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Theorem 10. If \( A\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}^{n} \), where \( {a}_{n} \geq 0 \) for all \( n \) and the series converges for \( 0 \leq z < 1 \), then we have\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{n + 1}\mathop{\sum }\limits_{{v = 0}}^{n}{a}_{v} = \math... | To get a feeling for this theorem, suppose all \( {a}_{n} = c > 0 \) . Then\n\n\[ A\left( z\right) = c\mathop{\sum }\limits_{{n = 0}}^{\infty }{z}^{n} = \frac{c}{1 - z} \]\n\nand the relation in (8.6.5) reduces to the trivial identity\n\n\[ \frac{1}{n + 1}\mathop{\sum }\limits_{{v = 0}}^{n}c = c = \left( {1 - z}\right)... | No |
Theorem 11. If \( I \) is finite and forms a single class (namely if there are only a finite number of states and they all communicate with each other), then the chain is necessarily recurrent. | Proof: Suppose the contrary; then each state is transient and so almost surely the particle can spend only a finite number of time units in it by Theorem 7. Since the number of states is finite, the particle can spend altogether only a finite number of time units in the whole space \( I \) . But time increases ad infin... | Yes |
Theorem 13. Suppose that we have, for every \( j \) ,\n\n\[ P\left\{ {{X}_{0} = j}\right\} = {w}_{j} \]\n\n(8.6.10)\n\nthen the same is true when \( {X}_{0} \) is replaced by any \( {X}_{n}, n \geq 1 \) . Furthermore the joint probability\n\n\[ P\left\{ {{X}_{n + v} = {j}_{v},0 \leq v \leq l}\right\} \]\n\n(8.6.11)\n\n... | Proof: We have by (8.6.9)\n\n\[ P\left\{ {{X}_{n} = j}\right\} = \mathop{\sum }\limits_{i}P\left\{ {{X}_{0} = i}\right\} {P}_{i}\left\{ {{X}_{n} = j}\right\} = \mathop{\sum }\limits_{i}{w}_{i}{p}_{ij}^{\left( n\right) } = {w}_{j}. \]\n\nSimilarly the probability in (8.6.11) is equal to\n\n\[ P\left\{ {{X}_{n} = {j}_{0}... | Yes |
A switch may be on off; call these two positions states 1 and 2. After each unit of time the state may hold or change, but the respective probabilities depend only on the present position. Thus we have a homogeneous Markov chain with \( I = \{ 1,2\} \) and\n\n\[ \n\Pi = \left\lbrack \begin{array}{ll} {p}_{11} & {p}_{12... | Clearly the second equation is just the negative of the first and may be discarded. Solving the first equation we get\n\n\[ \n{x}_{2} = \frac{1 - {p}_{11}}{{p}_{21}}{x}_{1} = \frac{{p}_{12}}{{p}_{21}}{x}_{1} \n\]\n\nThus\n\n\[ \n{w}_{1} = \frac{{p}_{21}}{{p}_{12} + {p}_{21}},\;{w}_{2} = \frac{{p}_{12}}{{p}_{12} + {p}_{... | Yes |
Let us find the stationary distribution for the Ehrenfest model. | We can proceed exactly as in Example 15, leading to the formula (8.6.15), but this time it stops at \( j = c \) . Substituting the numerical values from (8.3.16), we obtain\n\n\[ \n{c}_{j} = \frac{c\left( {c - 1}\right) \cdots \left( {c - j + 1}\right) }{1 \cdot 2\cdots j} = \left( \begin{array}{l} c \\ j \end{array}\r... | Yes |
Theorem 14. The \( \left\{ {y}_{i}\right\} \) above satisfies the system of equations\n\n\[ \n{x}_{i} = \mathop{\sum }\limits_{{j \in T}}{p}_{ij}{x}_{j} + \mathop{\sum }\limits_{{j \in C}}{p}_{ij},\;i \in T.\n\]\n\n(8.7.3)\n\nIf \( T \) is finite, it is the unique solution of this system. Hence it can be computed by st... | Proof: Let the particle start from \( i \), and consider its state \( j \) after one step. If \( j \in T \), then the Markov property shows that the conditional probability of absorption becomes \( {y}_{j} \) ; if \( j \in C \), then it is already absorbed; if \( j \in \) \( \left( {I - T}\right) - C \), then it can ne... | Yes |
Let us return to Problem 1 of \( §{8.1} \), where \( t = c - 1 \) . If \( p \neq q \) , then \( {v}_{i} = {\left( q/p\right) }^{i} \) is a nonconstant solution of (8.7.6). | This is trivial to verify but you may well demand to know how on earth did we discover such a solution? The answer in this case is easy (but motivated by knowledge of difference equations used in \( §{8.1} \) ): try a solution of the form \( {\lambda }^{i} \) and see what \( \lambda \) must be. Now if we substitute thi... | No |
The following model of random reproduction was introduced by S. Wright in his genetical studies (see, e.g., [Karlin] for further details). In a haploid organism the genes occur singly rather than in pairs as in the diploid case considered in \( §{5.6} \) . Suppose \( {2N} \) genes of types \( A \) and \( a \) (the alle... | It follows that [see (4.4.16) or (6.3.6)] the expected number of \( A \) genes is equal to\n\n\[ \n\mathop{\sum }\limits_{{j = 0}}^{{2N}}{p}_{ij}j = {2N}\frac{i}{2N} = i \n\]\n\n(8.7.12)\n\nThis means that the expected number of \( A \) genes in the next generation is equal to the actual (but random) number of these ge... | Yes |
What is the distribution of the number of particles of the second generation? | Let the generating function of \( {X}_{1} \) be \( g \) :\n\n\[ g\left( z\right) = \mathop{\sum }\limits_{{j = 0}}^{\infty }{a}_{j}{z}^{j} \]\n\nSuppose the number of particles in the first generation is equal to \( j \), and we denote the numbers of their descendants by \( {Z}_{1},\ldots ,{Z}_{j} \), respectively. The... | Yes |
Consider \( M = 2 \) assets. Assume that \( {\sigma }^{2}\left( {R}_{1}\right) = {\sigma }^{2}\left( {R}_{2}\right) > 0 \) , i.e., the two assets have the same risk. Assume further that \( - 1 \leq {\rho }_{12} < 1 \) , where \( {\rho }_{12} \) is the correlation coefficient between \( {R}_{1} \) and \( {R}_{2} \) (see... | Indeed,\n\n\[ \n{\sigma }^{2}\left( {\alpha {R}_{1} + \left( {1 - \alpha }\right) {R}_{2}}\right) = {\sigma }^{2}\left( {R}_{1}\right) + {2\alpha }\left( {1 - \alpha }\right) \left\lbrack {\operatorname{Cov}\left( {{R}_{1},{R}_{2}}\right) - \sigma \left( {R}_{1}\right) \sigma \left( {R}_{2}\right) }\right\rbrack \n\]\n... | Yes |
In the setting of Example 1, is there an allocation \( \left( {\alpha ,1 - \alpha }\right) \) , with \( 0 < \alpha < 1 \), that is least risky? In other words, is there \( \alpha \in \left( {0,1}\right) \) that minimizes the function \( V\left( \alpha \right) \) defined by\n\n\[ V\left( \alpha \right) = {\sigma }^{2}\l... | From (9.4.1) we see that \( V \) is a quadratic function in \( \alpha \), with second derivative\n\n\[ {V}^{\prime \prime }\left( \alpha \right) = 2{\sigma }^{2}\left( {R}_{1}\right) + 2{\sigma }^{2}\left( {R}_{2}\right) - 4{\rho }_{12}\sigma \left( {R}_{1}\right) \sigma \left( {R}_{2}\right) .\n\nUnder the assumptions... | Yes |
Consider again two assets such that \( \sigma \left( {R}_{1}\right) > 0 \) and \( \sigma \left( {R}_{2}\right) = \) 0 . The second asset is a riskless security as we saw previously. Then for \( 0 \leq \alpha \leq 1, V\left( \alpha \right) = {\sigma }^{2}\left( {\alpha {R}_{1} + \left( {1 - \alpha }\right) {R}_{2}}\righ... | Thus the minimand \( {\alpha }^{ * } \) of the return variance of the portfolio \( \left( {\alpha ,1 - \alpha }\right) \) is \( {\alpha }^{ * } = 0 \) and \( V\left( {\alpha }^{ * }\right) = 0 \) is the smallest risk. | Yes |
