h}^{\prime }\left( ... | Yes |
Lemma 8.28. Let\n\n\[ \nI\left( b\right) = {\int }_{1}^{b}{e}^{i\left( {t{\xi }_{1} + {t}^{2}{\xi }_{2}}\right) }{dt} \]\n\nThen for \( 1 < b < 2 \) and \( \left| {\xi }_{1}\right| > 1,\left| {I\left( b\right) }\right| \leq C{\left| {\xi }_{1}\right| }^{-1/2} \) . | Proof. Let \( h\left( t\right) = t{\xi }_{1} + {t}^{2}{\xi }_{2} \) . When \( \left| {\xi }_{1}\right| \geq 8\left| {\xi }_{2}\right| ,\left| {{h}^{\prime }\left( t\right) }\right| = \left| {{\xi }_{1} + {2t}{\xi }_{2}}\right| \geq \left| {\xi }_{1}\right| /2 \) , so by Van der Corput’s lemma, \( \left| {I\left( b\righ... | Yes |
Corollary 8.29. The operators \( {H}_{\Gamma } \) and \( {M}_{\Gamma } \) are bounded on \( {L}^{p}\left( {\mathbb{R}}^{2}\right) \) for \( 1 < p < \infty \) and \( 1 < p \leq \infty \) respectively. | Proof. Let \( {\sigma }_{j},{\mu }_{j} \) be as in (8.15) and (8.16). Then by Lemma 8.28, for \( \left| {\xi }_{1}\right| > 1 \) we have that\n\n\[ \left| {{\widehat{\mu }}_{0}\left( \xi \right) }\right| ,\left| {{\widehat{\sigma }}_{0}\left( \xi \right) }\right| \leq C{\left| {\xi }_{1}\right| }^{-1/2}. \]\n\n(To get ... | Yes |
Theorem 8.32. The characteristic function of a (Euclidean) ball in \( {\mathbb{R}}^{n} \) , \( n \geq 2 \), is not a multiplier on \( {L}^{p} \) if \( p \neq 2 \) . | This negative result was proved by C. Fefferman (The multiplier problem for the ball, Ann. of Math. 94 (1972), 330-336). | Yes |
Theorem 8.33. If for some \( p > 1 \) ,\n\n(8.28)\n\n\[ \parallel \widehat{f}{\parallel }_{{L}^{2}\left( {S}^{n - 1}\right) } \leq {C}_{p,2}\parallel f{\parallel }_{{L}^{p}\left( {\mathbb{R}}^{n}\right) } \]\n\nthen \( {T}^{a} \) is bounded on \( {L}^{p} \) for all a such that (8.26) holds. | C. Fefferman proved the restriction inequality for \( 1 < p < {4n}/\left( {{3n} + 1}\right) \) , which gives the condition \( a > \left( {n - 1}\right) /4 \) mentioned above. This result was improved by P. Tomas (A restriction theorem for the Fourier transform, Bull. Amer. Math. Soc. 81 (1975), 477-478), developing ear... | No |
Theorem 8.35. Let \( m \) be a bounded function which has uniformly bounded variation on each dyadic interval in \( \mathbb{R} \). Then if \( w \in {A}_{p},{T}_{m} \) is a bounded operator on \( {L}^{p}\left( {\mathbb{R}, w}\right) ,1 < p < \infty \) . | This result was proved by D. Kurtz; see the paper cited in Section 8.2 above. In it he also proved a weighted analogue of Theorem 8.14. | No |
Theorem 8.36. For \( n \geq 2 \), let \( k = \left\lbrack {n/2}\right\rbrack + 1 \) . Suppose \( n/k < p < \infty \) and \( w \in {A}_{{pk}/n} \), or \( 1 < p < {\left( n/k\right) }^{\prime } \) and \( {w}^{1 - {p}^{\prime }} \in {A}_{{p}^{\prime }k/n} \) . Then, if \( m \) is a multiplier which satisfies (8.10), \( {T... | Theorem 8.36 was proved by D. Kurtz and R. Wheeden (Results on weighted norm inequalities for multipliers, Trans. Amer. Math. Soc. 255 (1979), 343-362). | No |
Theorem 8.38. Given \( \Omega \in {L}^{\infty }\left( {S}^{n - 1}\right) \) such that \( \int \Omega \left( u\right) {d\sigma }\left( u\right) = 0 \), define the singular integral\n\n\[ \n{Tf}\left( x\right) = \text{ p. v. }{\int }_{{\mathbb{R}}^{n}}\frac{\Omega \left( {y}^{\prime }\right) }{{\left| y\right| }^{n}}f\le... | The proof of this result can be found in the paper by Duoandikoetxea and Rubio de Francia cited above. | No |
Lemma 9.1 (Cotlar’s Lemma). Let \( H \) be a Hilbert space, \( \left\{ {T}_{j}\right\} \) a sequence of bounded linear operators on \( H \) with adjoints \( \left\{ {T}_{j}^{ * }\right\} \), and \( \{ a\left( j\right) \} \) a sequence of non-negative numbers such that\n\n\[ \n{\begin{Vmatrix}{T}_{i}{T}_{j}^{ * }\end{Vm... | Proof. Let\n\n\[ \nS = \mathop{\sum }\limits_{{j = n}}^{m}{T}_{j} \n\]\n\nThen \( \parallel S\parallel = {\begin{Vmatrix}S{S}^{ * }\end{Vmatrix}}^{1/2} \) ; in fact, for any integer \( k > 0,\parallel S\parallel = {\begin{Vmatrix}{\left( S{S}^{ * }\right) }^{k}\end{Vmatrix}}^{1/{2k}} \) . But\n\n\[ \n{\left( S{S}^{ * }... | Yes |
Let \( H = {L}^{2}\left( \mathbb{R}\right) \) and for \( f \in {L}^{2}\left( \mathbb{R}\right) \) define\n\n\[ \n{T}_{j}f\left( x\right) = {\int }_{{2}^{j} < \left| t\right| \leq {2}^{j + 1}}\frac{f\left( {x - t}\right) }{t}{dt}.\n\]\n\nFor each integer \( j,\left| {{T}_{j}f\left( x\right) }\right| \leq {4Mf}\left( x\r... | Since \( {T}_{j}^{ * } = - {T}_{j} \), we only need to estimate \( \begin{Vmatrix}{{T}_{i}{T}_{j}}\end{Vmatrix} \) . If we let \( {K}_{j}\left( x\right) = \) \( {x}^{-1}{\chi }_{{\Delta }_{j}}\left( x\right) \), where \( {\Delta }_{j} = \left\{ {x \in \mathbb{R} : {2}^{j} < \left| x\right| < {2}^{j + 1}}\right\} \), th... | Yes |
Lemma 9.4. If \( \nu \) is a Carleson measure in \( {\mathbb{R}}_{ + }^{n + 1} \) and \( E \subset {\mathbb{R}}^{n} \) is open, then\n\n\[ \nu \left( \widehat{E}\right) \leq C\parallel \nu \parallel \left| E\right| \] | Proof. In \( {\mathbb{R}}^{n} \), form the Calderón-Zygmund decomposition of the characteristic function of \( E \) at height \( 1/2 \) . This yields a collection of disjoint dyadic cubes \( \left\{ {Q}_{j}\right\} \) such that\n\n\[ E \subset \mathop{\bigcup }\limits_{j}{Q}_{j}\;\text{ and }\;\left| E\right| \leq \mat... | Yes |
Theorem 9.5. Let \( \phi \) be a bounded, integrable function which is positive, radial and decreasing. For \( t > 0 \), let \( {\phi }_{t}\left( x\right) = {t}^{-n}\phi \left( {{t}^{-1}x}\right) \) . Then a measure \( \nu \) is a Carleson measure if and only if for every \( p,1 < p < \infty \) , \[ {\int }_{{\mathbb{R... | Proof. First suppose that \( \nu \) is a Carleson measure. Define the maximal operator \[ {M}_{\phi }f\left( x\right) = \sup \left\{ {\left| {{\phi }_{t} * f\left( y\right) }\right| : \left| {x - y}\right| < t}\right\} . \] It is easy to show (cf. Chapter 2, Section 8.7) that \[ {M}_{\phi }f\left( x\right) \leq {CMf}\l... | Yes |
