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Proposition 3.4 Let \( {\left( {X}_{n}\right) }_{n \geq 0} \) and \( {\left( {Y}_{n}\right) }_{n \geq 0} \) be supermartingales. Let a be a real number.\n\n1. If \( a \) is positive, then \( {\left( a{X}_{n}\right) }_{n \geq 0} \) is a supermartingale.\n\n2. If \( a \) is negative, then \( {\left( a{X}_{n}\right) }_{n ... | The proof of this proposition is a verification. For the last sentence, observe that a stochastic process is a martingale if and only if it is both a supermartingale and a submartingale. | No |
Proposition 3.5 Let \( {\left( {X}_{n}\right) }_{n \geq 0} \) be a martingale. Let \( \phi : \mathbb{R} \rightarrow \mathbb{R} \) be a convex function such that for all \( n \geq 0,\phi \left( {X}_{n}\right) \) is integrable. Then \( {\left( \phi \left( {X}_{n}\right) \right) }_{n \geq 0} \) is a submartingale.\n\nThe ... | Proof. Let us treat the case where \( X \) is a submartingale. The process \( {\left( \phi \left( {X}_{n}\right) \right) }_{n \geq 0} \) is adapted and integrable. For all \( n \geq 0 \), we have\n\n\[ \mathbb{E}\left\lbrack {\phi \left( {X}_{n + 1}\right) \mid {\mathcal{F}}_{n}}\right\rbrack \geq \phi \left( {\mathbb{... | Yes |
1. If \( X \) is a martingale and each random variable \( {H}_{n} \) is bounded, then \( H \bullet X \) is a martingale. | Proof. For all \( n \geq 1 \), the random variable \( {\left( H \bullet X\right) }_{n} \) is a function of the random variables \( {X}_{0},\ldots ,{X}_{n},{H}_{1},\ldots ,{H}_{n} \) which all are \( {\mathcal{F}}_{n} \) -measurable. Thus, \( {\left( H \bullet X\right) }_{n} \) is \( {\mathcal{F}}_{n} \) -measurable and... | Yes |
Corollary 3.11 A non-negative supermartingale converges almost surely towards a limit which is an integrable random variable. | Proof. For all \( n \geq 0 \), we have\n\n\[ \mathbb{E}\left\lbrack \left| {X}_{n}\right| \right\rbrack = \mathbb{E}\left\lbrack {X}_{n}\right\rbrack \leq \mathbb{E}\left\lbrack {X}_{0}\right\rbrack \]\n\nso that a non-negative supermartingale is bounded in \( {L}^{1} \) . The result now follows from Theorem 3.9. | Yes |
Proposition 3.12 (Doob’s upcrossing lemma) Let \( X \) be a supermartingale. Let \( a, b \) be two real numbers such that \( a < b \). For all \( n \geq 1 \), one has \[ \mathbb{E}\left\lbrack {{U}_{n}\left( {X;a, b}\right) }\right\rbrack \leq \frac{1}{b - a}\mathbb{E}\left\lbrack {\left( {X}_{n} - a\right) }^{ - }\rig... | Proof. Let us think in the same terms as before Definition 3.7. Here is the line of reasoning of a very clever gambler. \ | No |
Proposition 3.13 The process \( {\left( {m}^{-n}{X}_{n}\right) }_{n \geq 0} \) is a martingale with respect to the filtration \( {\left( {\mathcal{F}}_{n}\right) }_{n \geq 0} \) . | Proof. By definition, \( {X}_{0} \) is \( {\mathcal{F}}_{0} \) -measurable. Then, for all \( n \geq 1,{X}_{n} \) is a function of \( {X}_{n - 1} \) and \( \left\{ {{Z}_{n - 1, k} : k \geq 1}\right\} \) . By induction, and by definition of \( {\mathcal{F}}_{n} \), it follows that \( {X}_{n} \) is \( {\mathcal{F}}_{n} \)... | Yes |
Proposition 3.18 Let \( X \) be a supermartingale. Let \( T \) be a stopping time. The process \( {X}^{T} \) is a supermartingale. | Proof. In the picture of the gambler, stopping a process at time \( T \) amounts to playing 1 until time \( T \) and then playing 0 forever. Let us define accordingly, for all \( n \geq 1 \) ,\n\n\[ \n{H}_{n} = {\mathbb{1}}_{\{ T \geq n\} } = {\mathbb{1}}_{{\left\{ T \leq n - 1\right\} }^{c}}.\n\]\n\nWe play 1 during t... | Yes |
Theorem 3.19 Let \( X \) be a supermartingale. Let \( T \) be a stopping time. If one of the following conditions is satisfied :\n\n1. \( T \) is bounded,\n\n2. \( T \) is integrable and there exists \( M > 0 \) such that for all \( n \geq 0,\left| {{X}_{n + 1} - {X}_{n}}\right| \leq M \) ,\n\n3. \( T \) is almost sure... | Proof. 1. If \( T \) is bounded by an integer \( N \), then \( \mathbb{E}\left\lbrack {X}_{0}\right\rbrack \geq \mathbb{E}\left\lbrack {X}_{T \land N}\right\rbrack = \mathbb{E}\left\lbrack {X}_{T}\right\rbrack \) .\n\n2. Since \( T \) is integrable, it is almost surely finite, so that \( {X}_{T \land n} \) converges al... | Yes |
Lemma 3.20 Let \( X \) be a submartingale. Let \( S \) and \( T \) be two bounded stopping times such that \( S \leq T \) almost surely. Then \( \mathbb{E}\left\lbrack {X}_{S}\right\rbrack \leq \mathbb{E}\left\lbrack {X}_{T}\right\rbrack \) . | Proof. Let our gambler start playing at time \( S \) and stop at time \( T \) . This amounts to defining, for all \( n \geq 1 \) ,\n\n\[ \n{H}_{n} = {\mathbb{1}}_{\{ S < n \leq T\} } = {\mathbb{1}}_{\{ S \leq n - 1\} }{\mathbb{1}}_{\{ T \leq n - 1{\} }^{c}}.\n\]\n\nThe way we wrote it makes it clear that \( H = {\left(... | Yes |
Proposition 3.21 (Doob’s maximal inequality) Let \( X \) be a submartingale. For all integer \( n \geq 0 \) and all real number \( a \), the following inequalities hold :\n\n\[ \na\mathbb{P}\left( {\mathop{\sup }\limits_{{0 \leq k \leq n}}{X}_{k} \geq a}\right) \leq \mathbb{E}\left\lbrack {{X}_{n}{\mathbb{1}}_{\left\{ ... | Proof. Choose a real number \( a \) . Define \( T \) as the first hitting time of \( \lbrack a, + \infty ) \) for \( X \) :\n\n\[ \nT = \inf \left\{ {n \geq 0 : {X}_{n} \geq a}\right\} .\n\]\n\nConsider now an integer \( n \geq 0 \) . The random variable \( {X}_{T \land n} \) is equal to \( {X}_{T} \), hence greater or... | Yes |
Proposition 3.22 Let \( X \) be a martingale. Let \( p > 1 \) be a real number. For all integer \( n \geq 1 \), one has\n\n\[ \mathbb{E}\left\lbrack {\left( {X}_{n}^{ * }\right) }^{p}\right\rbrack \leq {\left( \frac{p}{p - 1}\right) }^{p}\mathbb{E}\left\lbrack {\left| {X}_{n}\right| }^{p}\right\rbrack \] | Proof. By the previous proposition applied to the submartingale \( {\left( \left| {X}_{n}\right| \right) }_{n \geq 0} \), we have, for all \( a > 0 \), \n\n\[ a\mathbb{P}\left( {{X}_{n}^{ * } \geq a}\right) \leq \mathbb{E}\left\lbrack {\left| {X}_{n}\right| {\mathbb{1}}_{\left\{ {X}_{n}^{ * } \geq a\right\} }}\right\rb... | Yes |
Theorem 3.23 Let \( X \) be a martingale. Let \( p > 1 \) be a real number. Assume that \( X \) is bounded in \( {L}^{p} \) . Then \( X \) converges almost surely and in \( {L}^{p} \) towards a random variable \( {X}_{\infty } \) which satisfies\n\n\[ E\left\lbrack {\left| {X}_{\infty }\right| }^{p}\right\rbrack = \sup... | Proof. Let us assume that \( X \) is bounded in \( {L}^{p} \) . In particular, \( X \) is bounded in \( {L}^{1} \), hence it converges almost surely to a random variable \( {X}_{\infty } \) .\n\nThe previous proposition, the fact that \( {\left( {X}_{n}^{ * }\right) }_{n \geq 0} \) is a non-decreasing sequence which co... | Yes |
