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Proposition 21.1.2. Assume that\n\n(1) \( \operatorname{char}\left( F\right) \) does not divide \( n \), and\n\n(2) \( F \) contains all \( n \) th roots of unity.\n\nThen \( K = F\left( \sqrt[n]{a}\right) \) is a cyclic extension of degree dividing \( n \) .
Proof. We may factor\n\n\[ {x}^{n} - a = \left( {x - \sqrt[n]{a}}\right) \left( {x - {\zeta }_{n}\sqrt[n]{a}}\right) \cdots \left( {x - {\zeta }_{n}^{n - 1}\sqrt[n]{a}}\right) \]\n\nSo \( K \) is the splitting field of \( {x}^{n} - a \) over \( F \) . As \( \left( {{x}^{n} - a, D\left( {{x}^{n} - a}\right) }\right) = \...
Yes
Proposition 21.1.3 (Kummer). If \( F \) is a field such that \( \operatorname{char}\left( F\right) \nmid n \) and \( F \) contains all \( n \) th roots of unity. Then any cyclic field extension \( K \) of \( F \) is of the form \( K = F\left( \sqrt[n]{a}\right) \) for some \( a \in {F}^{ \times } \) .
Proof. Write \( \operatorname{Gal}\left( {K/F}\right) \cong {\mathbf{Z}}_{n} = \langle \sigma \rangle \) . For each \( \alpha \in K \), we define\n\n(21.1.3.1)\n\n\[ b \mathrel{\text{:=}} \alpha + {\zeta }_{n}\sigma \left( \alpha \right) + \cdots + {\zeta }_{n}^{n - 1}{\sigma }^{n - 1}\left( \alpha \right) . \]\n\nBy l...
Yes
Proposition 21.2.3. An element \( \alpha \) can be expressed by radicals over a field \( F \) if \( \alpha \) is contained in a Galois extension \( K \) of \( F \) which admits a tower of subfields of the form (21.2.1.1).
Proof. By definition of expressing elements by radicals, there exists a finite extension \( {K}^{\prime } \) of \( F \) which admits a tower of subfields of the form (21.2.1.1). Let \( K \) be the Galois closure of \( {K}^{\prime } \) over \( F \) . This implies that for each \( \sigma \in {\operatorname{Hom}}_{F}\left...
No
An (irreducible) polynomial \( f\left( x\right) \) can be solved by radicals if and only if its Galois group (meaning the Galois group of its splitting field) is a solvable group.
Proof. \
No
The Galois group for \( {x}^{7} - 5 \) over \( \mathbb{Q} \) (irreducible by Eisenstein criterion).
The splitting field is \( \mathbb{Q}\left( {\sqrt[7]{5},{\zeta }_{7}}\right) \). The associated Galois group is \( {\mathbf{Z}}_{7} \rtimes {\mathbf{Z}}_{7}^{ \times } \).
No
Lemma 21.3.4. The group \( G \) acts transitively on the set \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{n}}\right\} \) .
Proof. Suppose now and suppose that \( \left\{ {{\alpha }_{1},\ldots ,{\alpha }_{r}}\right\} \) (with \( r < n \) ) is an orbit under \( G \), then \( \left( {x - {\alpha }_{1}}\right) \cdots \left( {x - {\alpha }_{r}}\right) \in F\left\lbrack x\right\rbrack \) is a factor of \( f\left( x\right) \) . Yet \( f\left( x\r...
No
Proposition 21.3.6. The field \( M = F\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) is a Galois extension over \( L = F\left( {{s}_{1},\ldots ,{s}_{n}}\right) \) with Galois group \( {S}_{n} \) .
Proof. Consider the polynomial\n\n\[ f\left( x\right) = \left( {x - {x}_{1}}\right) \cdots \left( {x - {x}_{n}}\right) = {x}^{n} - {s}_{1}{x}^{n - 1} + \cdots + {\left( -1\right) }^{n}{s}_{n} \in L\left\lbrack x\right\rbrack .\n\]\n\nThis \( M \) is the splitting field of \( f\left( x\right) \) over \( L \) . In partic...
Yes
Lemma 21.3.9. We have \( G = \operatorname{Gal}\left( {K/F}\right) \subseteq {A}_{n} \) if and only if \( D \) is a square in \( F \) . More precisely, we have the following diagram of Galois extensions:
Proof. The second statement implies the first one because its shows that \( F\left( \sqrt{D}\right) = F \) if and only if \( G \subseteq {A}_{n} \) . Indeed, \( \delta \mathrel{\text{:=}} \mathop{\prod }\limits_{{1 \leq i < j \leq n}}\left( {{\alpha }_{i} - {\alpha }_{j}}\right) \in K \) is a square root of \( D \), an...
Yes
Take one example: for \( p = 7 \), we show that 2 is invertible in \( {\mathbb{Z}}_{7} \) as follows:
\[ 2{x}_{1} \equiv 1{\;\operatorname{mod}\;7} \Rightarrow {x}_{1} \equiv 1{\;\operatorname{mod}\;7} \] \[ 2{x}_{2} \equiv 1{\;\operatorname{mod}\;{49}} \Rightarrow {x}_{2} \equiv {25}{\;\operatorname{mod}\;{49}}\left( { \equiv 4{\;\operatorname{mod}\;7}}\right) \] \[ \ldots \] We can always solve \( 2{x}_{i} \equiv 1{\...