Theorem 4. Let \( \left\{ {X}_{i}\right\} \) be independent, following the same stable distribution with characteristic function \( \varphi \left( \theta \right) = {e}^{-{\gamma }_{\alpha }{\left| \theta \right| }^{\alpha } + {id\theta }} \) . Then, for a given positive \( a,{Y}_{n} = \left( {1/a}\right) {n}^{-1/\alpha... | Proof: For \( 1 \leq k \leq n \), let \( {X}_{k}^{\prime } = \left( {{X}_{k} - d}\right) /\left( {a{n}^{1/\alpha }}\right) \) . Its characteristic function is\n\n\[ \n{\varphi }^{\prime }\left( \theta \right) = E\left\lbrack {e}^{{i\theta }{X}_{k}^{\prime }}\right\rbrack \n\]\n\n\[ \n= E\left\lbrack {e}^{{i\theta }\fra... | Yes |
Theorem 1. Under the same conditions of the theorem in Appendix 3, we have for a finite optional time \( T \)\n\n\[ E\left( {X}_{T}\right) \geq E\left( {X}_{0}\right) \text{ for a submartingale }\]\n\n\[ \text{and}\;E\left( {X}_{T}\right) \leq E\left( {X}_{0}\right) \text{for a supermartingale.} \] | Proof: The proof is exactly as in Appendix 3 with equalities replaced with the corresponding inequalities (left as an exercise). | No |
The time-0 price of a European option on a stock with price \( S \) and payoff \( g \) at time 1 is\n\n\[ \n{V}_{0} = \frac{1}{1 + r}\left\lbrack {\widetilde{p}g\left( H\right) + \left( {1 - \widetilde{p}}\right) g\left( T\right) }\right\rbrack \n\]\n\nwhere\n\n\[ \n\widetilde{p} = \frac{1 + r - d}{u - d} \n\] | Proof: We first show that \( {V}_{0} = {\beta }_{0}{S}_{0} + {\gamma }_{0}{B}_{0} \), where \( {\beta }_{0} \) and \( {\gamma }_{0} \) are defined by (10.2.2). Suppose \( \epsilon = {V}_{0} - \left( {{\beta }_{0}{S}_{0} + {\gamma }_{0}{B}_{0}}\right) > 0 \) . Consider the following portfolio \( \left( {-1,{\beta }_{0},... | Yes |
Proposition 2. The values \( {V}_{0},{V}_{1},\ldots ,{V}_{N} \) of a European option expiring at time \( {t}_{N} \) satisfy the recursive relations\n\n\[ \n{V}_{n - 1}\left( {\widehat{\omega }}_{n - 1}\right) = \frac{1}{1 + r}\left\lbrack {\widetilde{p}{V}_{n}\left( {{\widehat{\omega }}_{n - 1},0}\right) + \left( {1 - ... | Proof: As in the proof of Proposition 1, we need to first show that \( {V}_{n - 1} = \) \( {\beta }_{n - 1}{S}_{n - 1} + {\gamma }_{n - 1}{B}_{n - 1} \), where \( {\beta }_{n - 1} \) and \( {\gamma }_{n - 1} \) are defined in (10.3.1). This step is straightforward and is left as an exercise. | No |
Proposition 3. The time-0 price of a European call option on a stock with price process \( S \) defined above and payoff \( g \) at time \( N \) is\n\n\[ \n{V}_{0} = \frac{1}{{\left( 1 + r\right) }^{N}}\widetilde{E}\left\{ {g \mid \left( {{S}_{0},{B}_{0}}\right) }\right\} \]\n\nwhere \( \widetilde{E} \) refers to the e... | Proof: As in the one-period case (cf. Proposition 1) we can write, for \( 1 \leq i \leq N \)\n\n\[ \n{V}_{i - 1} = E\left\{ {\frac{1}{1 + r}{V}_{i} \mid \left( {{S}_{i - 1},{B}_{i - 1}}\right) }\right\} .\n\nThen\n\n\[ \n{V}_{i - 2} = E\left\{ {\frac{1}{1 + r}{V}_{i - 1} \mid \left( {{S}_{i - 2},{B}_{i - 2}}\right) }\r... | No |
Consider now\n\n(5)\n\[ f\left( x\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{b}_{n}{\delta }_{{a}_{n}}\left( x\right) \] | Since \( 0 \leq {\delta }_{{a}_{n}}\left( x\right) \leq 1 \) for every \( n \) and \( x \), the series in (5) is absolutely and uniformly convergent. Since each \( {\delta }_{{a}_{n}} \) is increasing, it follows that if \( {x}_{1} < {x}_{2} \), \n\n\[ f\left( {x}_{2}\right) - f\left( {x}_{1}\right) = \mathop{\sum }\li... | Yes |
Theorem 1.2.1. Let\n\n\[ \n{F}_{c}\left( x\right) = F\left( x\right) - {F}_{d}\left( x\right) \]\n\nthen \( {F}_{c} \) is positive, increasing, and continuous. | PROOF. Let \( x < {x}^{\prime } \), then we have\n\n\[ \n\text{4)}\;{F}_{d}\left( {x}^{\prime }\right) - {F}_{d}\left( x\right) = \mathop{\sum }\limits_{{x < {a}_{j} \leq {x}^{\prime }}}{b}_{j} = \mathop{\sum }\limits_{{x < {a}_{j} \leq {x}^{\prime }}}\left\lbrack {F\left( {a}_{j}\right) - F\left( {{a}_{j} - }\right) }... | Yes |
Theorem 1.2.2. Let \( F \) be a d.f. Suppose that there exist a continuous function \( {G}_{c} \) and a function \( {G}_{d} \) of the form\n\n\[ \n{G}_{d}\left( x\right) = \mathop{\sum }\limits_{j}{b}_{j}^{\prime }{\delta }_{{a}_{j}^{\prime }}\left( x\right)\n\]\n\n[where \( \left\{ {a}_{j}^{\prime }\right\} \) is a co... | PROOF. If \( {F}_{d} \neq {G}_{d} \), then either the sets \( \left\{ {a}_{j}\right\} \) and \( \left\{ {a}_{j}^{\prime }\right\} \) are not identical, or we may relabel the \( {a}_{j}^{\prime } \) so that \( {a}_{j}^{\prime } = {a}_{j} \) for all \( j \) but \( {b}_{j}^{\prime } \neq {b}_{j} \) for some \( j \) . In e... | Yes |
Theorem 2.1.1. A field is a B.F. if and only if it is also an M.C. | PROOF. The \ | No |
Theorem 2.1.2. Let \( {\mathcal{T}}_{0} \) be a field, \( \mathcal{G} \) the minimal M.C. containing \( {\mathcal{T}}_{0},\mathcal{F} \) the minimal B.F. containing \( {\mathcal{F}}_{0} \), then \( \mathcal{F} = \mathcal{G} \) . | PROOF. Since a B.F. is an M.C., we have \( \mathcal{F} \supset \mathcal{G} \) . To prove \( \mathcal{F} \subset \mathcal{G} \) it is sufficient to show that \( \mathcal{G} \) is a B.F. Hence by Theorem 2.1.1 it is sufficient to show that \( \mathcal{G} \) is a field. We shall show that it is closed under intersection a... | Yes |
Theorem 2.2.1. The axioms of finite additivity and of continuity together are equivalent to the axiom of countable additivity. | PROOF. Let \( {E}_{n} \downarrow \) . We have the obvious identity:\n\n\[ \n{E}_{n} = \mathop{\bigcup }\limits_{{k = n}}^{\infty }\left( {{E}_{k} \smallsetminus {E}_{k + 1}}\right) \cup \mathop{\bigcap }\limits_{{k = 1}}^{\infty }{E}_{k} \n\]\n\nIf \( {E}_{n} \downarrow \varnothing \), the last term is the empty set. H... | Yes |
Let \( \Omega \) be a countable set: \( \Omega = \left\{ {{\omega }_{j}, j \in J}\right\} \), where \( J \) is a countable index set, and let \( \mathcal{F} \) be the total B.F. of \( \Omega \) . Choose any sequence of numbers \( \left\{ {{p}_{j}, j \in J}\right\} \) satisfying\n\n\[ \forall j \in J : {p}_{j} \geq 0;\;... | In words, we assign \( {p}_{i} \) as the value of the \ | No |
Example 2. Let \( \mathcal{U} = (0,1\rbrack ,\ell \) the collection of intervals:\n\n\[ \mathcal{C} = \{ (a, b\rbrack : 0 < a < b \leq 1\}\]\n\n\( \mathcal{B} \) the minimal B.F. containing \( \mathcal{C}, m \) the Borel-Lebesgue measure on \( \mathcal{B} \) . Then \( \left( {\mathcal{U},\mathcal{B}, m}\right) \) is a ... | Let \( {\mathcal{B}}_{0} \) be the collection of subsets of \( \mathcal{U} \) each of which is the union of a finite number of members of \( \ell \) . Thus a typical set \( B \) in \( {\mathcal{B}}_{0} \) is of the form\n\n\[ B = \mathop{\bigcup }\limits_{{j = 1}}^{n}\left( {{a}_{j},{b}_{j}}\right\rbrack \;\text{ where... | No |