Proposition 3.24 The increments of a square-integrable martingale are orthogonal in \( {L}^{2} \) . More precisely, if \( X = {\left( {X}_{n}\right) }_{n \geq 0} \) is a martingale on \( \left( {\Omega ,\mathcal{A},{\left( {\mathcal{F}}_{n}\right) }_{n \geq 0},\mathbb{P}}\right) \) such that \( \mathbb{E}\left\lbrack {... | Proof. Consider three integers \( m \leq n \leq p \) . We have\n\n\[ \mathbb{E}\left\lbrack {\left( {{X}_{n} - {X}_{m}}\right) \left( {{X}_{p} - {X}_{n}}\right) }\right\rbrack = \mathbb{E}\left\lbrack {\mathbb{E}\left\lbrack {\left( {{X}_{n} - {X}_{m}}\right) \left( {{X}_{p} - {X}_{n}}\right) \mid {\mathcal{F}}_{n}}\ri... | Yes |
Proposition 3.25 Let \( X \) be a square-integrable martingale. For all \( n \geq 0 \), one has\n\n\[ \n\mathbb{E}\left\lbrack {X}_{n}^{2}\right\rbrack = \mathbb{E}\left\lbrack {X}_{0}^{2}\right\rbrack + \mathop{\sum }\limits_{{k = 0}}^{{n - 1}}\mathbb{E}\left\lbrack {\left( {X}_{k + 1} - {X}_{k}\right) }^{2}\right\rbr... | Proof. This follows immediately from the previous proposition and the Pythagorean theorem. | No |
Proposition 3.26 (Doob’s decomposition) Let \( \\left( {\\Omega ,\\mathcal{A},{\\left( {\\mathcal{F}}_{n}\\right) }_{n \\geq 0},\\mathbb{P}}\\right) \) be a filtered probability space. Let \( X = {\\left( {X}_{n}\\right) }_{n \\geq 0} \) be an adapted integrable stochastic process.\n\n1. There exists a martingale \( M ... | Proof. If such a decomposition exists, then we must have \( {A}_{0} = 0 \) and, for all \( n \\geq 0 \), \n\n\[ \n\\mathbb{E}\\left\\lbrack {{X}_{n + 1} - {X}_{n} \\mid {\\mathcal{F}}_{n}}\\right\\rbrack = {A}_{n + 1} - {A}_{n}. \n\]\n\nThis formula defines inductively a previsible process \( {\\left( {A}_{n}\\right) }... | Yes |
Theorem 3.28 Let \( X \) be a square-integrable martingale. The sequence \( {\left( {X}_{n}\right) }_{n \geq 0} \) converges almost surely on the event \( \left\{ {\langle X{\rangle }_{\infty } < \infty }\right\} \) . | Proof. Choose an integer \( k \geq 0 \) and define\n\n\[ \n{T}_{k} = \inf \left\{ {n \geq 0 : \langle X{\rangle }_{n + 1} > k}\right\} \n\] \n\nIt is a stopping time, because it is the hitting time of the Borel subset \( \left( {k, + \infty }\right) \) of \( \mathbb{R} \) by the adapted process \( {\left( \langle X{\ra... | Yes |
Proposition 3.30 Let \( \\left( {\\Omega ,\\mathcal{A},\\mathbb{P}}\\right) \) be a probability space. Let \( {\\left( {X}_{i}\\right) }_{i \\in I} \) be a family of random variables which is bounded in \( {L}^{1} \). The following two assertions are equivalent.\n\n1. The family \( {\\left( {X}_{i}\\right) }_{i \\in I}... | Proof. \( 1 \\Rightarrow 2 \) . Choose \( \\varepsilon > 0 \) . Since the family \( {\\left( {X}_{i}\\right) }_{i \\in I} \) is uniformly integrable, there exists a real \( M > 0 \) such that for all \( i \\in I,\\mathbb{E}\\left\\lbrack {\\left| {X}_{i}\\right| {\\mathbb{1}}_{\\left| {X}_{i}\\right| > M}}\\right\\rbra... | Yes |
Corollary 3.31 Let \( \\left( {\\Omega ,\\mathcal{A},\\mathbb{P}}\\right) \) be a probability space. Let \( Z \) be an integrable variable. The family \( \\{ \\mathbb{E}\\left\\lbrack {Z \\mid \\mathcal{B}}\\right\\rbrack : \\mathcal{B} \) sub- \( \\sigma \) -field of \( \\mathcal{A}\\}\\} \) is uniformly integrable. | Proof. For each sub- \( \\sigma \) -field \( \\mathcal{B} \) of \( \\mathcal{A} \), the random variable \( \\mathbb{E}\\left\\lbrack {Z \\mid \\mathcal{B}}\\right\\rbrack \) is integrable. Let us now choose \( \\varepsilon \) . Thanks to the uniform integrability of the family which consists in the single random variab... | Yes |
Lemma 3.35 Let \( S \) and \( T \) two stopping times. If \( S \leq T \), then \( {\mathcal{F}}_{S} \subset {\mathcal{F}}_{T} \) . | Proof. Consider \( A \in {\mathcal{F}}_{S} \) . Choose \( n \geq 0 \) . Then the event\n\n\[ A \cap \{ T = n\} = \mathop{\bigcup }\limits_{{k = 0}}^{\infty }\left( {A\cap \{ T = n\} \cap \{ S = k\} }\right) = \mathop{\bigcup }\limits_{{k = 0}}^{n}\left( {A\cap \{ S = k\} \cap \{ T = n\} }\right) \]\n\nbelongs to \( {\m... | Yes |
Lemma 3.36 Let \( X = {\left( {X}_{n}\right) }_{n \geq 0} \) be an adapted stochastic process. Let \( T \) be a stopping time. If one of the following assumptions is satisfied :\n\n1. \( T \) is finite almost surely,\n\n2. \( {X}_{n} \) converges almost surely to \( {X}_{\infty } \),\n\nthen \( {X}_{T} \) is well defin... | Proof. We know already that \( {X}_{T} \) is well defined under any of the two assumptions. Let us prove that \( {X}_{T} \) is \( {\mathcal{F}}_{T} \) measurable. For this, let us consider a Borel subset \( B \) of \( \mathbb{R} \) and an integer \( n \geq 0 \) . We have\n\n\[ \left\{ {{X}_{T} \in B}\right\} \cap \{ T ... | Yes |
Theorem 3.37 Let \( X \) be a uniformly integrable martingale. Let \( T \) be a stopping time. Then\n\n\[ \n{X}_{T} = \mathbb{E}\left\lbrack {{X}_{\infty } \mid {\mathcal{F}}_{T}}\right\rbrack \n\]\n\nIn particular, \( \mathbb{E}\left\lbrack {X}_{T}\right\rbrack = \mathbb{E}\left\lbrack {X}_{\infty }\right\rbrack = \ma... | Proof. Let us check that \( {X}_{T} \) is integrable. We have\n\n\[ \n\mathbb{E}\left\lbrack \left| {X}_{T}\right| \right\rbrack = \mathop{\sum }\limits_{{n = 0}}^{\infty }\mathbb{E}\left\lbrack {\left| {X}_{n}\right| {\mathbb{1}}_{\{ T = n\} }}\right\rbrack + \mathbb{E}\left\lbrack {\left| {X}_{\infty }\right| {\mathb... | Yes |
Lemma 3.39 A backward martingale is uniformly integrable. | Proof. Indeed, we have for all \( n \leq 0 \) the equality \( {X}_{n} = \mathbb{E}\left\lbrack {{X}_{0} \mid {\mathcal{F}}_{n}}\right\rbrack \), and the result is a consequence of Corollary 3.31. | Yes |
Theorem 3.40 Let \( X \) be a backward martingale. Set \( {\mathcal{F}}_{-\infty } = \mathop{\bigcap }\limits_{{n < 0}}{\mathcal{F}}_{n} \) . Then, as \( n \) tends to \( - \infty ,{X}_{n} \) converges almost surely and in \( {L}^{1} \) to \( \mathbb{E}\left\lbrack {{X}_{0} \mid {\mathcal{F}}_{-\infty }}\right\rbrack \... | Proof. The proof that \( X \) converges almost surely is the same as that for usual martingales. Indeed, for all \( N \leq 0 \), the sequence \( {X}_{N},{X}_{N + 1},\ldots ,{X}_{0} \) is a usual martingale and for all \( a < b \), the number of upcrossings of this martingale can be estimated by Doob’s upcrossing lemma.... | Yes |
Lemma 3.41 Let \( X,{Y}_{1},\ldots ,{Y}_{n} \) and \( {X}^{\prime },{Y}_{1}^{\prime },\ldots ,{Y}_{n}^{\prime } \) be random variables. Assume that the random vectors \( \left( {X,{Y}_{1},\ldots ,{Y}_{n}}\right) \) and \( \left( {{X}^{\prime },{Y}_{1}^{\prime },\ldots ,{Y}_{n}^{\prime }}\right) \) have the same distrib... | Proof. Let \( B \) be a Borel subset of \( {\mathbb{R}}^{n} \) . The assumptions imply that \[ {\int }_{\Omega }X{\mathbb{1}}_{B}\left( {{Y}_{1},\ldots ,{Y}_{n}}\right) {dP} = {\int }_{\Omega }h\left( {{Y}_{1},\ldots ,{Y}_{n}}\right) {\mathbb{1}}_{B}\left( {{Y}_{1},\ldots ,{Y}_{n}}\right) {dP}. \] Denoting by \( \mu \)... | Yes |