No
Why did we call this a limit? We can see this as\n\n\[ \mathbb{C}\llbracket x\rrbracket \mathrel{\text{:=}} \mathop{\lim }\limits_{n}\mathbb{C}\left\lbrack x\right\rbrack /\left( {x}^{n}\right) \]
Given a complex function \( f \) on \( \mathbb{C} \), holomorphic at 0, then the Taylor expansion at 0 :\n\n\[ f\left( 0\right) + x{f}^{\prime }\left( 0\right) + \frac{{f}^{\prime \prime }\left( 0\right) }{2}{x}^{2} + \cdots + \frac{{f}^{\left( n\right) }\left( 0\right) }{n!}{x}^{n} + \cdots \]\n\ndefines an element in...
No
We define \( \widehat{\mathbb{Z}} \mathrel{\text{:=}} \mathop{\lim }\limits_{{ \leftarrow n}}\mathbb{Z}/n\mathbb{Z} \) by divisibility (i.e. for \( m \mid n,\mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z} \) ).\n\nIt is fact that \( \widehat{\mathbb{Z}} \cong \mathop{\prod }\limits_{{p\text{ prime }}}{\mathbb...
We give the following proof: for each prime \( p \), we have\n\n\[ \n{\varphi }_{p} : \widehat{\mathbb{Z}} \rightarrow {\mathbb{Z}}_{p} \n\] \n\n\[ \n{\left( {a}_{n}\right) }_{n} \mapsto {\left( {a}_{{p}^{r}}\right) }_{r} \n\] \n\nThis together give a map \n\n\[ \n\varphi = \mathop{\prod }\limits_{p}{\varphi }_{p} : \w...
Yes
Lemma 22.3.2. The two above definitions of topology on \( \mathop{\lim }\limits_{{i \rightarrow i \in I}}{A}_{i} \) are equivalent.
Proof. Clearly, the open subset in (1) is clearly open in the sense of (2). Conversely, we note that an open subset as defined in (2) takes the form of \( {\pi }_{{i}_{1}}^{-1}\left( {a}_{{i}_{1}}\right) \cap \cdots {\pi }_{{i}_{n}}^{-1}\left( {a}_{{i}_{n}}\right) \), where \( {i}_{1},\ldots ,{i}_{n} \in I \) and \( {a...
Yes
Theorem 22.3.3. If each \( {A}_{i} \) is finite, then \( \mathop{\lim }\limits_{{i \in I}}{A}_{i} \) is compact and Hausdorff. In this case, we say that \( \mathop{\lim }\limits_{{i \in I}}{A}_{i} \) profinite.
Proof. We consider the subspace ![97650b70-8b1b-4cc6-91b2-de9112f1d8bc_145_1.jpg](images/97650b70-8b1b-4cc6-91b2-de9112f1d8bc_145_1.jpg)\n\nThe latter space is compact and Hausdorff (as the product of compact spaces is compact). The subspace is determined by taking the conditions \( {\varphi }_{ji}\left( {a}_{j}\right)...
Yes
Lemma 22.3.6. If \( H \leq G \) is an open subset of a topological group \( G \), then \( H \) is also closed!
Proof. Note that we have a disjoint union\n\n\[ G = \mathop{\coprod }\limits_{{{gH} \in G/H}}{gH} \]\n\nof subsets. But each \( {gH} \) is open. It implies that\n\n\[ H = G \smallsetminus \left( {\mathop{\coprod }\limits_{{{gH} \neq H}}{gH}}\right) \]\n\nis closed.
Yes
Lemma 22.3.7. If \( G \) is a compact topological group, then a subgroup \( H \leq G \) is open if and only if it is closed and of finite index in \( G \) .
Proof. \
No
Lemma 22.3.9. For a profinite group \( G \), we have\n\n\[ G \mathrel{\text{:=}} \mathop{\lim }\limits_{\substack{ \leftarrow \\ {H \leq G} }}\mathop{\lim }\limits_{\text{open normal }}G/H. \]
Proof. There is an obvious map \( G \rightarrow \mathop{\lim }\limits_{{H \leq G\text{ open normal }}}G/H = : {G}^{\prime } \) .\n\nBy definition, \( G = \mathop{\lim }\limits_{{i \leftarrow i \in I}}{G}_{i} \) . We want to construct the reserve arrow:\n\n\[ {G}^{\prime } = \mathop{\lim }\limits_{\substack{ \leftarrow ...
Yes
Example 22.4.2. (1) Write \( \mathbb{Q}\left( {\mu }_{{p}^{\infty }}\right) \mathrel{\text{:=}} \mathbb{Q}\left( {{\zeta }_{{p}^{n}};n \in \mathbb{N}}\right) \) . We have
\[ \operatorname{Gal}\left( {\mathbb{Q}\left( {\mu }_{{p}^{\infty }}\right) /\mathbb{Q}}\right) = \mathop{\lim }\limits_{n}\operatorname{Gal}\left( {\mathbb{Q}\left( {\zeta }_{{p}^{n}}\right) /\mathbb{Q}}\right) \cong \underset{n}{\underbrace{\lim }}{\left( \mathbb{Z}/{p}^{n}\mathbb{Z}\right) }^{ \times } = {\mathbb{Z}...