Theorem 9.9. An operator \( T : \mathcal{S}\left( {\mathbb{R}}^{n}\right) \rightarrow {\mathcal{S}}^{\prime }\left( {\mathbb{R}}^{n}\right) \), associated with a standard kernel \( K \), extends to a bounded operator on \( {L}^{2}\left( {\mathbb{R}}^{n}\right) \) if and only if the following conditions are true:\n\n(1)... | We have already seen the necessity of these three conditions: (1) and (2) follow from Theorem 6.6 and the remarks after it, and we noted the necessity of (3) immediately after Definition 9.8. | No |
Corollary 9.10. If \( K \) is a standard kernel which is anti-symmetric and \( T \) is the operator defined by (9.5), then \( T \) is bounded on \( {L}^{2}\left( {\mathbb{R}}^{n}\right) \) if and only if \( {T1} \in {BMO} \) . | This follows immediately from Theorem 9.9 since \( T \) always has the \( {WBP} \) and \( {T}^{ * }1 = - {T1} \) . | Yes |
Corollary 9.12. The operators \( {T}_{k} \) are bounded on \( {L}^{2} \) and there exists a positive constant \( C \) such that \( \begin{Vmatrix}{T}_{k}\end{Vmatrix} \leq {C}^{k}{\begin{Vmatrix}{A}^{\prime }\end{Vmatrix}}_{\infty }^{k} \) . | Proof. From the proof of Theorem 9.9 we will see that the norm of an operator \( T \) depends linearly on the constants involved in the hypotheses; in the special case of Corollary 9.10 these reduce to the constants of \( K \) as a standard kernel and the \( {BMO} \) norm of \( {T}_{k}1 \) . \n\nSince a straightforward... | No |
Lemma 9.14. Let\n\n\\[ \n{R}_{N} = \\mathop{\\sum }\\limits_{{j = - N}}^{N}\\left( {{S}_{j}T{\\Delta }_{j} + {\\Delta }_{j}T{S}_{j} - {\\Delta }_{j}T{\\Delta }_{j}}\\right) \n\\]\n\nThen for \\( f, g \\in {C}_{c}^{\\infty }\\left( {\\mathbb{R}}^{n}\\right) \\) ,\n\n\\[ \n\\mathop{\\lim }\\limits_{{N \\rightarrow \\inft... | Proof. If we expand the sum for \\( {R}_{N} \\) we get\n\n\\[ \n{R}_{N} = {S}_{-N}T{S}_{-N} - {S}_{N + 1}T{S}_{N + 1}. \n\\]\n\nSince \\( {S}_{-N}f \\) converges to \\( f \\) in \\( \\mathcal{S}\\left( {\\mathbb{R}}^{n}\\right) \\), and since \\( T \\) is a continuous map from \\( \\mathcal{S}\\left( {\\mathbb{R}}^{n}\... | Yes |
(1) For all \( x,{\int }_{{\mathbb{R}}^{n}}\left| {{A}_{j, k}\left( {x, y}\right) }\right| {dy} \leq C{2}^{-\delta \left| {j - k}\right| } \) ;\n\n(2) for all \( y,{\int }_{{\mathbb{R}}^{n}}\left| {{A}_{j, k}\left( {x, y}\right) }\right| {dx} \leq C{2}^{-\delta \left| {j - k}\right| } \) . | Proof. By Lemma 9.15\n\n\[ \left| {{A}_{j, k}\left( {x, y}\right) }\right| = \left| {{\int }_{{\mathbb{R}}^{n}}{K}_{j}\left( {x, z}\right) \left\lbrack {{K}_{k}\left( {y, z}\right) - {K}_{k}\left( {y, x}\right) }\right\rbrack {dz}}\right| \]\n\n(9.8)\n\n\[ \leq C{\int }_{{\mathbb{R}}^{n}}{p}_{j}\left( {x - z}\right) \m... | Yes |
Lemma 2.1 The risk set \( S \) is a convex set. | Proof Suppose \( {x}_{1} = \left( {R\left( {{\theta }_{1},{d}_{1}}\right) ,\ldots, R\left( {{\theta }_{t},{d}_{1}}\right) }\right) \) and \( {x}_{2} = \left( {R\left( {{\theta }_{1},{d}_{2}}\right) ,\ldots, R\left( {{\theta }_{t},{d}_{2}}\right) }\right) \) are two elements of \( S \), and suppose \( \lambda \in \left(... | Yes |
Theorem 2.2 An equaliser decision rule \( {\delta }_{0} \) which is extended Bayes must be minimax. | Proof of Theorem 2.2. The proof here is almost the same. If we suppose \( {\delta }_{0} \) is not minimax, then there exists a \( {\delta }^{\prime } \) for which \( \mathop{\sup }\limits_{\theta }R\left( {\theta ,{\delta }^{\prime }}\right) < C \), where \( C \) is the common value of \( R\left( {\theta ,{\delta }_{0}... | Yes |
Consider a binomial experiment in which \( X \sim \operatorname{Bin}\left( {n,\theta }\right) \) for known \( n \) and unknown \( \theta \) . Suppose the prior density is a Beta density on \( \left( {0,1}\right) \) , \[ \pi \left( \theta \right) = \frac{{\theta }^{a - 1}{\left( 1 - \theta \right) }^{b - 1}}{B\left( {a,... | Ignoring all components of \( \pi \) and \( f \) which do not depend on \( \theta \), we have \[ \pi \left( {\theta \mid x}\right) \propto {\theta }^{a + x - 1}{\left( 1 - \theta \right) }^{n - x + b - 1}. \] This is also of Beta form, with the parameters \( a \) and \( b \) replaced by \( a + x \) and \( b + n - x \),... | Yes |
Example 3.2 Suppose \( {X}_{1},\ldots ,{X}_{n} \) are independent, identically distributed from the normal distribution \( N\left( {\theta ,{\sigma }^{2}}\right) \), where the mean \( \theta \) is unknown and the variance \( {\sigma }^{2} \) is known. Let us also assume that the prior density for \( \theta \) is \( N\l... | \[ \pi \left( \theta \right) f\left( {x;\theta }\right) \propto \exp \left\{ {-\frac{{\left( \theta - {\mu }_{0}\right) }^{2}}{2{\sigma }_{0}^{2}} - \mathop{\sum }\limits_{{i = 1}}^{n}\frac{{\left( {x}_{i} - \theta \right) }^{2}}{2{\sigma }^{2}}}\right\} . \] Completing the square shows that \[ \frac{{\left( \theta - {... | Yes |
Example 3.3 Here is an extension of the previous example in which the normal variance as well as the normal mean is unknown. It is convenient to write \( \tau \) in place of \( 1/{\sigma }^{2} \) and \( \mu \) in place of \( \theta \), so that \( \theta = \left( {\tau ,\mu }\right) \) may be reserved for the two-dimens... | \[ \pi \left( {\tau ,\mu }\right) = \frac{{\beta }^{\alpha }}{\Gamma \left( \alpha \right) }{\tau }^{\alpha - 1}{e}^{-{\beta \tau }} \cdot {\left( 2\pi \right) }^{-1/2}{\left( k\tau \right) }^{1/2}\exp \left\{ {-\frac{k\tau }{2}{\left( \mu - v\right) }^{2}}\right\} ,\] or more simply \[ \pi \left( {\tau ,\mu }\right) \... | Yes |
find a minimax estimator of \( \theta \) based on a single observation \( X \sim \operatorname{Bin}\left( {n,\theta }\right) \) with \( n \) known, under squared error loss \( L\left( {\theta, d}\right) = {\left( \theta - d\right) }^{2} \). | We know by Theorem 2.2 of Chapter 2 that, if we can find a Bayes (or extended Bayes) estimator that has constant mean squared error (that is, risk), this will also be a minimax rule.\n\nWe do not know all the possible Bayes estimators for this problem, but we do know a very large class of them, namely all those that ar... | No |