Theorem 3.43 (Strong law of large numbers) Let \( {\left( {X}_{n}\right) }_{n \geq 1} \) be an i.i.d. sequence of integrable random variables. Then as \( n \) tends to infinity,\n\n\[ \n\frac{{X}_{1} + \ldots + {X}_{n}}{n} \rightarrow \mathbb{E}\left\lbrack {X}_{1}\right\rbrack \n\]\n\nalmost surely and in \( {L}^{1} \... | Proof. Set \( {S}_{0} = 0 \) and, for all \( n \geq 1,{S}_{n} = {X}_{1} + \ldots ,{X}_{n} \) . For all \( n \geq 0 \), set \( {\mathcal{F}}_{-n} = \) \( \sigma \left( {{S}_{n},{S}_{n + 1},\ldots }\right) = \sigma \left( {{S}_{k} : k \geq n}\right) \) . Then \( {\left( {\mathcal{F}}_{n}\right) }_{n \leq 0} \) is a backw... | Yes |
Proposition 4.3 With the notation of Definition 4.2, \( X \) is a Markov chain on \( E \) with transition kernel \( P \) if and only if the following condition holds:\n\n\[ \forall n \geq 0,\forall {x}_{0},\ldots ,{x}_{n} \in E,\mathbb{P}\left( {{X}_{0} = {x}_{0},\ldots ,{X}_{n} = {x}_{n}}\right) = \mathbb{P}\left( {{X... | Proof. Let us start by the ’only if’ part, that is, the implication \( \Rightarrow \) . We prove (3) by induction on \( n \) . For \( n = 0 \), it reduces to \( \mathbb{P}\left( {{X}_{0} = {x}_{0}}\right) = \mathbb{P}\left( {{X}_{0} = {x}_{0}}\right) \), which is true. Let us now assume that (3) has been proved up to r... | Yes |
Proposition 4.4 Let \( X \) be a Markov chain on \( E \) with transition kernel \( P \) . For all \( n \geq 0 \) and all \( y \in E \), one has\n\n\[ \mathbb{P}\left( {{X}_{n} = y \mid {X}_{0}}\right) = {P}^{n}\left( {{X}_{0}, y}\right) . \] | Proof. We need to prove that for every element \( {x}_{0} \) of \( E \) such that \( \mathbb{P}\left( {{X}_{0} = {x}_{0}}\right) > 0 \), the equality \( \mathbb{P}\left( {{X}_{n} = y \mid {X}_{0} = {x}_{0}}\right) = {P}^{n}\left( {{x}_{0}, y}\right) \) holds. But for such an \( {x}_{0} \), we have\n\n\[ \mathbb{P}\left... | Yes |
Proposition 4.5 Let \( X \) be a Markov chain on \( E \) with transition kernel \( P \) . Let \( N \geq 0 \) be an integer. For every \( n \geq 0 \), set \( {Y}_{n} = {X}_{N + n} \) . Then \( Y = {\left( {Y}_{n}\right) }_{n \geq 0} \) is a Markov chain on \( E \) with transition kernel \( P \) . | Proof. Let us consider \( n \geq 0 \) and \( {y}_{0},\ldots ,{y}_{n} \in E \) . Firstly, we know from the previous proposition that\n\n\[ \n\mathbb{P}\left( {{Y}_{0} = {y}_{0}}\right) = \mathop{\sum }\limits_{{{x}_{0} \in \mathbb{E}}}\mathbb{P}\left( {{X}_{0} = {x}_{0},{Y}_{0} = {y}_{0}}\right) \n\]\n\n\[ \n= \mathop{\... | Yes |
Proposition 4.7 There exists a probability space \( \left( {\Omega ,\mathcal{A},\mathbb{P}}\right) \) and a sequence \( U = {\left( {U}_{n}\right) }_{n \geq 0} \) of i.i.d. random variables with common distribution equal to the uniform distribution on the interval \( \left\lbrack {0,1}\right\rbrack \) . | Proof. Take \( \left( {\Omega ,\mathcal{A},\mathbb{P}}\right) = \left( {\left\lbrack {0,1}\right\rbrack ,{\mathcal{B}}_{\left\lbrack 0,1\right\rbrack },\lambda }\right) \), where \( \lambda \) is the Lebesgue measure. For all \( n \geq 1 \) , define a random variable \( {B}_{n} \) on this probability space by setting\n... | Yes |
Lemma 4.10 A map \( f \) from a measurable space \( \left( {\Omega ,\mathcal{A}}\right) \) into \( \left( {{E}^{\mathbb{N}},\mathcal{C}}\right) \) is measurable if and only if for all \( n \geq 0 \), the map \( {\widehat{X}}_{n} \circ f \) is measurable from \( \left( {\Omega ,\mathcal{A}}\right) \) to \( \left( {E,\ma... | Proof. The 'only if' part of the statement is a consequence of the fact that a composition of measurable maps is measurable. Let us prove the 'if' part: let us assume that for all \( n \geq 0 \), the map \( {\widetilde{X}}_{n} \circ f \) is measurable. Let us define the class\n\n\[ \mathcal{I} = \left\{ {C \in \mathcal... | Yes |
Theorem 4.16 (Strong Markov property) Let \( P \) be a transition kernel on the state space \( E \) . Let \( \left( {\Omega ,\mathcal{A},{\left( {\mathcal{F}}_{n}\right) }_{n \geq 0},{\left( {\mathbb{P}}_{x}\right) }_{x \in E}, X = {\left( {X}_{n}\right) }_{n \geq 0}}\right) \) be a Markov chain on \( E \) with transit... | Proof. Just as in the proof of the weak Markov property, the right-hand side of (7), which is a function of \( {X}_{T} \), is \( {\mathcal{F}}_{T} \) -measurable, and it suffices to prove that for every \( A \in {\mathcal{F}}_{T} \), both sides of (7) have the same \( {\mathbb{P}}_{x} \) -integral over \( A \) . Let us... | Yes |
Corollary 4.17 We use the notation of Theorem 4.16. Let \( x, y \) be elements of \( E \) and let \( T \) be a stopping time such that \( T < \infty \) and \( {X}_{T} = y{\mathbb{P}}_{x} \) -a.s. Then under \( {\mathbb{P}}_{x},{\theta }_{T}\left( X\right) \) is independent of \( {\mathcal{F}}_{T} \) and has the same di... | Proof. The statement is equivalent to saying that for all non-negative measurable function \( F \) on \( {E}^{\mathbb{N}} \) and all \( B \in {\mathcal{F}}_{T} \) ,\n\n\[ \n{\mathbb{E}}_{x}\left\lbrack {F\left( {{\theta }_{T}\left( X\right) }\right) {\mathbb{1}}_{B}}\right\rbrack = {\mathbb{P}}_{x}\left( B\right) {\mat... | Yes |
Proposition 4.18 Let \( x \) be an element of \( E \) . Exactly one of the following two situations occurs.\n\n1. \( {\mathbb{P}}_{x}\left( {{T}_{x} < \infty }\right) = 1 \) . In this case, \( {N}_{x} = \infty {\mathbb{P}}_{x} \) -a.s. and one says that \( x \) is recurrent.\n\n2. \( {\mathbb{P}}_{x}\left( {{T}_{x} < \... | Proof. Let \( k \geq 1 \) be an integer and let us compute \( {\mathbb{P}}_{x}\left( {{N}_{x} \geq k + 1}\right) \).\n\n\[ \n{\mathbb{P}}_{x}\left( {{N}_{x} \geq k + 1}\right) = {\mathbb{P}}_{x}\left( {{\widehat{N}}_{x}\left( X\right) \geq k + 1}\right) \n\]\n\n\[ \n= {\mathbb{P}}_{x}\left( {{T}_{x} < \infty ,{\widehat... | Yes |
Proposition 4.20 1. For all \( x, y \in E \), we have\n\n\[ G\left( {x, y}\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{P}^{n}\left( {x, y}\right) . \] | Proof. 1. By definition of \( G \) and the monotonce convergence theorem,\n\n\[ G\left( {x, y}\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\mathbb{E}}_{x}\left\lbrack {\mathbb{1}}_{\left\{ {X}_{n} = y\right\} }\right\rbrack = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\mathbb{P}}_{x}\left\lbrack {{X}_{n} = y}\rig... | Yes |
Lemma 4.21 On \( E \), the binary relation \( \rightarrow \) is reflexive and transitive. | Proof. For every \( x \in E \), one has \( G\left( {x, x}\right) \geq 1 \), so that \( x \rightarrow x \) . This means that \( \rightarrow \) is a reflexive relation.\n\nTo prove that it is transitive, consider three states \( x, y \) and \( z \) such that \( x \rightarrow y \) and \( y \rightarrow z \) . If any two of... | Yes |
Proposition 4.22 Let \( x \) and \( y \) be two states. Assume that \( x \) is recurrent and \( x \) leads to \( y \) . Then \( y \) is recurrent, \( y \) leads to \( x \), and \[ {\mathbb{P}}_{x}\left( {{T}_{y} < \infty }\right) = {\mathbb{P}}_{y}\left( {{T}_{x} < \infty }\right) = 1. \] | Proof. If \( y = x \), then there is nothing to prove. Let us assume that \( y \neq x \) . Then \[ 0 = {\mathbb{P}}_{x}\left( {{T}_{x} = \infty }\right) \geq {\mathbb{P}}_{x}\left( {{T}_{y} < \infty ,{\widehat{T}}_{x}\left( {{\theta }_{{T}_{y}}\left( X\right) }\right) = \infty }\right) \] \[ = {\mathbb{P}}_{x}\left( {{... | Yes |