Yes
Proposition 22.5.1. If \( G \) is a profinite group, then any continuous representation \( \rho : G \rightarrow \) \( {\mathrm{{GL}}}_{n}\left( \mathbb{C}\right) \) has finite image.
Proof. Take a very small open neighborhood \( U \) of \( {I}_{n} \in {\mathrm{{GL}}}_{n}\left( \mathbb{C}\right) \) . The preimage \( {\rho }^{-1}\left( U\right) \) is an open subset of \( G \) containing \( {I}_{n} \) . This implies that \( {\rho }^{-1}\left( U\right) \) contains an open subgroup \( H \) of \( G \) .\...
Yes
Theorem 6.1. Assume an irreducible Markov process with matrix \( Q \) has a stationary distribution \( \pi .\pi \) is detailed balanced if and only if for every sequence of distinct states \( {i}_{0},{i}_{1},\cdots ,{i}_{n - 1},{i}_{n} \in \mathcal{S} \) :\n\n\[ \n{q}_{{i}_{0}{i}_{1}}{q}_{{i}_{1}{i}_{2}}\cdots {q}_{{i}...
Proof. Necessity: From detailed balance, we have \n\n\[ \n1 = \left( {\mathop{\prod }\limits_{{k = 0}}^{{n - 1}}\frac{{\pi }_{{i}_{k}}{q}_{{i}_{k}{i}_{k + 1}}}{{\pi }_{{i}_{k + 1}}{q}_{{i}_{k + 1}{i}_{k}}}}\right) \frac{{\pi }_{{i}_{n}}{q}_{{i}_{n}{i}_{0}}}{{\pi }_{{i}_{0}}{q}_{{i}_{0}{i}_{n}}} = \left( {\mathop{\prod ...
Yes
Theorem 6.2. The following six statements regarding an irreducible stationary Markov process with matrix \( Q \) and stationary distribution \( \pi \) are equivalent [137].\n\n(i) Its stationary distribution satisfies detailed balance: \( {\pi }_{i}{q}_{ij} = {\pi }_{j}{q}_{ji},\forall i, j \in \mathcal{S} \).\n\n(ii) ...
Proof. \( \left( i\right) \Rightarrow \left( {ii}\right) \) :\n\nUsing \( \left( i\right) \), we have\n\n\[ \ln \left( \frac{{q}_{{i}_{0}{i}_{1}}}{{q}_{{i}_{1}{i}_{0}}}\right) + \ln \left( \frac{{q}_{{i}_{1}{i}_{2}}}{{q}_{{i}_{2}{i}_{1}}}\right) + \cdots + \ln \left( \frac{{q}_{{i}_{n - 1}{i}_{n}}}{{q}_{{i}_{n}{i}_{n -...
Yes
Lemma 2 For each \( \varepsilon \), assume\n\n\[ \n{k}^{\varepsilon }\left( {V,\mathbf{x}}\right) = {\pi }_{V}\left( {{B}_{\varepsilon }\left( \mathbf{x}\right) }\right) \exp \left\{ {V\mathop{\inf }\limits_{{\mathbf{y} \in {B}_{\varepsilon }\left( \mathbf{x}\right) }}{\varphi }^{ss}\left( \mathbf{y}\right) }\right\} \...
Proof. According to these assumptions, we know that for each \( \mathbf{x} \), given any \( \delta > 0 \) , there exists a constant \( {V}_{0} \) and \( {\varepsilon }_{0} \), such that when \( V > {V}_{0} \) and \( \varepsilon < {\varepsilon }_{0} \), \n\n\[ \n\left| {\frac{{k}^{\varepsilon }\left( {V,\mathbf{x}}\righ...
Yes
Under the assumptions in Theorem 7.2 and Lemma 2 and further assuming that the limit\n\n\[ \mathop{\lim }\limits_{{V \rightarrow \infty }}\frac{{r}_{\pm \ell }\left( {V\mathbf{x};V}\right) }{V} = {R}_{\pm \ell }\left( \mathbf{x}\right) \]\n\nis locally uniform for any sufficiently small neighborhood of each \( \mathbf{...
Given \( V \), denote \( \mathbf{n}\left( {\mathbf{x}, V}\right) \) as the nearest integer vector to point \( \mathbf{x}V \) . Then, the CME (7.5) at steady state can be rewritten as\n\n\[ \mathop{\sum }\limits_{{\ell = 1}}^{M}\frac{{p}_{V}^{ss}\left( {\mathbf{n}\left( {\mathbf{x}, V}\right) - {v}_{\ell }}\right) }{{p}...
Yes
Theorem 7.5. Assume Eqns. (7.38), (7.43), (7.44), and (7.47) as well as the assumptions of Lemma 2 hold, and assume that the limit\n\n\[ \mathop{\lim }\limits_{{V \rightarrow \infty }}\frac{{r}_{\pm \ell }\left( {V\mathbf{x};V}\right) }{V} = {R}_{\pm \ell }\left( \mathbf{x}\right) \]\n\nis locally uniform for any suffi...
Proof. According to the strong law of large numbers (Eqn. (7.38)), we know that given any small \( \varepsilon > 0 \), for sufficiently large \( V \), the probability is concentrated on the integers satisfying \( \frac{\mathbf{n}}{V} \in {B}_{\varepsilon }\left( {\mathbf{x}\left( t\right) }\right) \).\n\nGiven any smal...