Theorem 2.2.3. Let \( \mu \) and \( v \) be two measures defined on the same B.F. \( \mathcal{F} \) , which is generated by the field \( {\mathcal{T}}_{0} \). If either \( \mu \) or \( v \) is \( \sigma \) -finite on \( {\mathcal{T}}_{0} \), and \( \mu \left( E\right) = v\left( E\right) \) for every \( E \in {\mathcal{... | PROOF. We give the proof only in the case where \( \mu \) and \( v \) are both finite, leaving the rest as an exercise. Let\n\n\[ \mathcal{C} = \{ E \in \mathcal{F} : \mu \left( E\right) = v\left( E\right) \}\]\n\nthen \( \ell \supset {\mathcal{T}}_{0} \) by hypothesis. But \( \ell \) is also a monotone class, for if \... | No |
Theorem 3.1.1. For any function \( X \) from \( \Omega \) to \( {\mathcal{R}}^{1} \) (or \( {\mathcal{R}}^{ * } \) ), not necessarily an r.v., the inverse mapping \( {X}^{-1} \) has the following properties: | \[ {X}^{-1}\left( {A}^{c}\right) = {\left( {X}^{-1}\left( A\right) \right) }^{c} \] \[ {X}^{-1}\left( {\mathop{\bigcup }\limits_{\alpha }{A}_{\alpha }}\right) = \mathop{\bigcup }\limits_{\alpha }{X}^{-1}\left( {A}_{\alpha }\right) \] \[ {X}^{-1}\left( {\mathop{\bigcap }\limits_{\alpha }{A}_{\alpha }}\right) = \mathop{\... | Yes |
Theorem 3.1.2. \( X \) is an r.v. if and only if for each real number \( x \), or each real number \( x \) in a dense subset of \( {\mathcal{R}}^{1} \), we have\n\n\[ \n\{ \omega : X\left( \omega \right) \leq x\} \in \mathcal{F}.\n\] | PROOF. The preceding condition may be written as\n\n(3)\n\n\[ \n\forall x : {X}^{-1}(\left( {-\infty, x\rbrack }\right) \in \mathcal{F}.\n\]\n\nConsider the collection \( \mathcal{A} \) of all subsets \( S \) of \( {\mathcal{R}}^{1} \) for which \( {X}^{-1}\left( S\right) \in \mathcal{F} \) . From Theorem 3.1.1 and the... | Yes |
Theorem 3.1.3. Each r.v. on the probability space \( \left( {\Omega ,\mathcal{F},\mathcal{P}}\right) \) induces a probability space \( \left( {{\mathcal{R}}^{1},{\mathcal{B}}^{1},\mu }\right) \) by means of the following correspondence:\n\n(4)\n\n\[ \forall B \in {\mathcal{B}}^{1} : \;\mu \left( B\right) = \mathcal{P}\... | PROOF. Clearly \( \mu \left( B\right) \geq 0 \) . If the \( {B}_{n} \) ’s are disjoint sets in \( {\mathcal{B}}^{1} \), then the \( {X}^{-1}\left( {B}_{n}\right) \) ’s are disjoint by Theorem 3.1.1. Hence\n\n\[ \mu \left( {\mathop{\bigcup }\limits_{n}{B}_{n}}\right) = \mathcal{P}\left( {{X}^{-1}\left( {\mathop{\bigcup ... | Yes |
In this case an r.v. is by definition just a Borel measurable function. According to the usual definition, \( f \) on \( \mathcal{U} \) is Borel measurable iff \( {f}^{-1}\left( {\mathcal{B}}^{1}\right) \subset \mathcal{B} \) . | In particular, the function \( f \) given by \( f\left( \omega \right) \equiv \omega \) is an r.v. The two r.v.’s \( \omega \) and \( 1 - \omega \) are not identical but are identically distributed; in fact their common distribution is the underlying measure \( m \) . | Yes |
Theorem 3.1.4. If \( X \) is an r.v., \( f \) a Borel measurable function [on \( \left( {{\mathcal{R}}^{1},{\mathcal{B}}^{1}}\right) \) ], then \( f\left( X\right) \) is an r.v. | PROOF. The quickest proof is as follows. Regarding the function \( f\left( X\right) \) of \( \omega \) as the \ | No |
Theorem 3.1.5. If \( X \) and \( Y \) are r.v.’s and \( f \) is a Borel measurable function of two variables, then \( f\left( {X, Y}\right) \) is an r.v. | \[ {\left\lbrack f \circ \left( X, Y\right) \right\rbrack }^{-1}\left( {\mathcal{B}}^{1}\right) = {\left( X, Y\right) }^{-1} \circ {f}^{-1}\left( {\mathcal{B}}^{1}\right) \subset {\left( X, Y\right) }^{-1}\left( {\mathcal{B}}^{2}\right) \subset \mathcal{F}. \] The last inclusion says the inverse mapping \( {\left( X, Y... | Yes |
Theorem 3.1.6. If \( \left\{ {{X}_{j}, j \geq 1}\right\} \) is a sequence of r.v.’s, then\n\n\[ \mathop{\inf }\limits_{j}{X}_{j},\;\mathop{\sup }\limits_{j}{X}_{j},\;\mathop{\liminf }\limits_{j}{X}_{j},\;\mathop{\limsup }\limits_{j}{X}_{j} \]\n\nare r.v.'s, not necessarily finite-valued with probability one though ever... | PROOF. To see, for example, that \( \mathop{\sup }\limits_{j}{X}_{j} \) is an r.v., we need only observe\n\nthe relation\n\n\[ \forall x \in {\mathcal{R}}^{1} : \left\{ {\mathop{\sup }\limits_{j}{X}_{j} \leq x}\right\} = \mathop{\bigcap }\limits_{j}\left\{ {{X}_{j} \leq x}\right\} \]\n\nand use Theorem 3.1.2. Since\n\n... | Yes |
Theorem 3.2.1. We have\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\mathcal{P}\left( {\left| X\right| \geq n}\right) \leq \mathcal{E}\left( \left| X\right| \right) \leq 1 + \mathop{\sum }\limits_{{n = 1}}^{\infty }\mathcal{P}\left( {\left| X\right| \geq n}\right) \]\n\nso that \( \mathcal{E}\left( \left| X\right| \r... | PROOF. By the additivity property (iii), if \( {\Lambda }_{n} = \{ n \leq \left| X\right| < n + 1\} \), \n\n\[ \mathcal{E}\left( \left| X\right| \right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\int }_{{\Lambda }_{n}}\left| X\right| d\mathcal{P} \]\n\nHence by the mean value theorem (vi) applied to each set \( {\Lam... | Yes |
Theorem 3.2.2. Let \( X \) on \( \left( {\Omega ,\mathcal{F},\mathcal{P}}\right) \) induce the probability space \( \left( {{\mathcal{R}}^{1},{\mathcal{B}}^{1},\mu }\right) \) according to Theorem 3.1.3 and let \( f \) be Borel measurable. Then we have\n\n(11)\n\n\[{\int }_{\Omega }f\left( {X\left( \omega \right) }\rig... | PROOF. Let \( B \in {\mathcal{B}}^{1} \), and \( f = {1}_{B} \), then the left side in (11) is \( \mathcal{P}\left( {X \in B}\right) \) and the right side is \( \mu \left( B\right) \) . They are equal by the definition of \( \mu \) in (4) of Sec. 3.1. Now by the linearity of both integrals in (11), it will also hold if... | Yes |
Theorem 3.2.3. Let \( \\left( {X, Y}\\right) \) on \( \\left( {\\Omega ,\\mathcal{F},\\mathcal{P}}\\right) \) induce the probability space \( \\left( {{\\mathcal{R}}^{2},{\\mathcal{B}}^{2},{\\mu }^{2}}\\right) \) and let \( f \) be a Borel measurable function of two variables. Then we have\n\n(14)\n\n\[ \n{\\int }_{\\O... | Note that \( f\\left( {X, Y}\\right) \) is an r.v. by Theorem 3.1.5. | No |
Theorem 3.3.1. If \( \\left\\{ {{X}_{j},1 \\leq j \\leq n}\\right\\} \) are independent r.v.’s and \( \\left\\{ {{f}_{j},1 \\leq j \\leq }\\right. \) \( n\\} \) are Borel measurable functions, then \( \\left\\{ {{f}_{j}\\left( {X}_{j}\\right) ,1 \\leq j \\leq n}\\right\\} \) are independent r.v.'s. | PROOF. Let \( {A}_{j} \\in {\\mathcal{B}}^{1} \), then \( {f}_{j}^{-1}\\left( {A}_{j}\\right) \\in {\\mathcal{B}}^{1} \) by the definition of a Borel measurable function. By Theorem 3.1.1, we have\n\n\[ \n\\mathop{\\bigcap }\\limits_{{j = 1}}^{n}\\left\\{ {{f}_{j}\\left( {X}_{j}\\right) \\in {A}_{j}}\\right\\} = \\math... | Yes |