Proposition 4.23 Under the assumptions of the previous proposition, we have\n\n\[ \n{\mathbb{P}}_{x}\left( {{N}_{y} = \infty }\right) = {\mathbb{P}}_{y}\left( {{N}_{x} = \infty }\right) = 1.\n\]\n\nIn particular,\n\n\[ \nG\left( {x, y}\right) = G\left( {y, x}\right) = \infty .\n\] | Proof. Indeed, \( y \) is recurrent and for all \( k \geq 1 \) ,\n\n\[ \n{\mathbb{P}}_{x}\left( {{N}_{y} \geq k}\right) = {\mathbb{P}}_{x}\left( {{T}_{y} < \infty }\right) {\mathbb{P}}_{y}\left( {{N}_{y} \geq k}\right) = 1.\n\]\n\nHence, \( {\mathbb{P}}_{x}\left( {{N}_{y} = \infty }\right) = 1 \) . | Yes |
Theorem 4.24 (Classification of states) Let \( R \) be the subset of \( E \) consisting of all recurrent states. Let\n\n\[ R = \mathop{\bigsqcup }\limits_{{i \in I}}{R}_{i} \]\n\nbe the partition of \( R \) in equivalence classes for the relation \( \sim \) . Consider a state \( x \in E \) .\n\n1. If \( x \) is recurre... | Proof. 1. Consider \( y \in {R}_{i} \) . Then, according to Proposition 4.23, we have \( {\mathbb{P}}_{x}\left( {{N}_{y} = \infty }\right) = \) 1. Consider now \( z \in E \smallsetminus {R}_{i} \) . If \( z \) is transient, then \( G\left( {x, z}\right) = 0 \) by Proposition 4.22. If \( z \) is recurrent, then \( G\lef... | Yes |
Corollary 4.26 Let us assume that the Markov chain is irreducible. Then we are in exactly one of the following two situations.\n\n- All states are recurrent, there is only one recurrence class and\n\n\[ \n{\mathbb{P}}_{x}\left( {\forall y \in E,{N}_{y} = \infty }\right) = 1.\n\]\n\n- All states are transient and\n\n\[ ... | Proof. If there is one recurrent state, then by Proposition 4.22, all states are recurrent and there is only one class. Moreover, Proposition 4.23 implies that every state is visited infinitely often \( {\mathbb{P}}_{x} \) -almost surely for every \( x \in E \) .\n\nIf there is no recurrent state, then all states are t... | Yes |
Proposition 4.33 Assume that the chain is irreducible and recurrent. Then exactly one of the following two situations occurs.\n\n1. All invariant measures have infinite total mass and for all \( x \in E \), we have \( {\mathbb{E}}_{x}\left\lbrack {T}_{x}\right\rbrack = \infty \) . In this case, the Markov chain is call... | Proof. Let \( x \) be a state and let \( \mu \) be the unique invariant measure on \( E \) such that \( \mu \left( x\right) = 1 \) , which is given by Theorem 4.31. We have\n\n\[ \mu \left( E\right) = \mathop{\sum }\limits_{{y \in E}}{\mathbb{E}}_{x}\left\lbrack {\mathop{\sum }\limits_{{i = 0}}^{{{T}_{x} - 1}}{\mathbb{... | Yes |
Proposition 4.34 Assume that the chain is irreducible. If there exists an invariant probability measure, then the chain is recurrent. | Proof. Let \( y \) be a state such that \( \pi \left( y\right) > 0 \) . Let us compute \( G\left( {y, y}\right) \) . Remember that for every state \( x \), we have \( G\left( {x, y}\right) \leq G\left( {y, y}\right) \) . Thus,\n\n\[ \nG\left( {y, y}\right) = \mathop{\sum }\limits_{{x \in E}}\pi \left( x\right) G\left( ... | Yes |
Corollary 4.36 Suppose that the Markov chain is irreducible and recurrent.\n\n1. Assume that the chain is positive recurrent. Let \( \pi \) denote the unique invariant probability measure. Then for all probability measure \( \nu \) on \( E \) and all \( y \in E \), we have\n\n\[ \frac{1}{n}\mathop{\sum }\limits_{{i = 0... | Proof. It suffices to apply the ergodic theorem to the function \( g = 1 \) . | No |
Lemma 4.37 Let \( I \subset \mathbb{N} \) be a semigroup. Let \( d \) be the g.c.d. of the elements of \( I \) .\n\n1. The subgroup of \( \mathbb{Z} \) generated by \( I \) is the set\n\n\[ I - I = \{ x - y : x, y \in \mathbb{Z}\} .\n\]\n2. The equality \( I - I = d\mathbb{Z} \) holds.\n3. If \( d = 1 \), then there ex... | Proof. 1. The subgroup of \( \mathbb{Z} \) generated by \( I \) certainly contains \( I - I \) . Our claim is thus equivalent to the fact that \( I - I \) is a subgroup of \( \mathbb{Z} \) . Since \( I - I \) contains 0, it suffices to check that it is stable by subtraction. But for all \( n, m,{n}^{\prime },{m}^{\prim... | Yes |
Proposition 4.39 If the chain is irreducible, then all states have the same period. | Proof. Consider two states \( x \) and \( y \) . By irreducibility, there exists integers \( k \) and \( l \) such that \( {P}^{k}\left( {x, y}\right) > 0 \) and \( {P}^{l}\left( {y, x}\right) > 0 \) . It follows that\n\n\[ k + {I}_{y} + l \subset {I}_{x} \]\n\nThus, for all \( n, m \in {I}_{y} \), we have\n\n\[ n - m ... | Yes |
Proposition 4.41 Assume that the Markov chain is irreducible and aperiodic. For all \( x, y \in E \), there exists an integer \( {n}_{0} \) such that for all \( n \geq {n}_{0},{P}^{n}\left( {x, y}\right) > 0 \) . | Proof. Consider \( x, y \in E \) . Let \( {n}_{2} \geq 0 \) be such that \( {P}^{{n}_{2}}\left( {x, y}\right) > 0 \) . Such an integer \( {n}_{2} \) exists because the chain is irreducible. Then, since \( {d}_{x} = 1 \), there exists \( {n}_{1} \) such that \( {I}_{x} \) contains every integer larger than \( {n}_{1} \)... | Yes |
Theorem 2.1. (Gaussian tail distribution). If \( \xi \) is a Gaussian \( \mathcal{N}\left( {0,1}\right) \) random variable, then for any \( x > 0 \) ,\n\n\[ \frac{1}{\sqrt{2\pi }}\left( {\frac{1}{x} - \frac{1}{{x}^{3}}}\right) {\mathrm{e}}^{-{x}^{2}/2} \leq \mathbb{P}\left( {\xi > x}\right) \leq \frac{1}{\sqrt{2\pi }}\... | Proof of Theorem 2.1. Is an exercise. | No |
Proposition 2.4. (Convergence of sequences of Gaussian random variables). Let \( \left( {\xi }_{n}\right) \) be a sequence of random variables such that for any \( n,{\xi }_{n} \) is \( \mathcal{N}\left( {{\mu }_{n},{\sigma }_{n}^{2}}\right) \).\n\n(i) If \( {\xi }_{n} \rightarrow \xi \) in distribution, then \( \xi \)... | Proof. Is an exercise. | No |
Proposition 3.4. A process \( X = \left( {{X}_{t}, t \geq 0}\right) \) is Brownian motion with \( {X}_{0} = 0 \) a.s. if and only if it is centered Gaussian with covariance\n\n\[ \mathbb{E}\left( {{X}_{s}{X}_{t}}\right) = \min \{ s, t\} = : s \land t,\;s \geq 0,\;t \geq 0. \] | Proof. \ | No |
Theorem 4.3. (Kolmogorov’s criterion). Let \( X = \left( {{X}_{t}, t \in I}\right) \) be a process indexed by an interval \( I \subset \mathbb{R} \), taking values in a complete metric space \( \left( {E, d}\right) \). Suppose there exist \( p > 0 \), \( \varepsilon > 0 \) and \( C > 0 \) such that \[ \mathbb{E}\left\l... | Proof. The uniqueness is clear from the discussions in the previous paragraph. We need to prove the existence. For notational simplification, we assume \( I = \left\lbrack {0,1}\right\rbrack \). By assumption, for \( a > 0 \) and \( s, t \in \left\lbrack {0,1}\right\rbrack \), \[ \mathbb{P}\left\{ {d\left( {{X}_{s},{X}... | Yes |