Example 4.1 Suppose \( X \sim \operatorname{Bin}\left( {{10},\theta }\right) \) and we want to test \( {H}_{0} : \theta \leq \frac{1}{2} \) against \( {H}_{1} : \theta > \frac{1}{2} \) . The obvious test will be: reject \( {H}_{0} \) whenever \( X \geq {k}_{\alpha } \), where \( {k}_{\alpha } \) is chosen so that \( \m... | we get the answers \( {0.00098},{0.01074},{0.05469} \) etc. for \( k = {10},9,8,\ldots \) . So, if \( \alpha = {0.05} \), the test that takes \( {k}_{\alpha } = 8 \) has size \( {0.054}\ldots \), which is no good, so, for a non-randomised test, we would have to take \( {k}_{\alpha } = 9 \), which has actual size only 0... | Yes |
Test \( {H}_{0} : \theta = {\theta }_{0} \) against \( {H}_{1} : \theta > {\theta }_{0} \) . | Consider the test of \( \theta = {\theta }_{0} \) against \( \theta = {\theta }_{1} \) for some \( {\theta }_{1} > {\theta }_{0} \) . Since\n\n\[ f\left( {x;\theta }\right) = \frac{1}{{\theta }^{n}}\exp \left\{ {-\frac{1}{\theta }\sum {x}_{i}}\right\} \]\n\nwe have\n\n\[ \frac{f\left( {x;{\theta }_{1}}\right) }{f\left(... | Yes |
Consider a simple one-parameter exponential family in which observations \( {X}_{1},\ldots ,{X}_{n} \) are independent, identically distributed from the density\n\n\[ f\left( {x;\theta }\right) = c\left( \theta \right) h\left( x\right) {e}^{{\theta \tau }\left( x\right) }.\]\n\nIf we redefine \( X \) to be the vector \... | \[ \frac{f\left( {x;{\theta }_{2}}\right) }{f\left( {x;{\theta }_{1}}\right) } = {\left\{ \frac{c\left( {\theta }_{2}\right) }{c\left( {\theta }_{1}\right) }\right\} }^{n}\exp \left\{ {\left( {{\theta }_{2} - {\theta }_{1}}\right) t\left( x\right) }\right\} . \]\n\nThis is non-decreasing in \( t\left( x\right) \), and ... | Yes |
Example 4.4 Suppose \( {X}_{1},\ldots ,{X}_{n} \) are independent, identically distributed from the uniform distribution on \( \left( {0,\theta }\right) \) | Define \( t\left( x\right) = \max \left( {{x}_{1},\ldots ,{x}_{n}}\right) \) . Then, for \( 0 < {\theta }_{1} \leq {\theta }_{2} \) ,\n\n\[ \frac{f\left( {x;{\theta }_{2}}\right) }{f\left( {x;{\theta }_{1}}\right) } = \left\{ \begin{array}{ll} {\left( \frac{{\theta }_{1}}{{\theta }_{2}}\right) }^{n} & \text{ if }0 \leq... | Yes |
As an example of a one-parameter family which is not MLR, suppose we have one observation from the Cauchy density\n\n\[ f\left( {x;\theta }\right) = \frac{1}{\pi \left\{ {1 + {\left( x - \theta \right) }^{2}}\right\} }.\] | Then the likelihood ratio is\n\n\[ \frac{f\left( {x;{\theta }_{2}}\right) }{f\left( {x;{\theta }_{1}}\right) } = \frac{1 + {\left( x - {\theta }_{1}\right) }^{2}}{1 + {\left( x - {\theta }_{2}\right) }^{2}}.\]\n\nIt is easily seen that this likelihood ratio is not a monotonic function of \( x \), and it may readily be ... | Yes |
Example 4.6 Suppose \( {X}_{1},\ldots ,{X}_{n} \) are independent, identically distributed from \( N\left( {\theta ,{\sigma }^{2}}\right) \) , with \( {\sigma }^{2} \) known. Consider \( {H}_{0} : \theta = 0 \) against \( {H}_{1} : \theta \neq 0 \) . Also suppose the prior \( {g}_{1} \) for \( \theta \) under \( {H}_{1... | \[ B = \frac{{P}_{1}}{{P}_{2}} \] where \[ {P}_{1} = {\left( 2\pi {\sigma }^{2}\right) }^{-n/2}\exp \left( {-\frac{1}{2{\sigma }^{2}}\sum {X}_{i}^{2}}\right) , \] \[ {P}_{2} = {\left( 2\pi {\sigma }^{2}\right) }^{-n/2}{\int }_{-\infty }^{\infty }\exp \left\{ {-\frac{1}{2{\sigma }^{2}}\sum {\left( {X}_{i} - \theta \righ... | Yes |
Example 4.7 Here is another example which illustrates even more strikingly the difference between the classical significance testing and Bayes factor approaches. Suppose \( X \sim \operatorname{Bin}\left( {n,\theta }\right) \) and consider testing \( {H}_{0} : \theta = {\theta }_{0} \) against \( {H}_{1} : \theta \neq ... | \[ {P}_{1} = \left( \begin{array}{l} n \\ x \end{array}\right) {\theta }_{0}^{x}{\left( 1 - {\theta }_{0}\right) }^{n - x} \] \[ {P}_{2} = \left( \begin{array}{l} n \\ x \end{array}\right) {\int }_{0}^{1}{\theta }^{x}{\left( 1 - \theta \right) }^{n - x}{d\theta } \] \[ = \left( \begin{array}{l} n \\ x \end{array}\right... | Yes |
Example 5.2 The Beta density\n\n\[ f\left( {x;a, b}\right) = \frac{\Gamma \left( {a + b}\right) }{\Gamma \left( a\right) \Gamma \left( b\right) }{x}^{a - 1}{\left( 1 - x\right) }^{b - 1},\;0 < x < 1, a > 0, b > 0 \] | is of exponential family form with \( \theta = \left( {a, b}\right) \), if we define \( {\tau }_{1}\left( x\right) = \log x,\;{\tau }_{2}\left( x\right) = \) \( \log \left( {1 - x}\right) \) . | Yes |
For any \( S \subseteq \{ 1,2,\ldots, k\} \), the joint distribution of \( \left\{ {{t}_{i}\left( X\right), i \in S}\right\} \) conditionally on \( \left\{ {{t}_{i}\left( X\right), i \notin S}\right\} \), is of exponential family form, with a distribution depending only on \( \left\{ {{\pi }_{i}\left( \theta \right), i... | Write \( {T}_{1},\ldots ,{T}_{k} \) for \( {t}_{1}\left( X\right) ,\ldots ,{t}_{k}\left( X\right) \) . By Lemma 5.1,\n\n\[ \mathop{\Pr }\limits_{\theta }\left\{ {{T}_{1} = {y}_{1},\ldots ,{T}_{k} = {y}_{k}}\right\} = c{\left( \theta \right) }^{n}{h}_{0}\left( y\right) \exp \left\{ {\mathop{\sum }\limits_{{i = 1}}^{k}{\... | Yes |
Lemma 6.1 Let \( t\left( x\right) \) denote some function of \( x \) . Then the following are equivalent:\n\n(i) There exist functions \( h\left( x\right) \) and \( g\left( {t;\theta }\right) \) such that\n\n\[ f\left( {x;\theta }\right) = h\left( x\right) g\left( {t\left( x\right) ;\theta }\right) .\](6.1)\n\n(ii) For... | Proof That (i) implies (ii) is obvious.\n\nConversely, suppose (ii) holds. Fix some reference value \( {\theta }_{0} \) . For any \( \theta \),\n\n\[ \frac{f\left( {x;\theta }\right) }{f\left( {x;{\theta }_{0}}\right) } = {\Lambda }_{x}\left( {\theta ,{\theta }_{0}}\right) = {g}^{ * }\left( {t\left( x\right) ,\theta ,{... | Yes |
Lemma 6.2 If \( T \) and \( S \) are minimal sufficient statistics, then there exist injective functions \( {g}_{1} \) and \( {g}_{2} \) such that \( T = {g}_{1}\left( S\right) \) and \( S = {g}_{2}\left( T\right) \) . | Proof The definition of minimal sufficiency implies that there must exist some functions \( {g}_{1} \) and \( {g}_{2} \) such that \( T = {g}_{1}\left( S\right), S = {g}_{2}\left( T\right) \) . The task is to prove that \( {g}_{1} \) and \( {g}_{2} \) are injective on the ranges of \( S \) and \( T \) respectively.\n\n... | Yes |