Yes
Proposition 3 The three terms \( {\sigma }^{\text{tot }}\left\lbrack \mathbf{x}\right\rbrack ,{f}_{d}^{\text{macro }}\left\lbrack \mathbf{x}\right\rbrack \) and \( {E}_{\text{in }}^{\text{macro }}\left\lbrack \mathbf{x}\right\rbrack \) are all nonnegative. Furthermore,\n\n(a) \( {\sigma }^{\text{tot }}\left\lbrack \mat...
Proof. (a) is straightforward.\n\n(b) Note that \( \forall x \geq 0,\ln x \geq 1 - \frac{1}{x} \) and \( \ln x \leq x - 1 \) . Thus, we have\n\n\[ \n\ln \left( {\frac{{R}_{-\ell }\left( \mathbf{x}\right) }{{R}_{+\ell }\left( \mathbf{x}\right) }{\mathrm{e}}^{-{\mathbf{V}}_{\ell } \cdot {\nabla }_{\mathbf{x}}{\varphi }^{...
Yes
Theorem 7.6. \( \mathbf{q} \) is any stable fixed point of Eqn. (7.2). Assume \( {\varphi }^{ss}\left( \mathbf{x}\right) \) is at least twice differentiable. Define\n\n\[ \n{\Xi }_{ij} = \frac{{\partial }^{2}{\varphi }^{ss}\left( \mathbf{q}\right) }{\partial {\mathbf{x}}_{i}\partial {\mathbf{x}}_{j}},{A}_{ij} = \mathop...
Proof. Taking the second derivative of the left-hand side of Eqn. (7.47) and noticing that \( {v}_{\ell } \cdot {\nabla }_{\mathbf{x}}{\varphi }^{ss}\left( \mathbf{q}\right) = {B}_{i}\left( \mathbf{q}\right) = 0 \) for each \( \ell \) and \( i \), we can obtain\n\n\[ \n\mathop{\sum }\limits_{{\ell = 1}}^{M}\left\{ {\le...
Yes
Proposition 3 (Abel’s formula). The fundmental matrix (2.10) satisfies\n\n(2.12)\n\n\[ \frac{d}{dt}\left| {\Psi \left( t\right) }\right| = \operatorname{trace}\left( {A\left( t\right) }\right) \left| {\Psi \left( t\right) }\right| \]\n\n\[ \left( {Abel}\right) \]\n\nwhere \( \left| {\Psi \left( t\right) }\right| = \det...
Proof.\n\n\[ \frac{d}{dt}\left| {\Psi \left( t\right) }\right| = \frac{d}{dt}\left| \begin{array}{ll} {\psi }_{11}\left( t\right) & {\psi }_{21}\left( t\right) \\ {\psi }_{12}\left( t\right) & {\psi }_{22}\left( t\right) \end{array}\right| = \left| \begin{array}{ll} {\dot{\psi }}_{11}\left( t\right) & {\dot{\psi }}_{21...
Yes
Proposition 3 (Abel's formula). The fundmental matrix (2.10) satisfies \[ \frac{d}{dt}\left| {\Psi \left( t\right) }\right| = \operatorname{trace}\left( {A\left( t\right) }\right) \left| {\Psi \left( t\right) }\right| \] where \( \left| {\Psi \left( t\right) }\right| = \det \Psi \left( t\right) \).
Proof. \[ \frac{d}{dt}\left| {\Psi \left( t\right) }\right| = \frac{d}{dt}\left| \begin{array}{ll} {\psi }_{11}\left( t\right) & {\psi }_{21}\left( t\right) \\ {\psi }_{12}\left( t\right) & {\psi }_{22}\left( t\right) \end{array}\right| = \left| \begin{array}{ll} {\dot{\psi }}_{11}\left( t\right) & {\dot{\psi }}_{21}\l...
Yes
Proposition 7 (Gronwall's Inequality). Assume that for some constant \( C \geq 0 \) and non-negative integrable functions \( f \) and \( g \), we have\n\n(2.16)\n\n\[ f\left( t\right) \leq C + {\int }_{0}^{t}f\left( s\right) g\left( s\right) {ds} \]\n\nthen\n\n(2.17)\n\n\[ f\left( t\right) \leq C{e}^{{\int }_{0}^{t}g\l...
Proof. Let \( F\left( t\right) = C + {\int }_{0}^{t}f\left( s\right) g\left( s\right) {ds} \), then \( F\left( t\right) \) is differentiable\n\nand\n\n\[ {F}^{\prime }\left( t\right) = f\left( t\right) g\left( t\right) \]\n\nSince \( g\left( t\right) \geq 0 \) and \( f\left( t\right) \leq F\left( t\right) \), we have\n...
Yes
Proposition 9. If \( t \in \left\lbrack {{t}_{0} - h,{t}_{0} + h}\right\rbrack \), then\n\n\[ \left| {{r}_{n}\left( t\right) }\right| \leq M{L}^{n}{t}^{n + 1}/\left( {n + 1}\right) ! \]
Proof. We prove this by induction. When \( n = 0 \) ,\n\n\[ \left| {{r}_{0}\left( t\right) }\right| = \left| {{y}_{1}\left( t\right) - {y}^{0}}\right| = \left| {{\int }_{0}^{t}f\left( {s,{y}^{0}}\right) {ds}}\right| \leq {\int }_{0}^{t}\left| {f\left( {s,{y}^{0}}\right) }\right| {ds} \leq {Mt}. \]\n\nSuppose that \( \l...