Theorem 3.3.3. If \( X \) and \( Y \) are independent and both have finite expectations, then\n\n\[ \mathcal{E}\left( {XY}\right) = \mathcal{E}\left( X\right) \mathcal{E}\left( Y\right) \] | PROOF. We give two proofs in detail of this important result to illustrate the methods. Cf. the two proofs of (14) in Sec. 3.2, one indicated there, and one in Exercise 9 of Sec. 3.2.\n\nFirst proof. Suppose first that the two r.v.’s \( X \) and \( Y \) are both discrete belonging respectively to the weighted partition... | Yes |
Let \( n \geq 2 \) and \( \left( {{\Omega }_{j},{\mathcal{J}}_{j},{\mathcal{P}}_{j}}\right) \) be \( n \) discrete probability spaces. We define the product space\n\n\[ \n{\Omega }^{n} = {\Omega }_{1} \times \cdots \times {\Omega }_{n}\left( {n\text{ factors }}\right) \n\]\n\nto be the space of all ordered \( n \) -tup... | To see this, we observe that the left side is, by definition, equal to\n\n\[ \n\mathop{\sum }\limits_{{{\omega }_{1} \in {S}_{1}}}\cdots \mathop{\sum }\limits_{{{\omega }_{n} \in {S}_{n}}}{\mathcal{P}}^{n}\left( \left\{ {{\omega }_{1},\ldots ,{\omega }_{n}}\right\} \right) = \mathop{\sum }\limits_{{{\omega }_{1} \in {S... | Yes |
Can we construct r.v.’s on the probability space \( \left( {\mathcal{U},\mathcal{B}, m}\right) \) itself, without going to a product space? | Indeed we can, but only by imbedding a product structure in \( \mathcal{U} \). For each real number in \( (0,1\rbrack \), consider its binary digital expansion\n\n\[ x = \cdot {\epsilon }_{1}{\epsilon }_{2}\cdots {\epsilon }_{n}\cdots = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\epsilon }_{n}}{{2}^{n}},\;\operato... | Yes |
Theorem 3.3.4. Let a finite or infinite sequence of p.m.’s \( \left\{ {\mu }_{j}\right\} \) on \( \left( {{\mathcal{R}}^{1},{\mathcal{B}}^{1}}\right) \) , or equivalently their d.f.'s, be given. There exists a probability space \( \left( {\Omega ,\mathcal{F},\mathcal{P}}\right) \) and a sequence of independent r.v.’s \... | PROOF. Without loss of generality we may suppose that the given sequence is infinite. (Why?) For each \( n \), let \( \left( {{\Omega }_{n},{\mathcal{T}}_{n},{\mathcal{P}}_{n}}\right) \) be a probability space\nin which there exists an r.v. \( {X}_{n} \) with \( {\mu }_{n} \) as its p.m. Indeed this is possible if we t... | No |
Theorem 3.3.5. Let \( {\mathcal{T}}_{0} \) be a field of subsets of an abstract space \( \Omega \), and \( \mathcal{P} \) a p.m. on \( {\mathcal{T}}_{0} \) . There exists a unique p.m. on the B.F. generated by \( {\mathcal{T}}_{0} \) that agrees with \( \mathcal{P} \) on \( {\mathcal{T}}_{0} \) . | The uniqueness is proved in Theorem 2.2.3. | No |
Theorem 3.3.6. For each \( n \geq 1 \), let \( {\mu }^{n} \) be a p.m. on \( \left( {{\mathcal{R}}^{n},{\mathcal{B}}^{n}}\right) \) such that\n\n(16)\n\[ \forall m < n : {\mu }^{n} \circ {\pi }_{mn} = {\mu }^{m}. \]\n\nThen there exists a probability space \( \left( {\Omega ,\mathcal{F},\mathcal{P}}\right) \) and a seq... | For a proof of this first fundamental theorem in the theory of stochastic processes, see Kolmogorov [8]. | No |
Theorem 4.1.1. The sequence \( \left\{ {X}_{n}\right\} \) converges a.e. to \( X \) if and only if for every \( \epsilon > 0 \) we have\n\n(2)\n\n\[ \mathop{\lim }\limits_{{m \rightarrow \infty }}\mathcal{P}\left\{ {\left| {{X}_{n} - X}\right| \leq \epsilon \text{ for all }n \geq m}\right\} = 1 \]\n\nor equivalently\n\... | PROOF. Suppose there is convergence a.e. and let \( {\Omega }_{0} = \Omega \smallsetminus \mathbf{N} \) where \( \mathbf{N} \) is as in (1). For \( m \geq 1 \) let us denote by \( {A}_{m}\left( \epsilon \right) \) the event exhibited in (2), namely:\n\n(3)\n\n\[ {A}_{m}\left( \epsilon \right) = \mathop{\bigcap }\limits... | Yes |
Theorem 4.1.4. If \( {X}_{n} \) converges to 0 in \( {L}^{p} \), then it converges to 0 in pr. The converse is true provided that \( \left\{ {X}_{n}\right\} \) is dominated by some \( Y \) that belongs to \( {L}^{p} \) . | PROOF. By Chebyshev inequality with \( \varphi \left( x\right) \equiv {\left| x\right| }^{p} \), we have\n\n(9)\n\n\[ \mathcal{P}\left\{ {\left| {X}_{n}\right| \geq \epsilon }\right\} \leq \frac{\mathcal{E}\left( {\left| {X}_{n}\right| }^{p}\right) }{{\epsilon }^{p}} \]\n\nLetting \( n \rightarrow \infty \), the right ... | No |
Theorem 4.1.5. \( {X}_{n} \rightarrow 0 \) in pr. if and only if\n\n\[ \mathcal{E}\left( \frac{\left| {X}_{n}\right| }{1 + \left| {X}_{n}\right| }\right) \rightarrow 0 \] | PROOF. If \( \rho \left( {X, Y}\right) = 0 \), then \( \mathcal{E}\left( \left| {X - Y}\right| \right) = 0 \), hence \( X = Y \) a.e. by Exercise 1 of Sec. 3.2. To show that \( \rho \left( {\cdot , \cdot }\right) \) is metric it is sufficient to show that\n\n\[ \mathcal{E}\left( \frac{\left| X - Y\right| }{1 + \left| {... | No |
Convergence in pr. does not imply convergence in \( {L}^{p} \), and the latter does not imply convergence a.e. | Take the probability space \( \left( {\Omega ,\mathcal{F},\mathcal{P}}\right) \) to be \( \left( {\mathcal{U},\mathcal{B}, m}\right) \) as in Example 2 of Sec. 2.2. Let \( {\varphi }_{k.j} \) be the indicator of the interval\n\n\[ \left( {\frac{j - 1}{k},\frac{j}{k}}\right) ,\;k \geq 1,1 \leq j \leq k.\]\n\nOrder these... | Yes |
Example 2. Convergence a.e. does not imply convergence in \( {L}^{p} \) . | In \( \left( {\mathcal{U},\mathcal{B}, m}\right) \) define\n\n\[ \n{X}_{n}\left( \omega \right) = \left\{ \begin{array}{ll} {2}^{n}, & \text{ if }\omega \in \left( {0,\frac{1}{n}}\right) ; \\ 0, & \text{ otherwise. } \end{array}\right. \]\n\nThen \( \mathcal{E}\left( {\left| {X}_{n}\right| }^{p}\right) = {2}^{np}/n \ri... | Yes |
Theorem 4.2.1. We have for arbitrary events \( \left\{ {E}_{n}\right\} \) :\n\n(4)\n\[ \mathop{\sum }\limits_{n}\mathcal{P}\left( {E}_{n}\right) < \infty \Rightarrow \mathcal{P}\left( {{E}_{n}\text{ i.o. }}\right) = 0. \] | PROOF. By Boole's inequality for p.m.'s, we have\n\n\[ \mathcal{P}\left( {F}_{m}\right) \leq \mathop{\sum }\limits_{{n = m}}^{\infty }\mathcal{P}\left( {E}_{n}\right) \]\n\nHence the hypothesis in (4) implies that \( \mathcal{P}\left( {F}_{m}\right) \rightarrow 0 \), and the conclusion in (4) now follows by (2). | Yes |
Theorem 4.2.2. \( {X}_{n} \rightarrow 0 \) a.e. if and only if\n\n(5)\n\[ \forall \epsilon > 0 : \mathcal{P}\left\{ {\left| {X}_{n}\right| > \epsilon \text{ i.o. }}\right\} = 0. \] | PROOF. Using the notation \( {A}_{m} = \mathop{\bigcap }\limits_{{n = m}}^{\infty }\left\{ {\left| {X}_{n}\right| \leq \epsilon }\right\} \) as in (3) of Sec. 4.1\n\n(with \( X = 0 \) ), we have\n\n\[ \left\{ {\left| {X}_{n}\right| > \epsilon \text{ i.o. }}\right\} = \mathop{\bigcap }\limits_{{m = 1}}^{\infty }\mathop{... | Yes |