Consider \( {X}_{1},\ldots ,{X}_{n} \) independent, identically distributed from \( N\left( {\mu ,{\sigma }^{2}}\right) \). Then\n\n\[ f\left( {x;\mu ,{\sigma }^{2}}\right) = {\left( 2\pi {\sigma }^{2}\right) }^{-n/2}\exp \left\{ {-\frac{\sum {x}_{i}^{2}}{2{\sigma }^{2}} + \frac{\mu \sum {x}_{i}}{{\sigma }^{2}} - \frac... | Thus\n\n\[ \frac{f\left( {x;{\mu }_{1},{\sigma }_{1}^{2}}\right) }{f\left( {x;{\mu }_{2},{\sigma }_{2}^{2}}\right) } = b\left( {{\mu }_{1},{\sigma }_{1}^{2},{\mu }_{2},{\sigma }_{2}^{2}}\right) \exp \left\{ {\sum {x}_{i}^{2}\left( {\frac{1}{2{\sigma }_{2}^{2}} - \frac{1}{2{\sigma }_{1}^{2}}}\right) +\sum {x}_{i}\left( ... | Yes |
Example 6.2 The reasoning in Example 6.1 is not specific to the normal distribution but applies whenever we have a full exponential family. Specifically, if we have an exponential family in its natural parametrisation with natural statistics \( \left( {{T}_{1},\ldots ,{T}_{k}}\right) = \left( {{t}_{1}\left( X\right) ,\... | \[ f\left( {x;\theta }\right) = c\left( \theta \right) h\left( x\right) \exp \left\{ {\mathop{\sum }\limits_{{i = 1}}^{k}{\theta }_{i}{t}_{i}\left( x\right) }\right\} ,\] then \( \left( {{T}_{1},\ldots ,{T}_{k}}\right) \) is sufficient for \( \theta \) by the factorisation theorem. If, in addition, \( \Theta \) contain... | Yes |
Lemma 6.3 If \( T = \left( {{T}_{1},\ldots ,{T}_{k}}\right) \) is the natural statistic for a full exponential family in its natural parametrisation, and if \( \Theta \) contains an open rectangle in \( {\mathbb{R}}^{k} \), then \( T \) is complete. | Proof This follows from the first of the technical results given in Section 5.1.4. | No |
Suppose the density of the statistic \( T \) satisfies\n\n\[ f\left( {t;\theta }\right) = \left\{ \begin{array}{ll} h\left( t\right) c\left( \theta \right) & \text{ if }0 \leq t \leq \theta \\ 0 & \text{ if }t > \theta \end{array}\right. \]\n\nwith \( h\left( t\right) \neq 0 \). For example, the uniform density is of t... | proof of this: use the factorisation theorem | No |
Theorem 6.2 Suppose \( X \) has density \( f\left( {x;\theta }\right) \) and \( T\left( X\right) \) is sufficient and complete for \( \theta \) . Then \( T \) is minimal sufficient. | Proof We know from the Remark following Theorem 6.1 that there exists a minimal sufficient statistic. By Lemma 6.2 this is unique up to one-to-one transformations, so call this \( S \) . Then \( S = {g}_{1}\left( T\right) \) for some function \( {g}_{1} \) .\n\nDefine \( {g}_{2}\left( S\right) = \mathbb{E}\{ T \mid S\}... | Yes |
Theorem 6.3 Suppose we want to estimate a real-valued parameter \( \theta \) with an estimator \( d\left( X\right) \) say. Suppose the loss function \( L\left( {\theta, d}\right) \) is a convex function of \( d \) for each \( \theta \) . Let \( {d}_{1}\left( X\right) \) be an unbiased estimator for \( \theta \) and sup... | Proof That \( \chi \left( T\right) \) is unbiased follows from the iterated expectation formula,\n\n\[ {\mathbb{E}}_{\theta }\chi \left( T\right) = {\mathbb{E}}_{\theta }\left\{ {\mathbb{E}\left( {{d}_{1}\left( X\right) \mid T}\right) }\right\} = {\mathbb{E}}_{\theta }{d}_{1}\left( X\right) = \theta . \]\n\nFor the ris... | Yes |
Lemma 7.1 (the generalised Neyman-Pearson Theorem) Consider the test\n\n\[ \n{\phi }^{\prime }\left( x\right) = \left\{ \begin{array}{ll} 1 & \text{ if }{f}_{0}\left( x\right) > {k}_{1}{f}_{1}\left( x\right) + \cdots + {k}_{m}{f}_{m}\left( x\right) , \\ \gamma \left( x\right) & \text{ if }{f}_{0}\left( x\right) = {k}_{... | The proof is omitted, being an elementary extension of the ordinary Neyman-Pearson Theorem from Chapter 4. | No |
Lemma 7.2 Let \( {\theta }_{a} < {\theta }_{b} < {\theta }_{c} \) and consider the set\n\n\[ \nS\left( {{K}_{1},{K}_{2}}\right) = \left\{ {x \in \mathbb{R} : {K}_{1}{e}^{{\theta }_{a}x} + {K}_{2}{e}^{{\theta }_{c}x} > {e}^{{\theta }_{b}x}}\right\} .\n\]\n\nThen:\n\n(i) For any \( {K}_{1} > 0,{K}_{2} > 0 \), the set \( ... | (i) Let \( g\left( x\right) = {K}_{1}\exp \left\{ {\left( {{\theta }_{a} - {\theta }_{b}}\right) x}\right\} + {K}_{2}\exp \left\{ {\left( {{\theta }_{c} - {\theta }_{b}}\right) x}\right\} - 1 \) . Then \( g \) is a convex function, and \( g \rightarrow + \infty \) as \( x \rightarrow \pm \infty \) . Thus \( g \) has no... | Yes |
Example 7.1 Suppose \( {X}_{1},\ldots ,{X}_{n} \) are independent, identically distributed \( N\left( {\theta ,1}\right) \) . Then the minimal sufficient statistic \( T = \bar{X} \) has the \( N\left( {\theta ,1/n}\right) \) distribution. Consider the test | \[ \phi \left( x\right) = \left\{ \begin{array}{ll} 1 & \text{ if }\bar{x} < {t}_{1}\text{ or }\bar{x} > {t}_{2} \\ 0 & \text{ if }{t}_{1} \leq \bar{x} \leq {t}_{2} \end{array}\right. \] Then \( w\left( \theta \right) = {\mathbb{E}}_{\theta }\{ \phi \left( X\right) \} \) is given by \[ w\left( \theta \right) = \Phi \le... | Yes |
Example 7.3 Suppose \( X \) and \( Y \) are independent Poisson random variables with means \( \lambda \) and \( \mu \) respectively, and we want to test \( {H}_{0} : \lambda \leq \mu \) against \( {H}_{1} : \lambda > \mu \) . | The joint probability mass function is\n\n\[ \n f\left( {x, y}\right) = {e}^{-\left( {\lambda + \mu }\right) }\frac{{e}^{x\log \lambda + y\log \mu }}{x!y!}, \]\n\nwhich is of exponential family form with natural parameters \( \left( {\log \lambda ,\log \mu }\right) \) . We identify \( \left( {{T}_{1},{T}_{2},{\theta }_... | Yes |
Example 7.4 Let \( {X}_{1},\ldots ,{X}_{n} \) be independent, identically distributed \( N\left( {\theta ,{\sigma }^{2}}\right) \), where \( \theta \) and \( {\sigma }^{2} \) are both unknown but our interest is in \( \theta \) (so already we are making a slight extension of the above framework, by allowing the nuisanc... | So \( T = \sqrt{n}\left( {\bar{X} - \theta }\right) /{s}_{X} \) is pivotal. Usually we define \( {c}_{1} = - {c}_{2} \), where \( \Pr \left\{ {T > {c}_{2}}\right\} = \alpha /2 \) , calculated from tables of the \( {t}_{n - 1} \) distribution, and hence\n\n\[ \mathop{\Pr }\limits_{\theta }\left\{ {-{c}_{2} \leq \frac{\s... | Yes |