Yes
Proposition 11. Assume that \( f\left( x\right) \) is a continuous function If over any interval \( \left\lbrack {a, b}\right\rbrack \left( {a < b}\right) \), the integration \( {\int }_{a}^{b}f\left( x\right) \mathrm{d}x = 0 \) then, \( f\left( x\right) \equiv 0. \)
Proof. If at some point \( {x}_{0}, f\left( {x}_{0}\right) \neq 0 \), assume \( f\left( {x}_{0}\right) > 0 \) without loss of generality, then there is a \( \delta > 0 \) such that\n\n(3.4)\n\n\[ \left| {f\left( x\right) - f\left( {x}_{0}\right) }\right| \leq \frac{1}{2}f\left( {x}_{0}\right) \]\n\nfor \( x \in \left\l...
Yes
Proposition 16. \( \; \bullet {G}^{\mu } \) is symmetric and decreasing.
\[ {G}^{\mu }\left( x\right) = \left\{ \begin{array}{ll} \frac{1}{2\mu }{e}^{-\mu \left| x\right| } & d = 1, \\ \frac{1}{{4\pi }\left| x\right| }{e}^{-\mu \left| x\right| } & d = 3. \end{array}\right. \]
Yes
The probability space for the outcome of one trial can be defined as follows. The sample space \( \Omega = \{ H, T\} \) where \( H \) and \( T \) represent head and tail, respectively. The \( \sigma \) -algebra
\[ \mathcal{F} = \text{ all subsets of }\Omega = \{ \varnothing ,\{ H\} ,\{ T\} ,\Omega \} \] and \[ \mathbb{P}\left( \varnothing \right) = 0,\;\mathbb{P}\left( {\{ H\} }\right) = \mathbb{P}\left( {\{ T\} }\right) = \frac{1}{2},\;\mathbb{P}\left( \Omega \right) = 1. \]
Yes
Example 1.5 (Uniform orientation distribution on \( {\mathbb{S}}^{2} \) ). In this case, the sample space \( \Omega = {\mathbb{S}}^{2} \) . Let \( \mathcal{B} \) be the set of all open sets of \( {\mathbb{S}}^{2} \), defined as the intersection of any open set \( B \subset {\mathbb{R}}^{3} \) and \( {\mathbb{S}}^{2} \)...
Within this framework, the standard rules of set theory are used to answer probability questions. For instance, if both \( A, B \in \mathcal{F} \), the probalility that both \( A \) and \( B \) occurs is given by \( \mathbb{P}\left( {A \cap B}\right) \), the probability that either \( A \) or \( B \) occurs is given by...
Yes
Example 1.7 (Bernoulli distribution). The Bernoulli distribution has the form\n\n\[ \n\\mathbb{P}\\left( {X = j}\\right) = \\left\\{ \\begin{array}{ll} p, & j = 1 \\\\ q, & j = 0 \\end{array}\\right.\n\]\n\n\( p + q = 1 \) and \( p, q \\geq 0 \) . When \( p = q = 1/2 \), it corresponds to the toss of a fair coin. The m...
\n\[ \n\\mathbb{E}X = p,\\;\\operatorname{Var}\\left( X\\right) = {pq}.\n\]
Yes
Example 1.8 (Binomial distribution \( B\left( {n, p}\right) \) ). The binomial distribution \( B\left( {n, p}\right) \) has the form\n\n(1.9)\n\n\[ \mathbb{P}\left( {X = k}\right) = \left( \begin{array}{l} n \\ k \end{array}\right) {p}^{k}{q}^{n - k},\;k = 0,1,\ldots, n. \]
It is straightforward to obtain\n\n\[ \mathbb{E}X = {np},\;\operatorname{Var}\left( X\right) = {npq}. \]
No
Lemma 1.15 (Chebyshev’s inequality). Let \( \mathbf{X} \) be a random variable such that \( \mathbb{E}{\left| \mathbf{X}\right| }^{p} < \infty \) for some \( p > 0 \) . Then\n\n\[ \mathbb{P}\{ \left| \mathbf{X}\right| \geq \lambda \} \leq \frac{1}{{\lambda }^{p}}\mathbb{E}{\left| \mathbf{X}\right| }^{p} \]\n\nfor any p...
Proof. For any \( \lambda > 0 \) ,\n\n\[ \mathbb{E}{\left| \mathbf{X}\right| }^{p} = {\int }_{{\mathbb{R}}^{d}}{\left| \mathbf{x}\right| }^{p}\mu \left( {d\mathbf{x}}\right) \geq {\int }_{\left| \mathbf{x}\right| \geq \lambda }{\left| \mathbf{x}\right| }^{p}\mu \left( {d\mathbf{x}}\right) \geq {\lambda }^{p}{\int }_{\l...