Theorem 4.2.3. If \( {X}_{n} \rightarrow X \) in pr., then there exists a sequence \( \left\{ {n}_{k}\right\} \) of integers increasing to infinity such that \( {X}_{{n}_{k}} \rightarrow X \) a.e. Briefly stated: convergence in pr. implies convergence a.e. along a subsequence. | PROOF. We may suppose \( X \equiv 0 \) as explained before. Then the hypothesis may be written as\n\n\[ \forall k > 0 : \;\mathop{\lim }\limits_{{n \rightarrow \infty }}\mathcal{P}\left( {\left| {X}_{n}\right| > \frac{1}{{2}^{k}}}\right) = 0. \]\n\nIt follows that for each \( k \) we can find \( {n}_{k} \) such that\n\... | Yes |
Theorem 4.2.4. If the events \( \left\{ {E}_{n}\right\} \) are independent, then\n\n(6)\n\[ \mathop{\sum }\limits_{n}\mathcal{P}\left( {E}_{n}\right) = \infty \Rightarrow \mathcal{P}\left( {{E}_{n}\text{ i.o. }}\right) = 1. \] | PROOF. By (3) we have\n\n(7)\n\[ \mathcal{P}\left\{ {\mathop{\liminf }\limits_{n}{E}_{n}^{c}}\right\} = \mathop{\lim }\limits_{{m \rightarrow \infty }}\mathcal{P}\left( {\mathop{\bigcap }\limits_{{n = m}}^{\infty }{E}_{n}^{c}}\right) .\n\nThe events \( \left\{ {E}_{n}^{c}\right\} \) are independent as well as \( \left\... | No |
Let \( {X}_{n} = {c}_{n} \) where the \( {c}_{n} \)’s are constants tending to zero. Then \( {X}_{n} \rightarrow 0 \) deterministically. | For any interval \( I \) such that \( 0 \notin \overline{\mathrm{I}} \), where \( \overline{\mathrm{I}} \) is the closure of \( I \), we have \( \mathop{\lim }\limits_{n}{\mu }_{n}\left( I\right) = 0 = \mu \left( I\right) \) ; for any interval such that \( 0 \in {I}^{ \circ } \), where \( {I}^{ \circ } \) is the interi... | Yes |
Example 2. Let \( {X}_{n} = {c}_{n} \) where \( {c}_{n} \rightarrow + \infty \) . Then \( {X}_{n} \rightarrow + \infty \) deterministically. According to our definition of a r.v., the constant \( + \infty \) indeed qualifies. But for any finite interval \( \left( {a, b}\right) \) we have \( \mathop{\lim }\limits_{n}{\m... | This example can be easily ramified; e.g. let \( {a}_{n} \rightarrow - \infty \) , \( {b}_{n} \rightarrow + \infty \) and\n\n\[ \n{X}_{n} = \left\{ \begin{array}{ll} {a}_{n} & \text{ with probability }\alpha , \\ 0 & \text{ with probability }1 - \alpha - \beta , \\ {b}_{n} & \text{ with probability }\beta . \end{array}... | Yes |
Theorem 4.3.1. Let \( \left\{ {\mu }_{n}\right\} \) and \( \mu \) be s.p.m.’s. The following propositions are equivalent.\n\n(i) For every finite interval \( \left( {a, b}\right) \) and \( \epsilon > 0 \), there exists an \( {n}_{0}\left( {a, b,\epsilon }\right) \) such that if \( n \geq {n}_{0} \), then\n\n\( \mu \lef... | PROOF. To prove that (i) \( \Rightarrow \) (ii), let \( \left( {a, b}\right) \) be a continuity interval of \( \mu \) .\n\nIt follows from the monotone property of a measure that\n\n\[ \n\mathop{\lim }\limits_{{\epsilon \downarrow 0}}\mu \left( {a + \epsilon, b - \epsilon }\right) = \mu \left( {a, b}\right) = \mu \left... | Yes |
Theorem 4.3.4. If every vaguely convergent subsequence of the sequence of s.p.m.’s \( \left\{ {\mu }_{n}\right\} \) converges to the same \( \mu \), then \( {\mu }_{n}\overset{v}{ \rightarrow }\mu \) . | PROOF. To prove the theorem by contraposition, suppose \( {\mu }_{n} \) does not converge vaguely to \( \mu \) . Then by Theorem 4.3.1,(ii), there exists a continuity interval \( \left( {a, b}\right) \) of \( \mu \) such that \( {\mu }_{n}\left( {a, b}\right) \) does not converge to \( \mu \left( {a, b}\right) \) . By ... | Yes |
Theorem 4.4.2. Let \( \left\{ {\mu }_{n}\right\} \) and \( \mu \) be p.m.’s. Then \( {\mu }_{n}\overset{v}{ \rightarrow }\mu \) if and only if\n\n\[ \forall f \in {C}_{B} : {\int }_{{\mathcal{R}}^{1}}f\left( x\right) {\mu }_{n}\left( {dx}\right) \rightarrow {\int }_{{\mathcal{R}}^{1}}f\left( x\right) \mu \left( {dx}\ri... | PROOF. Suppose \( {\mu }_{n}\overset{v}{ \rightarrow }\mu \) . Given \( \epsilon > 0 \), there exist \( a \) and \( b \) in \( D \) such that\n\n\[ \text{(7)}\;\mu \left( {(a, b{\rbrack }^{c}}\right) = 1 - \mu (\left( {a, b\rbrack }\right) < \epsilon \text{.} \]\n\nIt follows from vague convergence that there exists \(... | Yes |
Theorem 4.4.3. Let a family of p.m.’s \( \left\{ {{\mu }_{\alpha },\alpha \in A}\right\} \) be given on an arbitrary index set \( A \) . In order that every sequence of them contains a subsequence which converges vaguely to a p.m., it is necessary and sufficient that the following condition be satisfied: for any \( \ep... | PROOF. Suppose (11) holds. For any sequence \( \left\{ {\mu }_{n}\right\} \) from the family, there exists a subsequence \( \left\{ {\mu }_{n}^{\prime }\right\} \) such that \( {\mu }_{n}^{\prime }\overset{v}{ \rightarrow }\mu \) . We show that \( \mu \) is a p.m. Let \( J \) be a continuity interval of \( \mu \) which... | Yes |
Theorem 4.4.4. If \( \left\{ {\mu }_{n}\right\} \) and \( \mu \) are p.m.’s, then \( {\mu }_{n}\overset{v}{ \rightarrow }\mu \) if and only if one of the two conditions below is satisfied:\n\n(13)\n\n\[ \forall f \in L : \mathop{\lim }\limits_{n}\int f\left( x\right) {\mu }_{n}\left( {dx}\right) \geq \int f\left( x\rig... | PROOF. We begin by observing that the two conditions above are equivalent by putting \( f = - g \) . Now suppose \( {\mu }_{n}\overset{v}{ \rightarrow }\mu \) and let \( {f}_{k} \in {C}_{B},{f}_{k} \uparrow f \) . Then we have\n\n(14)\n\n\[ \mathop{\lim }\limits_{n}\int f\left( x\right) {\mu }_{n}\left( {dx}\right) \ge... | Yes |
Theorem 4.4.5. Let \( \\left\\{ {F}_{n}\\right\\}, F \) be the d.f.’s of the r.v.’s \( \\left\\{ {X}_{n}\\right\\}, X \) . If \( {X}_{n} \\rightarrow X \) in pr., then \( {F}_{n}\\overset{v}{ \\rightarrow }F \) . More briefly stated, convergence in pr. implies convergence in dist. | PROOF. If \( {X}_{n} \\rightarrow X \) in pr., then for each \( f \\in {C}_{K} \), we have \( f\\left( {X}_{n}\\right) \\rightarrow f\\left( X\\right) \) in pr. as easily seen from the uniform continuity of \( f \) (actually this is true for any continuous \( f \), see Exercise 10 of Sec. 4.1). Since \( f \) is bounded... | No |
Theorem 4.4.6. If \( {X}_{n} \rightarrow X \) in dist, and \( {Y}_{n} \rightarrow 0 \) in dist., then\n\n(a) \( {X}_{n} + {Y}_{n} \rightarrow X \) in dist.\n\n(b) \( {X}_{n}{Y}_{n} \rightarrow 0 \) in dist. | PROOF. We begin with the remark that for any constant \( c,{Y}_{n} \rightarrow c \) in dist. is equivalent to \( {Y}_{n} \rightarrow c \) in pr. (Exercise 4 below). To prove (a), let \( f \in {C}_{K} \) , \( \left| f\right| \leq M \) . Since \( f \) is uniformly continuous, given \( \epsilon > 0 \) there exists \( \del... | No |
Theorem 4.5.1. If \( {X}_{n} \rightarrow X \) a.e., then for every \( r > 0 \) :\n\n(1)\n\[ \mathcal{E}\left( {\left| X\right| }^{r}\right) \leq \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathcal{E}\left( {\left| {X}_{n}\right| }^{r}\right) \] | PROOF. (1) is just a case of Fatou's lemma (see Sec. 3.2):\n\n\[ {\int }_{\Omega }{\left| X\right| }^{r}d\mathcal{P} = {\int }_{\Omega }\mathop{\lim }\limits_{n}{\left| {X}_{n}\right| }^{r}d\mathcal{P} \leq \mathop{\lim }\limits_{\frac{}{n}}{\int }_{\Omega }{\left| {X}_{n}\right| }^{r}d\mathcal{P} \] | Yes |