Example 7.5 Suppose \( {X}_{1},\ldots ,{X}_{n} \) are independent, identically distributed from the uniform distribution on \( \left( {0,\theta }\right) \) for unknown \( \theta > 0 \) . | A complete sufficient statistic is \( T = \max \left\{ {{X}_{1},\ldots ,{X}_{n}}\right\} \) and we calculate\n\n\[ \mathop{\Pr }\limits_{\theta }\left\{ {\frac{T}{\theta } \leq y}\right\} = \mathop{\Pr }\limits_{\theta }\left\{ {{X}_{1} \leq {y\theta },\ldots ,{X}_{n} \leq {y\theta }}\right\} = {y}^{n}, \]\n\nfor any \... | Yes |
Consider \( {X}_{1},\ldots ,{X}_{n} \) independent, identically distributed from \( N\left( {\mu ,\tau }\right) \), with \( \mu \) and \( \tau \) both unknown. Then | \[ L\left( {\mu ,\tau }\right) = {\left( 2\pi \tau \right) }^{-n/2}\exp \left\{ {-\frac{1}{2}\mathop{\sum }\limits_{i}\frac{{\left( {X}_{i} - \mu \right) }^{2}}{\tau }}\right\} . \] It is an easy exercise to check that \[ \widehat{\mu } = \bar{X},\;\left( { = \frac{1}{n}\sum {X}_{i}}\right) \] \[ \widehat{\tau } = \fra... | No |
Consider \( {X}_{1},\ldots ,{X}_{n} \), independent, identically distributed from the uniform distribution on \( (0,\theta \rbrack \), where \( \theta \) is the unknown parameter. The likelihood function is\n\n\[ L\left( \theta \right) = \frac{1}{{\theta }^{n}}I\left\{ {{X}_{1} \leq \theta ,\ldots ,{X}_{n} \leq \theta ... | where \( I\left( \cdot \right) \) is the indicator function. Equation (8.3) is strictly decreasing in \( \theta \) over the range for which it is non-zero, which is for \( \theta \geq \max \left\{ {{X}_{1},\ldots ,{X}_{n}}\right\} \) . Therefore the MLE in this case is \( \widehat{\theta } = \max \left\{ {{X}_{1},\ldot... | Yes |
Example 8.3 Suppose \( {X}_{1},{X}_{2} \) are independent, identically distributed from the Cauchy density\n\n\[ f\left( {x;\theta }\right) = \frac{1}{\pi \left\{ {1 + {\left( x - \theta \right) }^{2}}\right\} },\; - \infty < x < \infty . \]\n\nThe likelihood equation is\n\n\[ \frac{\left( {X}_{1} - \theta \right) }{1 ... | The following statements are left as an exercise for the reader:\n\n(a) If \( \left| {{X}_{1} - {X}_{2}}\right| \leq 2 \) then there is a unique solution to (8.4) given by \( \widehat{\theta } = \left( {{X}_{1} + {X}_{2}}\right) /2 \) and this maximises the likelihood function.\n\n(b) If \( \left| {{X}_{1} - {X}_{2}}\r... | No |
Example 8.4 Suppose \( {X}_{1},\ldots ,{X}_{n} \) are independent, identically distributed from an exponential distribution with mean \( \theta \), so that the common density is \( f\left( {x;\theta }\right) = \) \( {\theta }^{-1}\exp \left( {-x/\theta }\right) ,0 < x < \infty ,0 < \theta < \infty \) . Let \( {S}_{n} =... | \[ {l}_{n}\left( \theta \right) = - n\log \theta - \frac{{S}_{n}}{\theta } \] \[ {l}_{n}^{\prime }\left( \theta \right) = - \frac{n}{\theta } + \frac{{S}_{n}}{{\theta }^{2}} \] \[ {l}_{n}^{\prime \prime }\left( \theta \right) = \frac{n}{{\theta }^{2}} - 2\frac{{S}_{n}}{{\theta }^{3}} \] It then follows that the likelih... | Yes |
Example 9.1 We consider first the location model, which is the simplest example of a transformation model, the general theory of which was described in Chapter 5. We have \( {X}_{1},\ldots ,{X}_{n} \) independent random variables with\n\n\[ \n{X}_{j} = \theta + {\epsilon }_{j}, j = 1,\ldots, n,\n\]\n\nwhere \( {\epsilo... | Let \( a = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \), where \( {a}_{j} = {x}_{j} - \widehat{\theta } \) : it is readily shown that \( a \) is ancillary. We may write \( {x}_{j} = {a}_{j} + \widehat{\theta } \), so that the log-likelihood may be written\n\n\[ \nl\left( {\theta ;\widehat{\theta }, a}\right) = \sum g\lef... | No |
Example 9.2 As a further example, let \( {X}_{1},\ldots ,{X}_{n} \) be an independent sample from a full \( \left( {m, m}\right) \) exponential density\n\n\[ \exp \left\{ {{x}^{T}\theta - k\left( \theta \right) + D\left( x\right) }\right\} \]\n\nThe log-likelihood is, ignoring an additive constant,\n\n\[ l\left( \theta... | Since \( \widehat{\theta } \) satisfies the likelihood equation\n\n\[ \sum {x}_{j} - n{k}^{\prime }\left( \theta \right) = 0 \]\n\nthe log-likelihood may be written\n\n\[ l\left( {\theta ;\widehat{\theta }}\right) = n{k}^{\prime }{\left( \widehat{\theta }\right) }^{T}\theta - {nk}\left( \theta \right) . \] | No |
Example 10.1 Suppose for instance that we wish to predict a new observation \( Z \) from \( N\left( {\mu ,{\sigma }^{2}}\right) \), with known variance \( {\sigma }^{2} \) but unknown mean \( \mu \), on the basis of a set \( {X}_{1},\ldots ,{X}_{n} \) of independent identically distributed observations, with mean \( \b... | Then\n\n\[\n\left( {Z - \bar{X}}\right) /\{ \sigma \sqrt{1 + 1/n}\}\n\]\n\nis pivotal, being distributed as \( N\left( {0,1}\right) \), and can be used to construct a prediction interval for \( Z \) : details are worked out in Problem 10.1. | No |
Example 10.2 (Barndorff-Nielsen and Cox,1996) Suppose \( {X}_{1},\ldots ,{X}_{n + 1} \) are from a first-order autoregressive process\n\n\[ \n{X}_{1} \sim N\left( {\mu ,\frac{{\sigma }^{2}}{1 - {\rho }^{2}}}\right) \n\]\n\n(10.2)\n\n\[ \n{X}_{i + 1} = \mu + \rho \left( {{X}_{i} - \mu }\right) + {\epsilon }_{i + 1}, \n\... | We first calculate the maximum likelihood estimator (MLE) of \( \mu \) . The likelihood function for \( \mu \) is derived by writing the joint density of \( {X}_{1},\ldots ,{X}_{n} \) in the form\n\n\[ \nf\left( {X}_{1}\right) \mathop{\prod }\limits_{{i = 1}}^{{n - 1}}f\left( {{X}_{i + 1} \mid {X}_{i}}\right) = {\left(... | Yes |
Suppose \( {Z}_{1},\ldots ,{Z}_{n} \) are independent \( N\left( {\mu ,{\sigma }^{2}}\right) \) and, conditionally on \( \left( {{Z}_{1},\ldots ,{Z}_{n}}\right) \) , \( {X}_{1},\ldots ,{X}_{n} \) are independent with \( {X}_{i} \sim N\left( {{Z}_{i},{\tau }^{2}}\right) \) . We observe \( {X}_{1},\ldots ,{X}_{n} \) and ... | In this case the obvious estimator of \( \mu \) (which is also the MLE) is \( \widehat{\mu } = \bar{X} = {n}^{-1}\mathop{\sum }\limits_{{i = 1}}^{n}{X}_{i} \) , so it is natural to consider pivotal statistics of the form \[ {T}_{\lambda } = {Z}_{1} - \lambda {X}_{1} - \left( {1 - \lambda }\right) \bar{X} \] (10.5) wher... | No |