Yes
Lemma 1.16 (Jensen’s inequality). Let \( \mathbf{X} \) be a random variable such that \( \mathbb{E}\left| \mathbf{X}\right| < \infty \) and \( \phi : \mathbb{R} \rightarrow \mathbb{R} \) is a convex function such that \( \mathbb{E}\left| {\phi \left( \mathbf{X}\right) }\right| < \infty \) . Then\n\n(1.21)\n\n\[ \mathbb...
This follows directly from the definition of convex functions. Readers can also refer to \( \mathbf{{Chu01}} \) for the details.
No
Example 1.17 (Uniform distribution). The uniform distribution on a domain \( B \) (in \( {\mathbb{R}}^{d} \) ) is defined by the probability density function:\n\n\[ \rho \left( x\right) = \left\{ \begin{array}{ll} \frac{1}{\operatorname{vol}\left( B\right) }, & \text{ if }\mathbf{x} \in B, \\ 0, & \text{ otherwise. } \...
In one dimension if \( B = \left\lbrack {0,1}\right\rbrack \) (denoted as \( \mathcal{U}\left\lbrack {0,1}\right\rbrack \) later), this reduces to\n\n\[ \rho \left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ if }x \in \left\lbrack {0,1}\right\rbrack \\ 0, & \text{ otherwise. } \end{array}\right. \]\n\nFor the uni...
Yes
Example 1.18 (Exponential distribution). The exponential distribution \( \mathcal{E}\left( \lambda \right) \) is defined by the probability density function:\n\n\[ \rho \left( x\right) = \left\{ \begin{array}{ll} 0, & \text{ if }x < 0 \\ \lambda {e}^{-{\lambda x}}, & \text{ if }x \geq 0 \end{array}\right. \]
The mean and variance of \( E\left( \lambda \right) \) are\n\n(1.22)\n\n\[ \mathbb{E}X = \frac{1}{\lambda },\;\operatorname{Var}\left( X\right) = \frac{1}{{\lambda }^{2}}. \]
Yes
The one-dimensional normal distribution (also called Gaussian distribution) \( N\left( {\mu ,{\sigma }^{2}}\right) \) is defined by the probability density function:\n\n\[ \rho \left( x\right) = \frac{1}{\sqrt{{2\pi }{\sigma }^{2}}}\exp \left( {-\frac{1}{2{\sigma }^{2}}{\left( x - \mu \right) }^{2}}\right) \]
with mean \( \mu \) and variance \( {\sigma }^{2} \).
No
Example 1.20 (Gibbs distribution). In equilibrium statistical mechanics, we are concerned with a probability distribution \( \pi \) over a state space \( S \) . In the case of an \( n \) -particle system with continuous states, we have \( \mathbf{x} = \) \( \left( {{\mathbf{x}}_{1},\ldots ,{\mathbf{x}}_{n},{\mathbf{p}}...
\[ \pi \left( \mathbf{x}\right) = \frac{1}{Z}{e}^{-{\beta H}\left( \mathbf{x}\right) },\;\mathbf{x} \in {\mathbb{R}}^{6n},\beta = {\left( {k}_{B}T\right) }^{-1}, \] where \( H \) is the energy of the considered system, \( T \) is the absolute temperature, \( {k}_{B} \) is the Boltzmann constant, and \[ Z = {\int }_{{\m...
Yes
Proposition 1.26. Let \( g \) be a measurable function. Then\n\n\[ \mathbb{E}{\left( X - \mathbb{E}\left( X \mid Y\right) \right) }^{2} \leq \mathbb{E}{\left( X - g\left( Y\right) \right) }^{2}. \]
Proof. We have\n\n\[ \mathbb{E}{\left( X - g\left( Y\right) \right) }^{2} = \mathbb{E}{\left( X - E\left( X \mid Y\right) \right) }^{2} + \mathbb{E}{\left( E\left( X \mid Y\right) - g\left( Y\right) \right) }^{2} \]\n\n\[ + 2\mathbb{E}\left\lbrack {\left( {X - E\left( {X \mid Y}\right) }\right) \left( {E\left( {X \mid ...
Yes
(i) Almost sure convergence implies convergence in probability.
Note that\n\n\[ \mathbb{P}\left( {\left| {{X}_{n}\left( \omega \right) - X\left( \omega \right) }\right| > \epsilon }\right) = {\int }_{\Omega }{\chi }_{\left\{ \left| {X}_{n} - X\right| > \epsilon \right\} }\left( \omega \right) \mathbb{P}\left( {d\omega }\right) \rightarrow 0 \]\n\nby the almost sure convergence and ...
Yes
Proposition 1.33. The characteristic function has the following properties:\n\n(1) \( \forall \xi \in \mathbb{R},\left| {f\left( \xi \right) }\right| \leq 1, f\left( \xi \right) = \overline{f\left( {-\xi }\right) }, f\left( 0\right) = 1 \) ;\n\n(2) \( f \) is uniformly continuous on \( \mathbb{R} \) .
Proof. The proof of the first statements is straightforward. For the second statement, we have\n\n\[ \left| {f\left( {\xi }_{1}\right) - f\left( {\xi }_{2}\right) }\right| = \left| {\mathbb{E}\left( {{e}^{i{\xi }_{1}X} - {e}^{i{\xi }_{2}X}}\right) }\right| = \left| {\mathbb{E}\left( {{e}^{i{\xi }_{1}X}\left( {1 - {e}^{...