Theorem 4.5.2. If \( \left\{ {X}_{n}\right\} \) converges in dist. to \( X \), and for some \( p > 0 \) , \( \mathop{\sup }\limits_{n}{\mathcal{C}}^{c}\left\{ {\left| {X}_{n}\right| }^{p}\right\} = M < \infty \), then for each \( r < p \) : (2) \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathcal{E}\left( {\left|... | PROOF. We prove the second assertion since the first is similar. Let \( {F}_{n}, F \) be the d.f.’s of \( {X}_{n}, X \) ; then \( {F}_{n}\overset{v}{ \rightarrow }F \) . For \( A > 0 \) define \( {f}_{A} \) on \( {\mathcal{R}}^{1} \) as follows: (3) \[ {f}_{A}\left( x\right) = \left\{ \begin{array}{ll} {x}^{r}, & \text... | No |
Theorem 4.5.5. Suppose there is a unique d.f. \( F \) with the moments \( \left\{ {{m}^{\left( r\right) }, r \geq }\right. \) \( 1\} \), all finite. Suppose that \( \left\{ {F}_{n}\right\} \) is a sequence of d.f.’s, each of which has all its moments finite:\n\[ {m}_{n}^{\left( r\right) } = {\int }_{-\infty }^{\infty }... | PROOF. Let \( {\mu }_{n} \) be the p.m. corresponding to \( {F}_{n} \) . By Theorem 4.3.3 there exists a subsequence of \( \left\{ {\mu }_{n}\right\} \) that converges vaguely. Let \( \left\{ {\mu }_{{n}_{k}}\right\} \) be any subsequence converging vaguely to some \( \mu \) . We shall show that \( \mu \) is indeed a p... | Yes |
Theorem 5.1.1. If the \( {X}_{j} \) ’s are uncorrelated and their second moments have a common bound, then (1) is true in \( {L}^{2} \) and hence also in pr. | This simple theorem is actually due to Chebyshev, who invented his famous inequalities for its proof. | No |
Theorem 5.1.2. Under the same hypotheses as in Theorem 5.1.1, (1) holds also a.e. | PROOF. Without loss of generality we may suppose that \( {c}^{e}\left( {X}_{j}\right) = 0 \) for each \( j \), so that the \( {X}_{j} \) ’s are orthogonal. We have by (6):\n\n\[ \mathcal{E}\left( {S}_{n}^{2}\right) \leq {Mn} \]\n\nwhere \( M \) is a bound for the second moments. It follows by Chebyshev’s inequality tha... | Yes |
Theorem 5.1.3. Except for a Borel set of measure zero, every number in \( \\left\\lbrack {0,1}\\right\\rbrack \) is simply normal. | PROOF. Consider the probability space \( \\left( {\\mathcal{U},\\mathcal{B}, m}\\right) \) in Example 2 of Sec. 2.2. Let \( Z \) be the subset of the form \( m/{10}^{n} \) for integers \( n \\geq 1, m \\geq 1 \) , then \( m\\left( Z\\right) = 0 \) . If \( \\omega \\in \\mathcal{U} \\smallsetminus Z \), then it has a un... | No |
Theorem 5.2.2. Let \( \\left\\{ {X}_{n}\\right\\} \) be pairwise independent and identically distributed r.v.’s with finite mean \( m \) . Then we have\n\n(3)\n\[ \n\\frac{{S}_{n}}{n} \\rightarrow m\\;\\text{ in pr. } \n\] | PROOF. Let the common d.f. be \( F \) so that\n\n\[ \nm = \\mathcal{E}\\left( {X}_{n}\\right) = {\\int }_{-\\infty }^{\\infty }{xdF}\\left( x\\right) ,\\;\\mathcal{E}\\left( \\left| {X}_{n}\\right| \\right) = {\\int }_{-\\infty }^{\\infty }\\left| x\\right| {dF}\\left( x\\right) < \\infty .\n\]\n\nBy Theorem 3.2.1 the ... | No |
Theorem 5.2.3. Let \( \left\{ {X}_{n}\right\} \) be a sequence of independent r.v.’s with d.f.’s \( \left\{ {F}_{n}\right\} \) ; and \( {S}_{n} = \mathop{\sum }\limits_{{j = 1}}^{n}{X}_{j} \) . Let \( \left\{ {b}_{n}\right\} \) be a given sequence of real numbers increasing to \( + \infty \) .\n\nSuppose that we have\n... | PROOF OF SUFFICIENCY. Define for each \( n \geq 1 \) and \( 1 \leq j \leq n \) :\n\n\[ \n{Y}_{n, j} = \left\{ \begin{array}{ll} {X}_{j}, & \text{ if }\left| {X}_{j}\right| \leq {b}_{n}; \\ 0, & \text{ if }\left| {X}_{j}\right| > {b}_{n}; \end{array}\right.\n\]\n\nand write\n\n\[ \n{T}_{n} = \mathop{\sum }\limits_{{j = ... | Yes |
Theorem 5.3.1. Let \( \\left\\{ {X}_{n}\\right\\} \) be independent r.v.’s such that\n\n\[ \n\\forall n : \\mathcal{E}\\left( {X}_{n}\\right) = 0,\\;\\mathcal{E}\\left( {X}_{n}^{2}\\right) = {\\sigma }^{2}\\left( {X}_{n}\\right) < \\infty .\n\]\n\nThen we have for every \( \\epsilon > 0 \) :\n\n(1)\n\n\[ \n\\mathcal{P}... | PROOF. Fix \( \\epsilon > 0 \) . For any \( \\omega \) in the set\n\n\[ \n\\Lambda = \\left\\{ {\\omega : \\mathop{\\max }\\limits_{{1 \\leq j \\leq n}}\\left| {{S}_{j}\\left( \\omega \\right) }\\right| > \\epsilon }\\right\\}\n\]\n\nlet us define\n\n\[ \nv\\left( \\omega \\right) = \\min \\left\\{ {j : 1 \\leq j \\leq... | Yes |
Theorem 5.3.3. Let \( \\left\\{ {X}_{n}\\right\\} \) be independent r.v.’s and define for a fixed constant \( A > 0 \) :\n\n\[ \n{Y}_{n}\\left( \\omega \\right) = \\left\\{ \\begin{array}{ll} {X}_{n}\\left( \\omega \\right) , & \\text{ if }\\left| {{X}_{n}\\left( \\omega \\right) }\\right| \\leq A \\\\ 0, & \\text{ if ... | PROOF. Suppose that the three series converge. Applying Theorem 5.3.1\nto the sequence \( \\left\\{ {{Y}_{n} - \\mathcal{E}\\left( {Y}_{n}\\right) }\\right\\} \), we have for every \( m \\geq 1 \) :\n\n\[ \n\\mathcal{P}\\left\\{ {\\mathop{\\max }\\limits_{{n \\leq k \\leq {n}^{\\prime }}}\\left| {\\mathop{\\sum }\\limi... | Yes |
Theorem 5.3.4. If \( \left\{ {X}_{n}\right\} \) is a sequence of independent r.v.’s, then the convergence of the series \( \mathop{\sum }\limits_{n}{X}_{n} \) in pr. is equivalent to its convergence a.e. | PROOF. By Theorem 4.1.2, it is sufficient to prove that convergence of \( \mathop{\sum }\limits_{n}{X}_{n} \) in pr. implies its convergence a.e. Suppose the former; then, given \( \epsilon : 0 < \epsilon < 1 \), there exists \( {m}_{0} \) such that if \( n > m > {m}_{0} \), we have\n\n(8)\n\n\[ \mathcal{P}\left\{ {\le... | Yes |
Theorem 5.4.1. Then\n\n\[ \mathop{\sum }\limits_{n}\frac{\mathcal{E}\left( {X}_{n}^{2}\right) }{{a}_{n}^{2}} = \mathop{\sum }\limits_{n}\frac{{\sigma }_{n}^{2}}{{s}_{n}^{2}{\left( \log {s}_{n}\right) }^{1 + {2\epsilon }}} < \infty \] | by Dini's theorem, and consequently\n\n\[ \frac{{S}_{n}}{{s}_{n}{\left( \log {s}_{n}\right) }^{\left( {1/2}\right) + \epsilon }} \rightarrow 0\text{ a.e. } \] | No |
Theorem 5.4.2. Let \( \left\{ {X}_{n}\right\} \) be a sequence of independent and identically distributed r.v.'s. Then we have\n\n(8)\n\n\[ \n{\mathcal{E}}^{c}\left( \left| {X}_{1}\right| \right) < \infty \Rightarrow \frac{{S}_{n}}{n} \rightarrow \mathcal{E}\left( {X}_{1}\right) \text{ a.e.,} \n\] | PROOF. To prove (8) define \( \left\{ {Y}_{n}\right\} \) as in (4) with \( {a}_{n} = n \) . Since\n\n\[ \n\mathop{\sum }\limits_{n}\mathcal{P}\left\{ {{X}_{n} \neq {Y}_{n}}\right\} = \mathop{\sum }\limits_{n}\mathcal{P}\left\{ {\left| {X}_{n}\right| > n}\right\} = \mathop{\sum }\limits_{n}\mathcal{P}\left\{ {\left| {X}... | Yes |