Corollary 4.4. Let \( B = \left( {{B}_{t}, t \geq 0}\right) \) be Brownian motion. The process \( B \) admits a modification whose trajectories are locally Hölder continuous for exponent \( \frac{1}{2} - \varepsilon \), for all \( \varepsilon \in \left( {0,\frac{1}{2}}\right) \) . | Proof. Fix \( \varepsilon \in \left( {0,\frac{1}{2}}\right) \) . Let \( t, s \geq 0 \) . Since \( {B}_{t} - {B}_{s} \) is Gaussian \( \mathcal{N}\left( {0,\left| {t - s}\right| }\right) \), we have, for all \( p > 0,\mathbb{E}\left\lbrack {\left| {B}_{t} - {B}_{s}\right| }^{p}\right\rbrack = {C}_{p}{\left( t - s\right)... | Yes |
Corollary 4.7. (Paley, Wiener and Zygmund 1933). Almost surely, \( t \mapsto {B}_{t} \) is nowhere differentiable. | Since a function of finite variation is almost everywhere differentiable, this yields: | No |
Lemma 5.1. We have \( \sigma \left( {{X}_{t}, t \geq 0}\right) = \mathcal{C}\left( {{\mathbb{R}}_{ + },\mathbb{R}}\right) \) . | Proof. Is an exercise. | No |
Proposition 1.1. The following processes are Brownian motions:\n\n(i) \( {X}_{t} = - {B}_{t} \) . (symmetry)\n\n(ii) \( {X}_{t} = t{B}_{1/t},{X}_{0} = 0 \) . (time inversion)\n\n(iii) \( a > 0 \) fixed, \( {X}_{t} = \frac{1}{{a}^{1/2}}{B}_{at} \) . (scaling)\n\n(iv) \( T > 0 \) fixed, \( {X}_{t} = {B}_{T} - {B}_{T - t}... | Proof. Is trivial. It suffices to check, for each of the processes, that \( X \) is a centered Gaussian process with covariance \( s \land t \) . Only Part (ii) needs some special care because the trajectories are not necessarily continuous at 0 : this however, does not cause any trouble because \( X \) is, according t... | No |
Example 1.3. By continuity, \( \mathop{\lim }\limits_{{t \rightarrow 0 + }}{B}_{t} = 0 \), a.s., which, by time inversion, leads to:\n\n\[ \mathop{\lim }\limits_{{t \rightarrow \infty }}\frac{{B}_{t}}{t} = 0,\;\text{ a.s. } \] | It is possible to prove this directly, first using law of the large numbers (which says that \( \left. {\frac{{B}_{n}}{n} \rightarrow 0\text{, a.s.}}\right) \), and then proves that \( B \) \ | No |
Theorem 2.1. (Simple Markov property). Let \( s \geq 0 \) . The process \( \left( {{\widetilde{B}}_{t} \mathrel{\text{:=}} {B}_{t + s} - {B}_{s}, t \geq }\right. \) \( 0) \) is Brownian motion, independent of \( {\mathcal{F}}_{s} \) . | Proof. It is immediately checked that \( \widetilde{B} \) is a centered Gaussian process with a.s. continuous trajectories and with covariance \( \mathbb{E}\left( {{\widetilde{B}}_{t}{\widetilde{B}}_{{t}^{\prime }}}\right) = t \land {t}^{\prime } \) : it is Brownian motion.\n\nTo prove independence, it suffices to show... | Yes |
Proposition 2.2. Let \( s \geq 0 \), and define\n\n\[ \n{\mathcal{F}}_{s + } \mathrel{\text{:=}} \mathop{\bigcap }\limits_{{u > s}}{\mathcal{F}}_{u}\n\]\n\nThe process \( \left( {{\widetilde{B}}_{t} \mathrel{\text{:=}} {B}_{t + s} - {B}_{s}, t \geq 0}\right) \) is independent of \( {\mathcal{F}}_{s + } \) . | Proof. It suffices to check that for \( A \in {\mathcal{F}}_{s + },0 \leq {t}_{1} < {t}_{2} < \cdots < {t}_{n} \) and continuous and bounded \( F : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) ,\n\n(2.1)\n\n\[ \n\mathbb{E}\left\lbrack {{\mathbf{1}}_{A}F\left( {{\widetilde{B}}_{{t}_{1}},\cdots ,{\widetilde{B}}_{{t}_{n}}}\... | Yes |
Theorem 2.3. (Blumenthal 0-1 law). The \( \sigma \) -field \( {\mathcal{F}}_{0 + } \) is trivial, in the sense that \( \forall A \in {\mathcal{F}}_{0 + },\mathbb{P}\left( A\right) = 0 \) or 1 . | Proof. By Proposition 2.2, \( {\mathcal{F}}_{0 + } \) is independent of \( \sigma \left( {{B}_{t}, t \geq 0}\right) \), and thus of the completion of \( \sigma \left( {{B}_{t}, t \geq 0}\right) \) . Let \( A \in {\mathcal{F}}_{0 + } \) . Since \( A \in {\mathcal{F}}_{0 + } = \mathop{\bigcap }\limits_{{u > 0}}{\mathcal{... | Yes |
Let \( \tau \mathrel{\text{:=}} \inf \left\{ {t > 0 : {B}_{t} > 0}\right\} \). Then \( \tau = 0 \), a.s. | To prove this, we note that\n\n\[ \n\{ \tau = 0\} = \mathop{\bigcap }\limits_{{s > 0, s \in \mathbb{Q}}}\left\{ {\mathop{\sup }\limits_{{0 \leq u \leq s}}{B}_{u} > 0}\right\} \in {\mathcal{F}}_{0 + }.\n\]\n\nFor \( t > 0,\mathbb{P}\left( {\tau \leq t}\right) \geq \mathbb{P}\left( {{B}_{t} > 0}\right) = \frac{1}{2} \). ... | Yes |
Example 2.5. Let \( {\left( {t}_{n}\right) }_{n \geq 1} \) be a sequence decreasing to 0 . Then a.s. \( {B}_{{t}_{n}} > 0 \) for infinitely many \( n \), and \( {B}_{{t}_{n}} < 0 \) for infinitely many \( n \) . | Indeed, let \( {A}_{n} \mathrel{\text{:=}} \left\{ {{B}_{{t}_{n}} > 0}\right\} \), then \( \mathbb{P}\left( {\limsup {A}_{n}}\right) = \mathop{\lim }\limits_{{N \rightarrow \infty }}\mathbb{P}\left( {{ \cup }_{n = N}^{\infty }{A}_{n}}\right) \), which is \( \geq \) \( \mathop{\limsup }\limits_{{N \rightarrow \infty }}\... | Yes |
Example 2.6. We have\n\n\[ \mathop{\limsup }\limits_{{t \rightarrow \infty }}\frac{{B}_{t}}{{t}^{1/2}} = \infty ,\;\mathop{\liminf }\limits_{{t \rightarrow \infty }}\frac{{B}_{t}}{{t}^{1/2}} = - \infty ,\;\text{ a.s. } \] | In fact, fix a constant \( K > 0 \), and let \( {A}_{n} \mathrel{\text{:=}} \left\{ {\sqrt{n}{B}_{1/n} > K}\right\} \) . We have \( \mathbb{P}\left( {\limsup {A}_{n}}\right) \geq \) \( \lim \mathop{\sup }\limits_{{N \rightarrow \infty }}\mathbb{P}\left( {A}_{N}\right) = \mathbb{P}\left( {{B}_{1} > K}\right) > 0 \), so ... | Yes |
The constant time \( T = t \) is a stopping time. Another example is \( T \mathrel{\text{:=}} {T}_{a} \) , where \( {T}_{a} \mathrel{\text{:=}} \inf \left\{ {t > 0 : {B}_{t} = a}\right\} \) | indeed, for \( a \geq 0,\left\{ {{T}_{a} \leq t}\right\} = \left\{ {\mathop{\sup }\limits_{{s \in \left\lbrack {0, t}\right\rbrack }}{B}_{s} \geq a}\right\} \in {\mathcal{F}}_{t} \) | Yes |
We claim that \( T \) and \( {B}_{T}{\mathbf{1}}_{\{ T < \infty \} } \) are \( {\mathcal{F}}_{T} \) -measurable. | For \( {B}_{T}{\mathbf{1}}_{\{ T < \infty \} } \), it suffices to see that a.s., \[ {B}_{T}{\mathbf{1}}_{\{ T < \infty \} } = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sum }\limits_{{i = 0}}^{\infty }{\mathbf{1}}_{\left\{ \frac{i}{{2}^{n}} < T \leq \frac{i + 1}{{2}^{n}}\right\} }{B}_{\frac{i}{{2}^{n}}}, \... | Yes |
Theorem 4.7. (Reflection principle). Let \( {S}_{t} = \mathop{\sup }\limits_{{s \in \left\lbrack {0, t}\right\rbrack }}{B}_{s}, t > 0 \) . Then\n\n\[ \mathbb{P}\left( {{S}_{t} \geq a,{B}_{t} \leq b}\right) = \mathbb{P}\left( {{B}_{t} \geq {2a} - b}\right) ,\;a \geq 0, b \leq a. \] | Proof of Theorem 4.7. Recall that \( {T}_{a} < \infty \), a.s. We have\n\n\[ \mathbb{P}\left( {{S}_{t} \geq a,{B}_{t} \leq b}\right) = \mathbb{P}\left( {{T}_{a} \leq t,{B}_{t} \leq b}\right) = \mathbb{P}\left( {{T}_{a} \leq t,{\widetilde{B}}_{t - {T}_{a}} \leq b - a}\right) ,\]\n\nwhere \( {\widetilde{B}}_{s} \mathrel{... | Yes |