Yes
Theorem 1.35 (Lévy’s continuity theorem). Let \( {\left\{ {\mu }_{n}\right\} }_{n \in \mathbb{N}} \) be a sequence of probability measures, and let \( {\left\{ {f}_{n}\right\} }_{n \in \mathbb{N}} \) be their corresponding characteristic functions. Assume that:\n\n(1) \( {f}_{n} \) converges everywhere on \( \mathbb{R}...
For a proof, see [Chu01].
No
Theorem 1.37 (Bochner’s theorem). A function \( f \) is the characteristic function of a probability measure if and only if it is positive semidefinite and continuous at 0 with \( f\left( 0\right) = 1 \) .
Proof. We only prove the necessity part. The other part is less trivial and readers may consult Chu01. Assume that \( f \) is a characteristic function. Then\n\n(1.50)\n\n\[ \mathop{\sum }\limits_{{i, j = 1}}^{n}f\left( {{\xi }_{i} - {\xi }_{j}}\right) {v}_{i}{\bar{v}}_{j} = {\int }_{\mathbb{R}}{\left| \mathop{\sum }\l...
No
Theorem 1.40. Denote \( {M}_{X}\left( t\right) ,{M}_{Y}\left( t\right) \), and \( {M}_{X + Y}\left( t\right) \) the moment generating functions of the random variables \( X, Y \), and \( X + Y \), respectively. If \( X, Y \) are independent, then\n\n(1.56)\n\n\[ \n{M}_{X + Y}\left( t\right) = {M}_{X}\left( t\right) {M}...
Proof. The proof is straightforward by noticing\n\n\[ \n{M}_{X + Y}\left( t\right) = \mathbb{E}{e}^{t\left( {X + Y}\right) } = \mathbb{E}{e}^{tX}\mathbb{E}{e}^{tY} = {M}_{X}\left( t\right) {M}_{Y}\left( t\right) .\n\]
Yes
(1) If \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\mathbb{P}\left( {A}_{n}\right) < \infty \), then \( \mathbb{P}\left( \left\{ {{A}_{n}\text{i.o.}}\right\} \right) = 0 \) .
Proof. (1) We have\n\n\[ \mathbb{P}\left( \left\{ {\mathop{\bigcap }\limits_{{n = 1}}^{\infty }\mathop{\bigcup }\limits_{{k = n}}^{\infty }{A}_{k}}\right\} \right) \leq \mathbb{P}\left( \left\{ {\mathop{\bigcup }\limits_{{k = n}}^{\infty }{A}_{k}}\right\} \right) \leq \mathop{\sum }\limits_{{k = n}}^{\infty }\mathbb{P}...
Yes
Lemma 1.42. Let \( {\left\{ {X}_{n}\right\} }_{n \in \mathbb{N}} \) be a sequence of identically distributed (not necessarily independent) random variables, such that \( \mathbb{E}\left| {X}_{n}\right| < \infty \) . Then
Proof. For any \( \epsilon > 0 \), define\n\n\[ \n{A}_{n}^{\epsilon } = \left\{ {\omega \in \Omega : \left| {{X}_{n}\left( \omega \right) /n}\right| > \epsilon }\right\} \n\]\n\nThen\n\n\[ \n\mathop{\sum }\limits_{{n = 1}}^{\infty }\mathbb{P}\left( {A}_{n}^{\epsilon }\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\...
Yes
Theorem 2.1 (Weak law of large numbers (WLLN)). Let \( {\left\{ {X}_{j}\right\} }_{j = 1}^{\infty } \) be a sequence of i.i.d. random variables such that \( \mathbb{E}\left| {X}_{j}\right| < \infty \) . Then\n\n\[ \frac{{S}_{n}}{n} \rightarrow \eta \;\text{ in probability. } \]
Proving this result under the stated assumption is quite involved. We will give a proof of the WLLN under the stronger assumption that \( \mathbb{E}{\left| {X}_{j}\right| }^{2} < \) \( \infty \) .\n\nProof. Without loss of generality, we can assume \( \eta = 0 \) . From the Cheby-shev inequality, we have\n\n\[ \mathbb{...
Yes
Theorem 2.2 (Strong law of large numbers (SLLN)). Let \( {\left\{ {X}_{j}\right\} }_{j = 1}^{\infty } \) be a sequence of i.i.d. random variables such that \( \mathbb{E}\left| {X}_{j}\right| < \infty \) . Then\n\n\[ \frac{{S}_{n}}{n} \rightarrow \eta \text{ a.s. } \]
Proof. We will only give a proof of the SLLN under the stronger assumption that \( \mathbb{E}{\left| {X}_{j}\right| }^{4} < \infty \) . The proof under the stated assumption can be found in Chu01\n\nWithout loss of generality, we can assume \( \eta = 0 \) . Using the Chebyshev inequality, we have\n\n\[ \mathbb{P}\left(...
No
For the distribution with PDF\n\n\[ \n p\left( x\right) = \frac{1}{\pi \left( {1 + {x}^{2}}\right) },\;x \in \mathbb{R}, \]\n\nwe have \( \mathbb{E}\left| {X}_{j}\right| = \infty \) . For a sequence of i.i.d. random variables \( {\left\{ {X}_{j}\right\} }_{j = 1}^{\infty } \) with Cauchy-Lorentz distribution, one can e...