Example 10.4 Suppose \( g \) is the Gamma \( \left( {\alpha ,\theta }\right) \) density with known shape parameter \( \alpha \) :\n\n\[ g\left( {z;\theta }\right) = \frac{{\theta }^{\alpha }{z}^{\alpha - 1}{e}^{-{\theta z}}}{\Gamma \left( \alpha \right) }.\n\]\n\nWe may assume \( {X}_{1},\ldots ,{X}_{n} \) are independ... | To evaluate (10.9) based on (10.16), we must integrate with respect to both \( z \) and \( x \), using (10.12) and (10.13). As a side calculation, if \( Z \sim \operatorname{Gamma}\left( {\alpha ,\theta }\right) \), we have\n\n\[ \mathbb{E}\left( {Z}^{r}\right) = \frac{{\theta }^{-r}\Gamma \left( {\alpha + r}\right) }{... | No |
Suppose \( Y = \left( {{X}_{1},\ldots ,{X}_{n}}\right), Z = \left( {{X}_{n + 1},\ldots ,{X}_{n + m}}\right) \), where \( {X}_{1},\ldots ,{X}_{n + m} \) are independent Bernoulli random variables with \( \Pr \left\{ {{X}_{i} = 1}\right\} = 1 - \Pr \left\{ {{X}_{i} = 0}\right\} = \theta \) for all \( i \) . Then \( S = \... | \[ \Pr \{ S = s \mid R = s + t\} = \frac{\left( \begin{array}{l} n \\ s \end{array}\right) \left( \begin{matrix} m \\ t \end{matrix}\right) }{\left( \begin{matrix} n + m \\ s + t \end{matrix}\right) } \] | Yes |
Example 10.6 Suppose \( Y = \left( {{X}_{1},\ldots ,{X}_{n}}\right), Z = {X}_{n + 1} \), where \( {X}_{1},\ldots ,{X}_{n + 1} \) are independent from the uniform distribution on \( \left( {0,\theta }\right) \) for some unknown \( \theta > 0 \) . Let \( {M}_{j} = \max \left\{ {{X}_{1},\ldots ,{X}_{j}}\right\} \) . Then ... | The conditions of Definition 2 are now satisfied. When \( t = 0 \), the joint density of \( \left( {S, T}\right) \) , evaluated at \( \left( {s, t}\right) \), is \( n{s}^{n - 1}{\theta }^{-n} \cdot s{\theta }^{-1} \) (the first factor is the marginal density of \( S \) and the second is the conditional probability that... | Yes |
Example 10.7 Suppose \( {X}_{1},\ldots ,{X}_{n}, Z \) are independent \( N\left( {\mu ,{\sigma }^{2}}\right) \), where \( \mu \) and \( {\sigma }^{2} \) are both unknown. Recall from our earlier example 10.1 that, if \( {\sigma }^{2} \) is known, then \( T = \) \( \sqrt{\frac{n}{n + 1}}\left( {Z - \bar{X}}\right) \) is... | The usual estimator of \( {\sigma }^{2} \) is \( {s}^{2} = \frac{1}{n - 1}\mathop{\sum }\limits_{{i = 1}}^{n}{\left( {X}_{i} - \bar{X}\right) }^{2} \), for which \( \left( {n - 1}\right) \frac{{s}^{2}}{{\sigma }^{2}} \sim {\chi }_{n - 1}^{2} \) . Thus we suggest the estimator \( \widetilde{\theta } = {s}^{-1} \) and co... | Yes |
Suppose \( {X}_{1},\ldots ,{X}_{n}, Z \) are independent from a Poisson distribution with common mean \( \theta \) . The MLE is \( \widehat{\theta } = \bar{X} \) ; since \( n\bar{X} \) has a Poisson distribution with mean \( {n\theta } \) , the distribution of \( \widehat{\theta } \) is | \[ \Pr \{ \widehat{\theta } = t\} = \frac{{e}^{-{n\theta }}{\left( n\theta \right) }^{nt}}{\left( {nt}\right) !}, t = 0,\frac{1}{n},\frac{2}{n},\ldots \] | Yes |
Theorem 1.1.4 For any three events, \( A, B \), and \( C \), defined on a sample space \( S \), c. Distributive Laws \( \;A \cap \left( {B \cup C}\right) = \left( {A \cap B}\right) \cup \left( {A \cap C}\right) \) | Proof: The proof of much of this theorem is left as Exercise 1.3. Also, Exercises 1.9 and 1.10 generalize the theorem. To illustrate the technique, however, we will prove the Distributive Law:\n\n\[ A \cap \left( {B \cup C}\right) = \left( {A \cap B}\right) \cup \left( {A \cap C}\right) . \]\n\n(You might be familiar w... | No |
Theorem 1.2.8 If \( P \) is a probability function and \( A \) is any set in \( \mathcal{B} \), then\n\na. \( P\left( \varnothing \right) = 0 \), where \( \varnothing \) is the empty set;\n\nb. \( P\left( A\right) \leq 1 \) ;\n\nc. \( P\left( {A}^{\mathrm{c}}\right) = 1 - P\left( A\right) \) . | Proof: It is easiest to prove (c) first. The sets \( A \) and \( {A}^{\mathrm{c}} \) form a partition of the sample space, that is, \( S = A \cup {A}^{\mathrm{c}} \) . Therefore,\n\n(1.2.4)\n\n\[ P\left( {A \cup {A}^{c}}\right) = P\left( S\right) = 1 \]\n\nby the second axiom. Also, \( A \) and \( {A}^{c} \) are disjoi... | Yes |
Theorem 1.2.9 If \( P \) is a probability function and \( A \) and \( B \) are any sets in \( \mathcal{B} \), then\n\na. \( P\left( {B \cap {A}^{\mathrm{c}}}\right) = P\left( B\right) - P\left( {A \cap B}\right) \) ;\n\nb. \( P\left( {A \cup B}\right) = P\left( A\right) + P\left( B\right) - P\left( {A \cap B}\right) \)... | Proof: To establish (a) note that for any sets \( A \) and \( B \) we have\n\n\[ B = \{ B \cap A\} \cup \left\{ {B \cap {A}^{c}}\right\} \]\n\nand therefore\n\n(1.2.6)\n\n\[ P\left( B\right) = P\left( {\{ B \cap A\} \cup \left\{ {B \cap {A}^{\mathrm{c}}}\right\} }\right) = P\left( {B \cap A}\right) + P\left( {B \cap {A... | No |
Example 1.2.10 (Bonferroni’s Inequality) Bonferroni’s Inequality is particularly useful when it is difficult (or even impossible) to calculate the intersection probability, but some idea of the size of this probability is desired. Suppose \( A \) and \( B \) are two events and each has probability .95 . Then the probab... | \[ P\left( {A \cap B}\right) \geq P\left( A\right) + P\left( B\right) - 1 = {.95} + {.95} - 1 = {.90}. \] | Yes |
Theorem 1.2.14 If a job consists of \( k \) separate tasks, the ith of which can be done in \( {n}_{i} \) ways, \( i = 1,\ldots, k \), then the entire job can be done in \( {n}_{1} \times {n}_{2} \times \cdots \times {n}_{k} \) ways. | Proof: It suffices to prove the theorem for \( k = 2 \) (see Exercise 1.15). The proof is just a matter of careful counting. The first task can be done in \( {n}_{1} \) ways, and for each of these ways we have \( {n}_{2} \) choices for the second task. Thus, we can do the job in\n\n\[ \underset{{n}_{1}\text{ terms }}{\... | No |
What is the probability of having four aces? | If we specify that four of the cards are aces, then there are 48 different ways of specifying the fifth card. Thus, \[ P\left( \text{ four aces }\right) = \frac{48}{2,{598},{960}} \] | Yes |