Example 4.10. By Theorem 4.7, for any \( t > 0 \) ,\n\n\[ \n\mathbb{P}\left( {{T}_{a} \leq t}\right) = \mathbb{P}\left( {{S}_{t} \geq a}\right) = \mathbb{P}\left( {\left| {B}_{t}\right| \geq a}\right) = \mathbb{P}\left( {{t}^{1/2}\left| {B}_{1}\right| \geq a}\right) = \mathbb{P}\left( {\frac{{a}^{2}}{{B}_{1}^{2}} \leq ... | As a consequence, for \( a \neq 0 \) ,\n\n\[ \n{f}_{{T}_{a}}\left( t\right) = \frac{\left| a\right| }{{\left( 2\pi {t}^{3}\right) }^{1/2}}\exp \left( {-\frac{{a}^{2}}{2t}}\right) {\mathbf{1}}_{\{ t > 0\} }.\n\]\n\nIn particular, \( \mathbb{E}\left( {T}_{a}\right) = \infty \) if \( a \neq 0 \) . | Yes |
Theorem 4.11. (Strong Markov property) Let \( T \) be a stopping time. Let \( F : \Omega \rightarrow {\mathbb{R}}_{ + } \) be measurable. For \( x \in {\mathbb{R}}^{d} \) , \[ {\mathbb{E}}_{x}\left\lbrack {{\mathbf{1}}_{\{ T < \infty \} }\left( {F \circ {\theta }_{T}}\right) \mid {\mathcal{F}}_{T}}\right\rbrack = {\mat... | Proof. On the set \( \{ T\left( \mathrm{w}\right) < \infty \} \) , \[ \left( {{\theta }_{T}\mathrm{w}}\right) \left( t\right) = \mathrm{w}\left( {T + t}\right) = \mathrm{w}\left( T\right) + \left( {\mathrm{w}\left( {T + t}\right) - \mathrm{w}\left( T\right) }\right) = {B}_{T}\left( \mathrm{w}\right) + {\widetilde{B}}_{... | Yes |
Example 1.5. If \( \left( {X}_{t}\right) \) is continuous and adapted, then for any closed set \( F \) , \[ {T}_{F} \mathrel{\text{:=}} \inf \left\{ {t \geq 0 : {X}_{t} \in F}\right\} \] is a stopping time. | In fact, \( \left\{ {{T}_{F} \leq t}\right\} = \left\{ {\mathop{\inf }\limits_{{s \in \left\lbrack {0, t}\right\rbrack }}d\left( {{X}_{s}, F}\right) = 0}\right\} \) (by continuity of the trajectories), which is \( \left\{ {\mathop{\inf }\limits_{{s \in \left\lbrack {0, t}\right\rbrack \cap \mathbb{Q}}}d\left( {{X}_{s},... | Yes |
Theorem 1.9. (Doob’s \( {L}^{p} \) inequality). Let \( p > 1 \) . Let \( \left( {X}_{s}\right) \) be a right-continuous martingale. Then for any \( t \geq 0 \) , \[ {\begin{Vmatrix}\mathop{\sup }\limits_{{s \in \left\lbrack {0, t}\right\rbrack }}\left| {X}_{s}\right| \end{Vmatrix}}_{p} \leq q{\begin{Vmatrix}{X}_{t}\end... | Proof. Let \( 0 \leq {t}_{1} < {t}_{2} < \cdots < {t}_{k} = t \) . Then \( {Y}_{n} \mathrel{\text{:=}} {X}_{{t}_{n \land k}} \) is a discrete-time martingale. By Doob’s \( {L}^{p} \) inequality for discrete-time martingales, \[ {\begin{Vmatrix}\mathop{\max }\limits_{{0 \leq i \leq k}}\left| {X}_{{t}_{i}}\right| \end{Vm... | Yes |
Theorem 1.11. Let \( \left( {{X}_{t}, t \geq 0}\right) \) be a right-continuous submartingale satisfying \( {}^{2} \)\n\n\[ \mathop{\sup }\limits_{{t \geq 0}}\mathbb{E}\left( {X}_{t}^{ + }\right) < \infty \]\n\nThen \( {X}_{\infty } \mathrel{\text{:=}} \mathop{\lim }\limits_{{t \rightarrow \infty }}{X}_{t} \) exists a.... | Proof. (i) Let \( D \subset {\mathbb{R}}_{ + } \) be countable and dense. Let \( a < b \) and let \( X\left( {\left\lbrack {0, t}\right\rbrack \cap D}\right) \) be the number of crossings of \( \left( {{X}_{s}, s \in \left\lbrack {0, t}\right\rbrack \cap D}\right) \) over \( \left\lbrack {a, b}\right\rbrack \) . Then\n... | Yes |
Theorem 1.13. Let \( p > 1 \) . If \( \left( {{X}_{t}, t \geq 0}\right) \) is a right-continuous martingale satisfying\n\n\[ \mathop{\sup }\limits_{{t \geq 0}}\mathbb{E}\left( {\left| {X}_{t}\right| }^{p}\right) < \infty \]\n\nthen \( {X}_{t} \rightarrow {X}_{\infty } \) a.s. and in \( {L}^{p} \) . | Proof. (i) By Jensen’s (or more generally, Hölder’s) inequality, \( \mathop{\sup }\limits_{{t \geq 0}}\mathbb{E}\left( \left| {X}_{t}\right| \right) < \infty \) . Theorem 1.11 tells us that \( {X}_{t} \rightarrow {X}_{\infty } \) a.s.\n\n(ii) By Doob’s inequality, \( \mathbb{E}\left( {\mathop{\sup }\limits_{{t \geq 0}}... | Yes |
Theorem 1.14. Let \( \left( {{X}_{t}, t \geq 0}\right) \) be a uniformly integrable and right-continuous martingale. Then\n\n(i) \( {X}_{t} \rightarrow {X}_{\infty } \) in \( {L}^{1} \) ;\n\n(ii) \( {X}_{t} \rightarrow {X}_{\infty } \) a.s.;\n\n(iii) \( {X}_{t} = \mathbb{E}\left( {{X}_{\infty } \mid {\mathcal{F}}_{t}}\... | Proof. (ii) The uniform integrability implies \( \mathop{\sup }\limits_{{t \geq 0}}\mathbb{E}\left( \left| {X}_{t}\right| \right) < \infty \) . So by Theorem 1.11, \( {X}_{t} \rightarrow {X}_{\infty } \) a.s., and \( \mathbb{E}\left( {X}_{\infty }\right) < \infty \) .\n\n(i) The \( {L}^{1} \) convergence, being a conse... | Yes |
Theorem 1.15. (Doob’s optional sampling theorem). Let \( \left( {{X}_{t}, t \geq 0}\right) \) be a right-continuous martingale and let \( S \leq T \) be stopping times. If \( \left( {{X}_{t}, t \geq 0}\right) \) is uniformly integrable, then\n\n\[ \mathbb{E}\left( {{X}_{T} \mid {\mathcal{F}}_{S}}\right) = {X}_{S}\;\tex... | Proof. (i) Let us prove first \( \mathbb{E}\left( \left| {X}_{T}\right| \right) < \infty \) .\n\nDefine\n\n\[ {T}_{n} \mathrel{\text{:=}} \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{k}{{2}^{n}}{\mathbf{1}}_{\left\{ \frac{k - 1}{{2}^{n}} < T \leq \frac{k}{{2}^{n}}\right\} } + \left( {+\infty }\right) {\mathbf{1}}_{\{... | Yes |
Theorem 1.16. (Doob’s optional sampling theorem). Let \( \\left( {{X}_{t}, t \\geq 0}\\right) \) be a right-continuous martingale and let \( S \\leq T \) be stopping times. If \( T \) is bounded, then\n\n\[ \n\\mathbb{E}\\left( {{X}_{T} \\mid {\\mathcal{F}}_{S}}\\right) = {X}_{S}\\;\\text{ a.s. } \n\]\n\nIn particular,... | Proof. Let \( a > 0 \) be such that \( S \\leq T \\leq a \) . The proof of Theorem 1.15 remains valid. The only problem being the uniform integrability of the discrete-time \( \\left( {\\mathcal{F}}_{\\frac{i}{{2}^{n + 1}}}\\right) \) -martingale \( \\left( {{X}_{\\frac{i}{{2}^{n + 1}}}, i \\geq 0}\\right) \), it suffi... | Yes |
Let \( {T}_{a} \mathrel{\text{:=}} \inf \left\{ {t \geq 0 : {B}_{t} = a}\right\} \) . Let \( \theta > 0 \) . We know that \( \left( {{\mathrm{e}}^{\theta {B}_{t} - \frac{{\theta }^{2}}{2}t}, t \geq 0}\right) \) is a martingale. If \( a > 0 \), then \( \theta {B}_{t \land {T}_{a}} - \frac{{\theta }^{2}}{2}\left( {t \lan... | We have, by the optional sampling theorem,\n\n\[1 = \mathbb{E}\left\lbrack {\mathrm{e}}^{{\theta a} - \frac{{\theta }^{2}}{2}{T}_{a}}\right\rbrack\]\n\nIn other words, \( \mathbb{E}\left\lbrack {\mathrm{e}}^{-\frac{{\theta }^{2}}{2}{T}_{a}}\right\rbrack = {\mathrm{e}}^{-{\theta a}} \) . (In particular, \( \mathbb{P}\le... | Yes |
Example 2.3. Let \( \\left( {\\left( {{X}_{t},{Y}_{t}}\\right), t \\geq 0}\\right) \) be \( {\\mathbb{R}}^{2} \) -valued Brownian motion with \( \\left( {{X}_{0},{Y}_{0}}\\right) = \\left( {0,1}\\right) \). Let \( T \\mathrel{\\text{:=}} \\inf \\left\\{ {t \\geq 0 : {Y}_{t} = 0}\\right\\} \). What is the law of \( {X}_... | By the previous example, we have, for all \( \\theta \\geq 0,\\mathbb{E}\\left\\lbrack {\\mathrm{e}}^{-\\frac{{\\theta }^{2}}{2}T}\\right\\rbrack = {\\mathrm{e}}^{-\\theta } \). Since \( T \) is independent of \( \\sigma \\left( {{X}_{t}, t \\geq 0}\\right) \), we obtain: for all \( a \\in \\mathbb{R} \), \[ \\mathbb{E... | Yes |