Theorem 5.5.2. We have\n\n(11)\n\[ \mathop{\lim }\limits_{{t \rightarrow \infty }}\frac{N\left( t\right) }{t} = \frac{1}{m}\text{ a.e. } \]\n\nand\n\[ \mathop{\lim }\limits_{{t \rightarrow \infty }}\frac{\mathcal{E}\{ N\left( t\right) \} }{t} = \frac{1}{m} \]\n\nboth being true even if \( m = + \infty \), provided we t... | PROOF. It follows from (6) that for every \( \omega \) :\n\n\[ \begin{matrix} {S}_{N\left( {t,\omega }\right) }\left( \omega \right) \leq t < {S}_{N\left( {t,\omega }\right) + 1}\left( \omega \right) \end{matrix} \]\n\nand consequently, as soon as \( t \) is large enough to make \( N\left( {t,\omega }\right) > 0 \) ,\n... | Yes |
Theorem 5.5.3. Let \( \\left\\{ {{X}_{n}, n \\geq 1}\\right\\} \) be a sequence of independent and identically distributed r.v.’s with finite mean. For \( k \\geq 1 \) let \( {\\mathcal{T}}_{k},1 \\leq k < \\infty \), be the Borel field generated by \( \\left\\{ {{X}_{j},1 \\leq j \\leq k}\\right\\} \). Suppose that \(... | PROOF. Since \( {S}_{0} = 0 \) as usual, we have\n\n(14)\n\[ \n\\text{()}\\;\\mathcal{E}\\left( {S}_{N}\\right) = {\\int }_{\\Omega }{S}_{N}d\\mathcal{P} = \\mathop{\\sum }\\limits_{{k = 1}}^{\\infty }{\\int }_{\\left( N = k\\right) }{S}_{k}d\\mathcal{P} = \\mathop{\\sum }\\limits_{{k = 1}}^{\\infty }\\mathop{\\sum }\\... | Yes |
Theorem 6.1.1. Let \( {X}_{1} \) and \( {X}_{2} \) be independent r.v.’s with d.f.’s \( {F}_{1} \) and \( {F}_{2} \) , respectively. Then \( {X}_{1} + {X}_{2} \) has the d.f. \( {F}_{1} * {F}_{2} \) . | PROOF. We wish to show that\n\n(4)\n\n\[ \forall x : \mathcal{P}\left\{ {{X}_{1} + {X}_{2} \leq x}\right\} = \left( {{F}_{1} * {F}_{2}}\right) \left( x\right) . \]\n\nFor this purpose we define a function \( f \) of \( \left( {{x}_{1},{x}_{2}}\right) \) as follows, for fixed \( x \) :\n\n\[ f\left( {{x}_{1},{x}_{2}}\ri... | Yes |
Theorem 6.1.2. The convolution of two absolutely continuous d.f.'s with densities \( {p}_{1} \) and \( {p}_{2} \) is absolutely continuous with density \( {p}_{1} * {p}_{2} \) . | PROOF. We have by Fubini's theorem:\n\n\[ \n{\int }_{-\infty }^{x}p\left( u\right) {du} = {\int }_{-\infty }^{x}{du}{\int }_{-\infty }^{\infty }{p}_{1}\left( {u - v}\right) {p}_{2}\left( v\right) {dv} \n\]\n\n\[ \n= {\int }_{-\infty }^{\infty }\left\lbrack {{\int }_{-\infty }^{x}{p}_{1}\left( {u - v}\right) {du}}\right... | Yes |
Theorem 6.1.3. For each \( B \in \mathcal{B} \), we have\n\n\[ \left( {{\mu }_{1} * {\mu }_{2}}\right) \left( B\right) = {\int }_{{\mathcal{R}}^{1}}{\mu }_{1}\left( {B - y}\right) {\mu }_{2}\left( {dy}\right) . \] | PROOF. It is easy to verify that the set function \( \left( {{\mu }_{1} * {\mu }_{2}}\right) \left( \cdot \right) \) defined by (7) is a p.m. To show that its d.f. is \( {F}_{1} * {F}_{2} \), we need only verify that its value for \( B = ( - \infty, x\rbrack \) is given by the \( F\left( x\right) \) defined in (3). Thi... | Yes |
Corollary. If \( f \) is a ch.f., then so is \( {\left| f\right| }^{2} \) . | To prove the corollary, let \( X \) have the ch.f. \( f \) . Then there exists on some \( \Omega \) (why?) an r.v. \( Y \) independent of \( X \) and having the same d.f., and so also the same ch.f. \( f \) . The ch.f. of \( X - Y \) is\n\n\[ \mathcal{E}\left( {e}^{{it}\left( {X - Y}\right) }\right) = \mathcal{E}\left(... | Yes |
Theorem 6.1.5. For each \( \delta > 0 \), we have \( {f}_{\delta } \in {C}_{B}^{\infty } \) . Furthermore if \( f \in {C}_{U} \) , then \( {f}_{\delta } \rightarrow f \) uniformly in \( {\mathcal{R}}^{1} \) . | PROOF. It is easily verified that \( {n}_{\delta } \in {C}_{B}^{\infty } \) . Moreover its \( k \) th derivative \( {n}_{\delta }^{\left( k\right) } \) is dominated by \( {c}_{k,\delta }{n}_{2\delta } \) where \( {c}_{k,\delta } \) is a constant depending only on \( k \) and \( \delta \) so that\n\n\[ \left| {{\int }_{... | Yes |
Theorem 6.1.6. If \( \left\{ {\mu }_{n}\right\} \) and \( \mu \) are s.p.m.’s such that\n\n\[ \n\forall f \in {C}_{B}^{\infty } : {\int }_{{\mathcal{R}}^{1}}f\left( x\right) {\mu }_{n}\left( {dx}\right) \rightarrow {\int }_{{\mathcal{R}}^{1}}f\left( x\right) \mu \left( {dx}\right) ,\n\]\n\nthen \( {\mu }_{n}\overset{v}... | This is an immediate consequence of Theorem 4.4.1, and Theorem 6.1.5, if we observe that \( {C}_{0} \subset {C}_{U} \) . The reduction of the class of \ | No |
Theorem 6.2.1. If \( {x}_{1} < {x}_{2} \), then we have\n\n\[ \mu \left( \left( {{x}_{1},{x}_{2}}\right) \right) + \frac{1}{2}\mu \left( \left\{ {x}_{1}\right\} \right) + \frac{1}{2}\mu \left( \left\{ {x}_{2}\right\} \right) \]\n\n\[ = \mathop{\lim }\limits_{{T \rightarrow \infty }}\frac{1}{2\pi }{\int }_{-T}^{T}\frac{... | PROOF. Observe first that the integrand above is bounded by \( \left| {{x}_{1} - {x}_{2}}\right| \) everywhere and is \( O\left( {\left| t\right| }^{-1}\right) \) as \( \left| t\right| \rightarrow \infty \) ; yet we cannot assert that the \ | No |
Theorem 6.2.2. If two p.m.'s or d.f.'s have the same ch.f., then they are the same. | PROOF. If neither \( {x}_{1} \) nor \( {x}_{2} \) is an atom of \( \mu \), the inversion formula (4) shows that the value of \( \mu \) on the interval \( \left( {{x}_{1},{x}_{2}}\right) \) is determined by its ch.f. It follows that two p.m.'s having the same ch.f. agree on each interval whose endpoints are not atoms fo... | Yes |
Theorem 6.2.3. If \( f \in {L}^{1}\left( {-\infty , + \infty }\right) \), then \( F \) is continuously differentiable, and we have\n\n\[ \n{F}^{\prime }\left( x\right) = \frac{1}{2\pi }{\int }_{-\infty }^{\infty }{e}^{-{ixt}}f\left( t\right) {dt} \n\] | PROOF. Applying (4) for \( {x}_{2} = x \) and \( {x}_{1} = x - h \) with \( h > 0 \) and using \( F \) instead of \( \mu \), we have\n\n\[ \n\frac{F\left( x\right) + F\left( {x - }\right) }{2} - \frac{F\left( {x - h}\right) + F\left( {x - h - }\right) }{2} = \frac{1}{2\pi }{\int }_{-\infty }^{\infty }\frac{{e}^{ith} - ... | Yes |
Theorem 6.2.4. For each \( {x}_{0} \), we have\n\n(7)\n\n\[ \mathop{\lim }\limits_{{T \rightarrow \infty }}\frac{1}{2T}{\int }_{-T}^{T}{e}^{-{it}{x}_{0}}f\left( t\right) {dt} = \mu \left( \left\{ {x}_{0}\right\} \right) . \] | PROOF. Proceeding as in the proof of Theorem 6.2.1, we obtain for the integral average on the left side of (7):\n\n(8)\n\n\[ {\int }_{{\mathcal{R}}^{1} - \left\{ {x}_{0}\right\} }\frac{\sin T\left( {x - {x}_{0}}\right) }{T\left( {x - {x}_{0}}\right) }\mu \left( {dx}\right) + {\int }_{\left\{ {x}_{0}\right\} }{1\mu }\le... | Yes |
Theorem 6.2.5. We have\n\n\[ \mathop{\lim }\limits_{{T \rightarrow \infty }}\frac{1}{2T}{\int }_{-T}^{T}{\left| f\left( t\right) \right| }^{2}{dt} = \mathop{\sum }\limits_{{x \in {\mathcal{R}}^{1}}}\mu {\left( \{ x\} \right) }^{2}. \] | PROOF. Since the set of atoms is countable, all but a countable number of terms in the sum above vanish, making the sum meaningful with a value bounded by 1 . Formula (9) can be established directly in the manner of (5) and (7), but the following proof is more illuminating. As noted in the proof of the corollary to The... | Yes |