Example 1.2.19 (Sampling with replacement) Consider sampling \( r = 2 \) items from \( n = 3 \) items, with replacement. The outcomes in the ordered and unordered sample spaces are these. | <table><thead><tr><th>Unordered</th><th>\( \{ 1,1\} \)</th><th>\( \{ 2,2\} \)</th><th>\( \{ 3,3\} \)</th><th>\( \{ 1,2\} \)</th><th>\( \{ 1,3\} \)</th><th>\( \{ 2,3\} \)</th></tr></thead><tr><td>Ordered</td><td>\( \left( {1,1}\right) \)</td><td>\( \left( {2,2}\right) \)</td><td>\( \left( {3,3}\right) \)</td><td>\( \lef... | Yes |
Example 1.2.20 (Calculating an average) As an illustration of the distinguishable/indistinguishable approach, suppose that we are going to calculate all possible averages of four numbers selected from ## \(2,4,9,{12}\) where we draw the numbers with replacement. For example, possible draws are \( \{ 2,4,4,9\} \) with a... | The total number of distinct samples is \( \left( \begin{matrix} n + n - 1 \\ n \end{matrix}\right) \) . But now, to calculate the probability distribution of the sampled averages, we must count the different ways that a particular average can occur. The value 4.75 can occur only if the sample contains one 2, two 4s, a... | Yes |
Although the probability of getting all four aces is quite small, let us see how the conditional probabilities change given that some aces have already been drawn. Four cards will again be dealt from a well-shuffled deck, and we now calculate \( P\left( {4\text{ aces in }4\text{ cards } \mid i\text{ aces in }i\text{ ca... | The event \( \{ 4 \) aces in \( 4 \) cards \( \} \) is a subset of the event \( \{ i \) aces in \( i \) cards \( \} \) . Thus, from the definition of conditional probability, (1.3.1), we know that \[ P\left( {4\text{ aces in }4\text{ cards } \mid i\text{ aces in }i\text{ cards }}\right) = \frac{P\left( {\{ 4\text{ aces... | Yes |
Example 1.3.4 (Three prisoners) Three prisoners, A, B, and C, are on death row. The governor decides to pardon one of the three and chooses at random the prisoner to pardon. He informs the warden of his choice but requests that the name be kept secret for a few days.\n\nThe next day, A tries to get the warden to tell h... | It should be clear that the warden's reasoning is correct, but let us see why. Let \( A, B \), and \( C \) denote the events that \( \mathrm{A},\mathrm{B} \), or \( \mathrm{C} \) is pardoned, respectively. We know\n\nthat \( P\left( A\right) = P\left( B\right) = P\left( C\right) = \frac{1}{3}. \) Let \( \mathcal{W} \) ... | Yes |
Theorem 1.3.5 (Bayes’ Rule) Let \( {A}_{1},{A}_{2},\ldots \) be a partition of the sample space, and let \( B \) be any set. Then, for each \( i = 1,2,\ldots \) , | \[ P\left( {{A}_{i} \mid B}\right) = \frac{P\left( {B \mid {A}_{i}}\right) P\left( {A}_{i}\right) }{\mathop{\sum }\limits_{{j = 1}}^{\infty }P\left( {B \mid {A}_{j}}\right) P\left( {A}_{j}\right) }.\] | Yes |
Theorem 1.5.3 The function \( F\left( x\right) \) is a cdf if and only if the following three conditions hold:\n\na. \( \mathop{\lim }\limits_{{x \rightarrow - \infty }}F\left( x\right) = 0 \) and \( \mathop{\lim }\limits_{{x \rightarrow \infty }}F\left( x\right) = 1 \) .\n\nb. \( F\left( x\right) \) is a nondecreasing... | Outline of proof: To prove necessity, the three properties can be verified by writing \( F \) in terms of the probability function (see Exercise 1.48). To prove sufficiency, that if a function \( F \) satisfies the three conditions of the theorem then it is a cdf for some random variable, is much harder. It must be est... | No |
Example 1.5.4 (Tossing for a head) Suppose we do an experiment that consists of tossing a coin until a head appears. Let \( p = \) probability of a head on any given toss, and define a random variable \( X = \) number of tosses required to get a head. Then, for any \( x = 1,2,\ldots \) , | \[ P\left( {X = x}\right) = {\left( 1 - p\right) }^{x - 1}p \] since we must get \( x - 1 \) tails followed by a head for the event to occur and all trials are independent. From (1.5.2) we calculate, for any positive integer \( x \) ,\[ P\left( {X \leq x}\right) = \mathop{\sum }\limits_{{i = 1}}^{x}P\left( {X = i}\righ... | Yes |
An example of a continuous cdf is the function\n\n\[ \left( {1.5.5}\right) \]\n\n\[ {F}_{X}\left( x\right) = \frac{1}{1 + {e}^{-x}} \] | which satisfies the conditions of Theorem 1.5.3. For example,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow - \infty }}{F}_{X}\left( x\right) = 0\;\text{ since }\;\mathop{\lim }\limits_{{x \rightarrow - \infty }}{e}^{-x} = \infty \]\n\nand\n\n\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}{F}_{X}\left( x\right) = 1\;\... | Yes |
Example 1.5.9 (Identically distributed random variables) Consider the experiment of tossing a fair coin three times as in Example 1.4.3. Define the random variables \( X \) and \( Y \) by | The distribution of \( X \) is given in Example 1.4.3, and it is easily verified that the distribution of \( Y \) is exactly the same. That is, for each \( k = 0,1,2,3 \), we have \( P(X = \) \( k) = P\left( {Y = k}\right) \) . So \( X \) and \( Y \) are identically distributed. However, for no sample points do we have... | Yes |
Theorem 1.5.10 The following two statements are equivalent:\na. The random variables \( X \) and \( Y \) are identically distributed.\nb. \( {F}_{X}\left( x\right) = {F}_{Y}\left( x\right) \) for every \( x \) . | Proof: To show equivalence we must show that each statement implies the other. We first show that (a) \( \Rightarrow \) (b).\n\nBecause \( X \) and \( Y \) are identically distributed, for any set \( A \in {\mathcal{B}}^{1}, P\left( {X \in A}\right) = \) \( P\left( {Y \in A}\right) \) . In particular, for every \( x \)... | No |
For the geometric distribution of Example 1.5.4, we have the pmf\n\n\[ \n{f}_{X}\left( x\right) = P\left( {X = x}\right) = \left\{ \begin{array}{ll} {\left( 1 - p\right) }^{x - 1}p & \text{ for }x = 1,2,\ldots \\ 0 & \text{ otherwise. } \end{array}\right. \n\]\n\nRecall that \( P\left( {X = x}\right) \) or, equivalentl... | A widely accepted convention, which we will adopt, is to use an uppercase letter for the cdf and the corresponding lowercase letter for the pmf or pdf.\n\nWe must be a little more careful in our definition of a pdf in the continuous case. If we naively try to calculate \( P\left( {X = x}\right) \) for a continuous rand... | Yes |