Example 2.4. Let \( a > 0 \) and \( b > 0 \), and let \( {T}_{a, b} \mathrel{\text{:=}} \inf \left\{ {t \geq 0 : {B}_{t} = - a}\right. \) ou \( \left. {{B}_{t} = b}\right\} = \) \( {T}_{-a} \land {T}_{b} \), which is a stopping time. We are interested in the law of \( {T}_{a, b} \) . | Let \( \theta \in \mathbb{R} \), and consider the following continuous martingale:\n\n\[ \n{M}_{t} \mathrel{\text{:=}} \sinh \left( {\theta \left( {{B}_{t} + a}\right) }\right) {\mathrm{e}}^{-\frac{{\theta }^{2}}{2}t}.\n\]\n\nSince \( \left( {{M}_{t \land {T}_{a, b}}, t \geq 0}\right) \) is a continuous and bounded mar... | Yes |
Theorem 4.1. (Lévy). Fix \( t > 0 \) . Let \( \Pi \mathrel{\text{:=}} \left\{ {0 = {t}_{0} < {t}_{1} < \cdots < {t}_{p} = t}\right\} \) be a sequence of subdivisions of \( \left\lbrack {0, t}\right\rbrack \) . Write \( \parallel \Pi \parallel \mathrel{\text{:=}} \mathop{\max }\limits_{{1 \leq i \leq p}}\left( {{t}_{i} ... | Proof. Let us first prove \( {L}^{2} \) convergence. Define\n\n\[ {Y}_{i} \mathrel{\text{:=}} {\left( {B}_{{t}_{i}} - {B}_{{t}_{i - 1}}\right) }^{2} - \left( {{t}_{i} - {t}_{i - 1}}\right) ,\;1 \leq i \leq p. \]\n\nThen \( \left( {{Y}_{i},1 \leq i \leq p}\right) \) are i.i.d. centered, with (writing \( a \mathrel{\text... | Yes |
The map \( \alpha : R \rightarrow {R}^{2} \) given by \( \alpha \left( t\right) = \left( {t,\left| t\right| }\right), t \in R \), is not a parametrized differentiable curve. | since \( \left| t\right| \) is not differentiable at \( t = 0 \) (Fig. 1-4). | Yes |
Let us show that the unit sphere\n\n\[ \n{S}^{2} = \\left\\{ {\\left( {x, y, z}\\right) \\in {R}^{3};{x}^{2} + {y}^{2} + {z}^{2} = 1}\\right\\} \n\]\n\nis a regular surface. | We first verify that the map \( {\\mathbf{x}}_{1} : U \\subset {R}^{2} \\rightarrow {R}^{3} \) given by\n\n\[ \n{\\mathbf{x}}_{1}\\left( {x, y}\\right) = \\left( {x, y, + \\sqrt{1 - \\left( {{x}^{2} + {y}^{2}}\\right) }}\\right) ,\\;\\left( {x, y}\\right) \\in U, \n\]\n\nwhere \( {R}^{2} = \\left\\{ {\\left( {x, y, z}\... | Yes |
The ellipsoid\n\n\[ \n\frac{{x}^{2}}{{a}^{2}} + \frac{{y}^{2}}{{b}^{2}} + \frac{{z}^{2}}{{c}^{2}} = 1 \n\]\n\nis a regular surface. | In fact, it is the set \( {f}^{-1}\left( 0\right) \) where\n\n\[ \nf\left( {x, y, z}\right) = \frac{{x}^{2}}{{a}^{2}} + \frac{{y}^{2}}{{b}^{2}} + \frac{{z}^{2}}{{c}^{2}} - 1 \n\]\n\nis a differentiable function and 0 is a regular value of \( f \) . This follows from the fact that the partial derivatives \( {f}_{x} = {2... | Yes |
The one-sheeted cone \( C \), given by\n\n\[ z = + \sqrt{{x}^{2} + {y}^{2}},\;\left( {x, y}\right) \in {R}^{2}, \]\n\nis not a regular surface. | To show that this is not the case, we use Prop. 3. If \( C \) were a regular surface, it would be, in a neighborhood of \( \left( {0,0,0}\right) \in C \), the graph of a differentiable function having one of three forms: \( y = h\left( {x, z}\right), x = g\left( {y, z}\right), z = f\left( {x, y}\right) \) . The two fir... | Yes |
To prove that the parametrization \(\mathbf{x}(u, v) = \left( \left( r \cos u + a \right) \cos v, \left( r \cos u + a \right) \sin v, r \sin u \right)\) for the torus \( T \) is one-to-one. | To prove that \(\mathbf{x}\) is one-to-one, we first observe that \(\sin u = z/r\); also, if \(\sqrt{x^2 + y^2} \leq a\), then \(\pi /2 \leq u \leq 3\pi /2\), and if \(\sqrt{x^2 + y^2} \geq a\), then either \(0 < u \leq \pi /2\) or \(3\pi /2 \leq u < 2\pi\). Thus, given \((x, y, z)\), this determines \(u\), \(0 < u < 2... | Yes |
Let \( S \) be a regular surface and \( V \subset {R}^{3} \) be an open set such that \( S \subset V \). Let \( f : V \subset {R}^{3} \rightarrow R \) be a differentiable function. Then the restriction of \( f \) to \( S \) is a differentiable function on \( S \). | In fact, for any \( p \in S \) and any parametrization \( \mathbf{x} : U \subset {R}^{2} \rightarrow S \) in \( p \), the function \( f \circ \mathbf{x} : U \rightarrow R \) is differentiable. | Yes |
If \( \mathbf{x} : U \subset {R}^{2} \rightarrow S \) is a parametrization, \( {\mathbf{x}}^{-1} : \mathbf{x}\left( U\right) \rightarrow {R}^{2} \) is differentiable. | In fact, for any \( p \in \mathbf{x}\left( U\right) \) and any parametrization \( \mathbf{y} : V \subset \) \( {R}^{2} \rightarrow S \) in \( p \), we have that \( {\mathbf{x}}^{-1} \circ \mathbf{y} : {\mathbf{y}}^{-1}\left( W\right) \rightarrow {\mathbf{x}}^{-1}\left( W\right) \), where\n\n\[ W = \mathbf{x}\left( U\ri... | Yes |
Let \( {S}_{1} \) and \( {S}_{2} \) be regular surfaces. Assume that \( {S}_{1} \subset V \subset {R}^{3} \) , where \( V \) is an open set of \( {R}^{3} \), and that \( \varphi : V \rightarrow {R}^{3} \) is a differentiable map such that \( \varphi \left( {S}_{1}\right) \subset {S}_{2} \) . Then the restriction \( \va... | In fact, given \( p \in {S}_{1} \) and parametrizations \( {\mathbf{x}}_{1} : {U}_{1} \rightarrow {S}_{1},{\mathbf{x}}_{2} : {U}_{2} \rightarrow {S}_{2} \) , with \( p \in {\mathbf{x}}_{1}\left( {U}_{1}\right) \) and \( \varphi \left( {{\mathbf{x}}_{1}\left( {U}_{1}\right) }\right) \subset {\mathbf{x}}_{2}\left( {U}_{2... | Yes |
Let \( S \subset {R}^{3} \) be the set obtained by rotating a regular connected plane curve \( C \) about an axis in the plane which does not meet the curve; we shall take the \( {xz} \) plane as the plane of the curve and the \( z \) axis as the rotation axis. Let\n\n\[ x = f\left( v\right) ,\;z = g\left( v\right) ,\;... | To show that \( \mathbf{x} \) is a parametrization of \( S \) we must check conditions 1,2, and 3 of Def. 1, Sec. 2-2. Conditions 1 and 3 are straightforward, and we leave them to the reader. To show that \( \mathbf{x} \) is a homeomorphism, we first show that \( \mathbf{x} \) is one-to-one. In fact, since \( \left( {f... | No |
Let \( v \in {R}^{3} \) be a unit vector and let \( h : S \rightarrow R, h\left( p\right) = v \cdot p \) , \( p \in S \), be the height function defined in Example 1 of Sec. 2-3. To compute \( d{h}_{p}\left( w\right), w \in {T}_{p}\left( S\right) \) | choose a differentiable curve \( \alpha : \left( {-\epsilon ,\epsilon }\right) \rightarrow S \) with \( \alpha \left( 0\right) = p,{\alpha }^{\prime }\left( 0\right) = w \) . Since \( h\left( {\alpha \left( t\right) }\right) = \alpha \left( t\right) \cdot v \), we obtain\n\n\[ d{h}_{p}\left( w\right) = {\left. \frac{d}... | Yes |
Let \( {S}^{2} \subset {R}^{3} \) be the unit sphere and let \( {R}_{z,\theta } : {R}^{3} \rightarrow {R}^{3} \) be the rotation of angle \( \theta \) about the \( z \) axis. Then \( {R}_{z,\theta } \) restricted to \( {S}^{2} \) is a differentiable map of \( {S}^{2} \). We shall compute \( {\left( d{R}_{z,\theta }\rig... | Let \( \alpha : \left( {-\epsilon ,\epsilon }\right) \rightarrow {S}^{2} \) be a differentiable curve with \( \alpha \left( 0\right) = p,{\alpha }^{\prime }\left( 0\right) = w \) . Then, since \( {R}_{z,\theta } \) is linear, \[ {\left( d{R}_{z,\theta }\right) }_{p}\left( w\right) = \frac{d}{dt}{\left( {R}_{z,\theta } ... | Yes |