Theorem 6.2.6. \( X \) or \( \mu \) is symmetric if and only if its ch.f. is real-valued\n\n(for all \( t \) ). | PROOF. If \( X \) and \( - X \) have the same distribution, they must \ | No |
Theorem 6.3.1. Let \( \\left\\{ {{\\mu }_{n},1 \\leq n \\leq \\infty }\\right\\} \) be p.m.’s on \( {\\mathcal{R}}^{1} \) with ch.f.’s \( \\left\\{ {{f}_{n},1 \\leq }\\right. \) \( n \\leq \\infty \\} \) . If \( {\\mu }_{n} \) converges vaguely to \( {\\mu }_{\\infty } \), then \( {f}_{n} \) converges to \( {f}_{\\inft... | PROOF. Since \( {e}^{itx} \) is a bounded continuous function on \( {\\mathcal{R}}^{1} \), although complex-valued, Theorem 4.4.2 applies to its real and imaginary parts and yields (1) at once, apart from the asserted uniformity. Now for every \( t \) and \( h \) , we have, as in (ii) of Sec. 6.1:\n\n\[ \n\\left| {{f}_... | Yes |
Let \( {\mu }_{n} \) have mass \( \frac{1}{2} \) at 0 and mass \( \frac{1}{2} \) at \( n \) . Then \( {\mu }_{n} \rightarrow {\mu }_{\infty } \), where \( {\mu }_{\infty } \) has mass \( \frac{1}{2} \) at 0 and is not a p.m. | We have\n\n\[ \n{f}_{n}\left( t\right) = \frac{1}{2} + \frac{1}{2}{e}^{\mathrm{{int}}} \n\]\n\nwhich does not converge as \( n \rightarrow \infty \), except when \( t \) is equal to a multiple of \( {2\pi } \). | Yes |
Let \( {\mu }_{n} \) be the uniform distribution \( \left\lbrack {-n, n}\right\rbrack \) . Then \( {\mu }_{n} \rightarrow {\mu }_{\infty } \), where \( {\mu }_{\infty } \) is identically zero. | We have\n\n\[ \n{f}_{n}\left( t\right) = \left\{ \begin{matrix} \frac{\sin {nt}}{nt}, & \text{ if }t \neq 0 \\ 1, & \text{ if }t = 0 \end{matrix}\right. \n\]\n\nand\n\n\[ \n{f}_{n}\left( t\right) \rightarrow f\left( t\right) = \left\{ \begin{array}{ll} 0, & \text{ if }t \neq 0 \\ 1, & \text{ if }t = 0 \end{array}\right... | Yes |
Theorem 6.3.4. The topologies induced by the two metrics \( \langle {\rangle }_{1} \) and \( \langle {\rangle }_{2} \) on the space of p.m.’s on \( {\mathcal{R}}^{1} \) are equivalent. | This means that for each \( \mu \) and given \( \epsilon > 0 \), there exists \( \delta \left( {\mu ,\epsilon }\right) \) such that: \[ \langle \mu, v{\rangle }_{1} \leq \delta \left( {\mu ,\epsilon }\right) \Rightarrow \langle \mu, v{\rangle }_{2} \leq \epsilon \] \[ \langle \mu, v{\rangle }_{2} \leq \delta \left( {\m... | No |
Theorem 6.4.2. If \( F \) has a finite absolute moment of order \( k, k \) an integer \( \geq 1 \), then \( f \) has the following expansion in the neighborhood of \( t = 0 \) :\n\n(3)\n\n\[ f\left( t\right) = \mathop{\sum }\limits_{{j = 0}}^{k}\frac{{i}^{j}}{j!}{m}^{\left( j\right) }{t}^{j} + o\left( {\left| t\right| ... | PROOF. According to a theorem in calculus (see, e.g., Hardy [1], p. 290]), if \( f \) has a finite \( k \) th derivative at the point \( t = 0 \), then the Taylor expansion below is valid:\n\n(4)\n\[ f\left( t\right) = \mathop{\sum }\limits_{{j = 0}}^{k}\frac{{f}^{\left( j\right) }\left( 0\right) }{j!}{t}^{j} + o\left(... | Yes |
Theorem 6.4.3. If \( F \) has a finite mean \( m \), then\n\n\[ \frac{{S}_{n}}{n} \rightarrow m\\text{ in pr. } \] | PROOF. Since convergence to the constant \( m \) is equivalent to that in dist. to \( {\\delta }_{m} \) (Exercise 4 of Sec. 4.4), it is sufficient by Theorem 6.3.2 to prove that the ch.f. of \( {S}_{n}/n \) converges to \( {e}^{imt} \) (which is continuous). Now, by (2) of Sec. 6.1 we have\n\n\[ E\\left( {e}^{{it}\\lef... | No |
Theorem 6.4.4. If \( F \) has mean \( m \) and finite variance \( {\sigma }^{2} > 0 \), then\n\n\[ \frac{{S}_{n} - {mn}}{\sigma \sqrt{n}} \rightarrow \Phi \text{ in dist. } \]\n\nwhere \( \Phi \) is the normal distribution with mean 0 and variance 1 . | PROOF. We may suppose \( m = 0 \) by considering the r.v.’s \( {X}_{j} - m \), whose\n\nsecond moment is \( {\sigma }^{2} \) . As in the preceding proof, we have\n\n\[ \mathcal{E}\left( {\exp \left( {{it}\frac{{S}_{n}}{\sigma \sqrt{n}}}\right) }\right) = f{\left( \frac{t}{\sigma \sqrt{n}}\right) }^{n} \]\n\n\[ = {\left... | Yes |
Theorem 6.4.5. In the notation of Theorem 4.5.5, if (8) there holds together with the following condition:\n\n(6)\n\n\[ \forall t \in {\mathcal{R}}^{1} : \mathop{\lim }\limits_{{k \rightarrow \infty }}\frac{{m}^{\left( k\right) }{t}^{k}}{k!} = 0 \] \nthen \( {F}_{n}\overset{v}{ \rightarrow }F \) . | PROOF. Let \( {f}_{n} \) be the ch.f. of \( {F}_{n} \). For fixed \( t \) and an odd \( k \) we have by the Taylor expansion for \( {e}^{itx} \) with a remainder term:\n\n\[ {f}_{n}\left( t\right) = \int {e}^{itx}d{F}_{n}\left( x\right) = \int \left\{ {\mathop{\sum }\limits_{{j = 0}}^{k}\frac{{\left( itx\right) }^{j}}{... | No |
Theorem 6.4.6. Let \( X \) and \( Y \) be independent, identically distributed r.v.’s with mean 0 and variance 1 . If \( X + Y \) and \( X - Y \) are independent then the common distribution of \( X \) and \( Y \) is \( \Phi \) . | PROOF. Let \( f \) be the ch.f., then by \( \left( 1\right) ,{f}^{\prime }\left( 0\right) = 0,{f}^{\prime \prime }\left( 0\right) = - 1 \) . The ch.f. of \( X + Y \) is \( f{\left( t\right) }^{2} \) and that of \( X - Y \) is \( f\left( t\right) f\left( {-t}\right) \) . Since these two r.v.’s are independent, the ch.f.... | Yes |
Theorem 6.4.7. A ch.f. is that of a lattice distribution if and only if there exists a \( {t}_{0} \neq 0 \) such that \( \left| {f\left( {t}_{0}\right) }\right| = 1 \) . | PROOF. The \ | No |
Theorem 6.5.1. If \( f \) is positive definite, then for each \( t \in {\mathcal{R}}^{1} \) :\n\n\[ f\left( {-t}\right) = \overline{f\left( t\right) },\;\left| {f\left( t\right) }\right| \leq f\left( 0\right) . \] | PROOF. Taking \( n = 1,{t}_{1} = 0,{z}_{1} = 1 \) in (1), we see that\n\n\[ f\left( 0\right) \geq 0\text{.} \]\n\nTaking \( n = 2,{t}_{1} = 0,{t}_{2} = t,{z}_{1} = {z}_{2} = 1 \), we have\n\n\[ {2f}\left( 0\right) + f\left( t\right) + f\left( {-t}\right) \geq 0 \]\n\nchanging \( {z}_{2} \) to \( i \), we have\n\n\[ f\l... | Yes |
Theorem 6.5.3. Let \( f \) on \( {\mathcal{R}}^{1} \) satisfy the following conditions, for each \( t \) :\n\n(9)\n\[ f\left( 0\right) = 1,\;f\left( t\right) \geq 0,\;f\left( t\right) = f\left( {-t}\right) ,\]\n\n\( f \) is decreasing and continuous convex in \( {\mathcal{R}}_{ + } = \lbrack 0,\infty ) \) . Then \( f \... | PROOF. Without loss of generality we may suppose that\n\n\[ f\left( \infty \right) = \mathop{\lim }\limits_{{t \rightarrow \infty }}f\left( t\right) = 0 \]\n\notherwise we consider \( \left\lbrack {f\left( t\right) - f\left( \infty \right) }\right\rbrack /\left\lbrack {f\left( 0\right) - f\left( \infty \right) }\right\... | Yes |
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