Example 1.6.4 (Logistic probabilities) For the logistic distribution of Example 1.5.5 we have\n\n\[ \n{F}_{X}\left( x\right) = \frac{1}{1 + {e}^{-x}} \n\]\n\nand, hence,\n\n\[ \n{f}_{X}\left( x\right) = \frac{d}{dx}{F}_{X}\left( x\right) = \frac{{e}^{-x}}{{\left( 1 + {e}^{-x}\right) }^{2}}. \n\]\n\nThe area under the c... | \[ \nP\left( {a < X < b}\right) = {F}_{X}\left( b\right) - {F}_{X}\left( a\right) \n\]\n\n\[ \n= {\int }_{-\infty }^{b}{f}_{X}\left( x\right) {dx} - {\int }_{-\infty }^{a}{f}_{X}\left( x\right) {dx} \n\]\n\n\[ \n= {\int }_{a}^{b}{f}_{X}\left( x\right) {dx} \n\] | Yes |
Theorem 1.6.5 A function \( {f}_{X}\left( x\right) \) is a pdf (or pmf) of a random variable \( X \) if and only if\n\na. \( {f}_{X}\left( x\right) \geq 0 \) for all \( x \) .\n\nb. \( \mathop{\sum }\limits_{x}{f}_{X}\left( x\right) = 1\left( {pmf}\right) \; \) or \( \;{\int }_{-\infty }^{\infty }{f}_{X}\left( x\right)... | Proof: If \( {f}_{X}\left( x\right) \) is a pdf (or pmf), then the two properties are immediate from the definitions. In particular, for a pdf, using (1.6.3) and Theorem 1.5.3, we have that\n\n\[ 1 = \mathop{\lim }\limits_{{x \rightarrow \infty }}{F}_{X}\left( x\right) = {\int }_{-\infty }^{\infty }{f}_{X}\left( t\righ... | Yes |
Example 2.1.1 (Binomial transformation) A discrete random variable \( X \) has a binomial distribution if its pmf is of the form\n\n\[ \n{f}_{X}\left( x\right) = P\left( {X = x}\right) = \left( \begin{array}{l} n \\ x \end{array}\right) {p}^{x}{\left( 1 - p\right) }^{n - x},\;x = 0,1,\ldots, n, \]\n\nwhere \( n \) is a... | \n\[ \n{f}_{Y}\left( y\right) = \mathop{\sum }\limits_{{x \in {g}^{-1}\left( y\right) }}{f}_{X}\left( x\right) \]\n\n\[ \n= {fx}\left( {n - y}\right) \]\n\n\[ \n= \left( \begin{matrix} n \\ n - y \end{matrix}\right) {p}^{n - y}{\left( 1 - p\right) }^{n - \left( {n - y}\right) } \]\n\n\[ \n= \;\left( \begin{array}{l} n ... | Yes |
Example 2.1.2 (Uniform transformation) Suppose \( X \) has a uniform distribution on the interval \( \left( {0,{2\pi }}\right) \), that is,\n\n\[ \n{f}_{X}\left( x\right) = \left\{ \begin{array}{ll} 1/\left( {2\pi }\right) & 0 < x < {2\pi } \\ 0 & \text{ otherwise. } \end{array}\right.\n\]\n\nConsider \( Y = {\sin }^{2... | From the symmetry of the function \( {\sin }^{2}\left( x\right) \) and the fact that \( X \) has a uniform distribution, we have\n\n\[ \nP\left( {X \leq {x}_{1}}\right) = P\left( {X \geq {x}_{4}}\right) \;\text{ and }\;P\left( {{x}_{2} \leq X \leq {x}_{3}}\right) = {2P}\left( {{x}_{2} \leq X \leq \pi }\right) ,\n\]\n\n... | Yes |
Example 2.1.4 (Uniform-exponential relationship-I) Suppose \( X \sim {f}_{X}\left( x\right) = \) 1 if \( 0 < x < 1 \) and 0 otherwise, the uniform \( \left( {0,1}\right) \) distribution. It is straightforward to check that \( {F}_{X}\left( x\right) = x,0 < x < 1 \) . We now make the transformation \( Y = g\left( X\righ... | Since\n\n\[ \frac{d}{dx}g\left( x\right) = \frac{d}{dx}\left( {-\log x}\right) = \frac{-1}{x} < 0,\;\text{ for }\;0 < x < 1, \]\n\n\( g\left( x\right) \) is a decreasing function. As \( X \) ranges between 0 and \( 1, - \log x \) ranges between 0 and \( \infty \), that is, \( \mathcal{Y} = \left( {0,\infty }\right) \) ... | Yes |
Theorem 2.1.5 Let \( X \) have pdf \( {f}_{X}\left( x\right) \) and let \( Y = g\left( X\right) \), where \( g \) is a monotone function. Let \( \mathcal{X} \) and \( \mathcal{Y} \) be defined by (2.1.7). Suppose that \( {f}_{X}\left( x\right) \) is continuous on \( \mathcal{X} \) and that \( {g}^{-1}\left( y\right) \)... | Proof: From Theorem 2.1.3 we have, by the chain rule, \[ {f}_{Y}\left( y\right) = \frac{d}{dy}{F}_{Y}\left( y\right) = \left\{ \begin{array}{ll} {f}_{X}\left( {{g}^{-1}\left( y\right) }\right) \frac{d}{dy}{g}^{-1}\left( y\right) & \text{ if }g\text{ is increasing } \\ - {f}_{X}\left( {{g}^{-1}\left( y\right) }\right) \... | Yes |
Example 2.1.6 (Inverted gamma pdf) Let \( {f}_{X}\left( x\right) \) be the gamma pdf\n\n\[ f\left( x\right) = \frac{1}{\left( {n - 1}\right) !{\beta }^{n}}{x}^{n - 1}{e}^{-x/\beta },\;0 < x < \infty ,\]\n\nwhere \( \beta \) is a positive constant and \( n \) is a positive integer. Suppose we want to find the pdf of \( ... | Applying the above theorem, for \( y \in \left( {0,\infty }\right) \), we get\n\n\[ {f}_{Y}\left( y\right) = {f}_{X}\left( {{g}^{-1}\left( y\right) }\right) \left| {\frac{d}{dy}{g}^{-1}\left( y\right) }\right|\]\n\n\[ = \frac{1}{\left( {n - 1}\right) !{\beta }^{n}}{\left( \frac{1}{y}\right) }^{n - 1}{e}^{-1/\left( {\be... | Yes |
Theorem 2.1.8 Let \( X \) have pdf \( {f}_{X}\left( x\right) \), let \( Y = g\left( X\right) \), and define the sample space \( \mathcal{X} \) as in (2.1.7). Suppose there exists a partition, \( {A}_{0},{A}_{1},\ldots ,{A}_{k} \), of \( \mathcal{X} \) such that \( P\left( {X \in {A}_{0}}\right) = 0 \) and \( {f}_{X}\le... | The important point in Theorem 2.1.8 is that \( \mathcal{X} \) can be divided into sets \( {A}_{1},\ldots ,{A}_{k} \) such that \( g\left( x\right) \) is monotone on each \( {A}_{i} \) . We can ignore the \ | No |
Example 2.1.9 (Normal-chi squared relationship) Let \( X \) have the standard normal distribution,\n\n\[ \n{f}_{X}\left( x\right) = \frac{1}{\sqrt{2\pi }}{e}^{-{x}^{2}/2},\; - \infty < x < \infty .\n\]\n\nConsider \( Y = {X}^{2} \) . The function \( g\left( x\right) = {x}^{2} \) is monotone on \( \left( {-\infty ,0}\ri... | \[ \n{f}_{Y}\left( y\right) = \frac{1}{\sqrt{2\pi }}{e}^{-{\left( -\sqrt{y}\right) }^{2}/2}\left| {-\frac{1}{2\sqrt{y}}}\right| + \frac{1}{\sqrt{2\pi }}{e}^{-{\left( \sqrt{y}\right) }^{2}/2}\left| \frac{1}{2\sqrt{y}}\right| \n\]\n\n\[ \n= \frac{1}{\sqrt{2\pi }}\frac{1}{\sqrt{y}}{e}^{-y/2},\;0 < y < \infty .\n\]\n\nThe ... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.