A coordinate system for a plane \( P \subset {R}^{3} \) passing through \( {p}_{0} = \left( {{x}_{0},{y}_{0},{z}_{0}}\right) \) and containing the orthonormal vectors \( {w}_{1} = \left( {{a}_{1},{a}_{2},{a}_{3}}\right) \) , \( {w}_{2} = \left( {{b}_{1},{b}_{2},{b}_{3}}\right) \) is given as follows:\n\n\[ \mathbf{x}\l... | To compute the first fundamental form for an arbitrary point of \( P \) we observe that \( {\mathbf{x}}_{u} = {w}_{1},{\mathbf{x}}_{v} = {w}_{2} \) ; since \( {w}_{1} \) and \( {w}_{2} \) are unit orthogonal vectors, the functions \( E, F, G \) are constant and given by\n\n\[ E = 1,\;F = 0,\;G = 1. \]\n\nIn this trivia... | Yes |
The right cylinder over the circle \( {x}^{2} + {y}^{2} = 1 \) admits the parametrization \( \mathbf{x} : U \rightarrow {R}^{3} \), where (Fig. 2-26)\n\n\[ \mathbf{x}\left( {u, v}\right) = \left( {\cos u,\sin u, v}\right) ,\]\n\n\[ U = \left\{ {\left( {u, v}\right) \in {R}^{2};\;0 < u < {2\pi },\; - \infty < v < \infty... | To compute the first fundamental form, we notice that\n\n\[ {\mathbf{x}}_{u} = \left( {-\sin u,\cos u,0}\right) ,\;{\mathbf{x}}_{v} = \left( {0,0,1}\right) ,\]\n\nand therefore\n\n\[ E = {\sin }^{2}u + {\cos }^{2}u = 1,\;F = 0,\;G = 1. \] | Yes |
Consider a helix that is given by (see Example 1, Sec. 1-2) \( \left( {\cos u,\sin u,{au}}\right) \) . Through each point of the helix, draw a line parallel to the \( {xy} \) plane and intersecting the \( z \) axis. The surface generated by these lines is called a helicoid and admits the following parametrization:\n\n\... | The verification that the helicoid is a regular surface is straightforward and left to the reader. | No |
We shall compute the first fundamental form of a sphere at a point of the coordinate neighborhood given by the parametrization (cf. Example 1, Sec. 2-2)\n\n\[ \mathbf{x}\left( {\theta ,\varphi }\right) = \left( {\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta }\right) . \] | First, observe that\n\n\[ {\mathbf{x}}_{\theta }\left( {\theta ,\varphi }\right) = \left( {\cos \theta \cos \varphi ,\cos \theta \sin \varphi , - \sin \theta }\right) ,\]\n\n\[ {\mathbf{x}}_{\varphi }\left( {\theta ,\varphi }\right) = \left( {-\sin \theta \sin \varphi ,\sin \theta \cos \varphi ,0}\right) . \]\n\nHence,... | Yes |
Let us compute the area of the torus of Example 6, Sec. 2-2. For that, we consider the coordinate neighborhood corresponding to the parametrization\n\n\[ \n\\mathbf{x}\\left( {u, v}\\right) = \\left( {\\left( {a + r\\cos u}\\right) \\cos v,\\left( {a + r\\cos u}\\right) \\sin v, r\\sin u}\\right) ,\n\]\n\n\[ \n0 < u < ... | Now, consider the region \( {R}_{\\epsilon } \) obtained as the image by \( \\mathbf{x} \) of the region \( {Q}_{\\epsilon } \) (Fig. 2-29) given by \( \\left( {\\epsilon > 0\\text{and small}}\\right) \) ,\n\n\[ \n{Q}_{\\epsilon } = \\left\\{ {\\left( {u, v}\\right) \\in {R}^{2};0 + \\epsilon \\leq u \\leq {2\\pi } - \... | Yes |
A surface which is the graph of a differentiable function (cf. Sec. 2-2, Prop. 1) is an orientable surface. | In fact, all surfaces which can be covered by one coordinate neighborhood are trivially orientable. | No |
Example 2. The sphere is an orientable surface. | Instead of proceeding to a direct calculation, let us resort to a general argument. The sphere can be covered by two coordinate neighborhoods (using stereographic projection; see Exercise 16 of Sec. 2-2), with parameters \( \left( {u, v}\right) \) and \( \left( {\bar{u},\bar{v}}\right) \), in such a way that the inters... | No |
Let \( F : {R}^{3} \rightarrow {R}^{3} \) be the map which assigns to each \( p \in {R}^{3} \) the point which is symmetric to \( p \) with respect to the origin \( O \in {R}^{3} \) . Then \( F\left( p\right) = - p \) | \[ F\left( {x, y, z}\right) = \left( {-x, - y, - z}\right) ,\] and the component functions of \( F \) are \[ {f}_{1}\left( {x, y, z}\right) = - x,\;{f}_{2}\left( {x, y, z}\right) = - y,\;{f}_{3}\left( {x, y, z}\right) = - z. \] | Yes |
Let \( F : {R}^{2} - \{ \left( {0,0}\right) \} \rightarrow {R}^{2} \) be defined as follows. Denote by \( \left| p\right| \) the distance to the origin \( \left( {0,0}\right) = O \) of a point \( p \in {R}^{2} \) . By definition, \( F\left( p\right), p \neq 0 \), belongs to the half-line \( {Op} \) and is such that \( ... | \[ F\left( {x, y}\right) = \left( {\frac{x}{{x}^{2} + {y}^{2}},\frac{y}{{x}^{2} + {y}^{2}}}\right) ,\;\left( {x, y}\right) \neq \left( {0,0}\right) ,\] and the component functions of \( F \) are \[ {f}_{1}\left( {x, y}\right) = \frac{x}{{x}^{2} + {y}^{2}},\;{f}_{2}\left( {x, y}\right) = \frac{y}{{x}^{2} + {y}^{2}}. \] | Yes |
PROPOSITION 1. F: \( \mathrm{U} \subset {\mathrm{R}}^{\mathrm{n}} \rightarrow {\mathrm{R}}^{\mathrm{m}} \) is continuous if and only if each component function \( {\mathrm{f}}_{\mathrm{i}} : \mathrm{U} \subset {\mathrm{R}}^{\mathrm{n}} \rightarrow \mathrm{R},\mathrm{i} = 1,\ldots ,\mathrm{m} \), is continuous. | Proof. Assume that \( F \) is continuous at \( p \in U \) . Then given \( \epsilon > 0 \), there exists \( \delta > 0 \) such that \( F\left( {{B}_{\delta }\left( p\right) }\right) \subset {B}_{\epsilon }\left( {F\left( p\right) }\right) \) . Thus, if \( q \in {B}_{\delta }\left( p\right) \), then\n\n\[ F\left( q\right... | Yes |
PROPOSITION 2. A map \( \mathrm{F} : \mathrm{U} \subset {\mathrm{R}}^{\mathrm{n}} \rightarrow {\mathrm{R}}^{\mathrm{m}} \) is continuous at \( \mathrm{p} \in \mathrm{U} \) if and only if, given a neighborhood \( \mathrm{V} \) of \( \mathrm{F}\left( \mathrm{p}\right) \) in \( {\mathrm{R}}^{\mathrm{m}} \) there exists a ... | Proof. Assume that \( F \) is continuous at \( p \) . Since \( V \) is an open set containing \( F\left( p\right) \), it contains a ball \( {B}_{\epsilon }\left( {F\left( p\right) }\right) \) for some \( \epsilon > 0 \) . By continuity, there exists a ball \( {B}_{\delta }\left( p\right) = W \) such that\n\n\[ F\left( ... | Yes |
Example 7. Let \( F : U \subset {R}^{2} \rightarrow {R}^{3} \) be given by \[ F\left( {u, v}\right) = \left( {\cos u\cos v,\cos u\sin v,{\cos }^{2}v}\right) ,\;\left( {u, v}\right) \in U. \] The component functions of \( F \), namely, \[ {f}_{1}\left( {u, v}\right) = \cos u\cos v,\;{f}_{2}\left( {u, v}\right) = \cos u\... | have continuous partial derivatives of all orders in \( U \) . Thus, \( F \) is differentiable in \( U \) . | Yes |
Given a vector \( w \in {R}^{m} \) and a point \( {p}_{0} \in U \subset {R}^{m} \), we can always find a differentiable curve \( \alpha : \left( {-\epsilon ,\epsilon }\right) \rightarrow U \) with \( \alpha \left( 0\right) = {p}_{0} \) and \( {\alpha }^{\prime }\left( 0\right) = \) \( w \) . | Simply define \( \alpha \left( t\right) = {p}_{0} + {tw}, t \in \left( {-\epsilon ,\epsilon }\right) \) . By writing \( {p}_{0} = \left( {{x}_{1}^{0},\ldots ,{x}_{m}^{0}}\right) \) and \( w = \left( {{w}_{1},\ldots ,{w}_{m}}\right) \), the component functions of \( \alpha \) are \( {x}_{i}\left( t\right) = {x}_{i}^{0} ... | Yes |
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