In fact, the converse statement for SLLN is also true: If \( {S}_{n}/n \rightarrow a < \) \( \infty \) a.s. holds, then \( \mathbb{E}\left| {X}_{j}\right| < \infty \) . The proof can be found in Chu01.
No
Theorem 2.4 (Lindeberg-Lévy central limit theorem (CLT)). Let \( {\left\{ {X}_{j}\right\} }_{j = 1}^{\infty } \) be a sequence of i.i.d. random variables. Assume that \( \mathbb{E}{X}_{j}^{2} < \infty \) and let \( {\sigma }^{2} = \operatorname{Var}\left( {X}_{j}\right) \) . Then\n\n\[ \frac{{S}_{n} - {n\eta }}{\sqrt{n...
Outline of proof. Assume without loss of generality that \( \eta = 0 \) and \( \sigma = 1 \) ; otherwise we can shift and rescale \( {X}_{j} \) . Let \( f \) be the characteristic function of \( {X}_{1} \) and let \( {g}_{n} \) be the characteristic function of \( {S}_{n}/\sqrt{n} \) . Then\n\n\[ {g}_{n}\left( \xi \rig...
Yes
Lemma 2.9. The rate function \( I\left( x\right) \) has the following properties:\n\n(i) \( I\left( x\right) \) is convex and lower semicontinuous.\n\n(ii) \( I\left( x\right) \) is nonnegative and \( I\left( \eta \right) = 0 \).\n\n(iii) \( I\left( x\right) \) is nondecreasing in \( \left\lbrack {\eta ,\infty )\text{a...
Proof. (i) The convexity of \( \Lambda \left( \lambda \right) \) follows from Hölder’s inequality. For any \( 0 \leq \theta \leq 1, \)\n\n\[ \Lambda \left( {\theta {\lambda }_{1} + \left( {1 - \theta }\right) {\lambda }_{2}}\right) = \log \mathbb{E}\left( {\exp \left( {\theta {\lambda }_{1}{X}_{j}}\right) \exp \left( {...
Yes
Example 2.10 (Cramér's theorem applied to the Bernoulli distribution with parameter \( p\left( {0 < p < 1}\right) \) ). We have \( \Lambda \left( \lambda \right) = \log \left( {p{e}^{\lambda } + q}\right) \) where \( q = 1 - p \) . The rate function
\[ I\left( x\right) = \left\{ \begin{array}{ll} x\log \frac{x}{p} + \left( {1 - x}\right) \log \frac{1 - x}{q}, & x \in \left\lbrack {0,1}\right\rbrack , \\ \infty , & \text{ otherwise. } \end{array}\right. \] It is obvious that \( I\left( x\right) \geq 0 \), and \( I\left( x\right) \) achieves its global minimum 0 at ...
Yes
Assume that \( {X}_{j} \) is exponentially distributed; i.e., if we denote by \( \rho \left( x\right) \) the probability density function of \( {X}_{j} \), then\n\n\[ \rho \left( x\right) = \left\{ \begin{array}{ll} {e}^{-x}, & \text{ if }x > 0 \\ 0, & \text{ if }x \leq 0 \end{array}\right. \]\n\nThen \( \mathbb{P}\lef...
\[ = \mathop{\prod }\limits_{{j = 1}}^{n}\mathbb{P}\left( {{X}_{j} \leq x}\right) = {\left( 1 - {e}^{-x}\right) }^{n}. \]\n\nThis remains true even if \( x \) depends on \( n \) . We will choose \( x = {x}_{n} \) such that \( {\left( 1 - {e}^{-{x}_{n}}\right) }^{n} \) has a nontrivial limit. For this purpose, we let\n\...
Yes
Assume that \( {X}_{j} \) is uniformly distributed on \( \left\lbrack {0,1}\right\rbrack \) ; i.e., \n\n\[ \n\rho \left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ if }x \in \left\lbrack {0,1}\right\rbrack \\ 0, & \text{ otherwise. } \end{array}\right. \n\] \n\nWe expect that \( {M}_{n} \rightarrow 1 \) as \( n \...
Notice that if we let \( {x}_{n} = 1 - x/n \), then \n\n\[ \n\mathbb{P}\left( {{M}_{n} \leq {x}_{n}}\right) = {\left( 1 - \frac{x}{n}\right) }^{n} \rightarrow {e}^{-x} \n\] \nas \( n \rightarrow \infty \), or equivalently \n\n\[ \n\mathbb{P}\left( {n\left( {{M}_{n} - 1}\right) \leq x}\right) \rightarrow {e}^{-\left| x\...
Yes
Example 3.1 (Symmetric random walk). Consider the sequence of random variables \( {\left\{ {\xi }_{j}\right\} }_{j = 1}^{\infty } \), where the \( {\left\{ {\xi }_{j}\right\} }^{\prime } \) s are i.i.d and \( {\xi }_{j} = \pm 1 \) with probability \( 1/2 \) . Let\n\n\[ \n{X}_{n} = \mathop{\sum }\limits_{{j = 1}}^{n}{\x...
\[ \n\mathbb{P}\left( {{X}_{n + 1} = i \pm 1 \mid {X}_{n} = i}\right) = \mathbb{P}\left( {{\xi }_{n + 1} = \pm 1}\right) = \frac{1}{2} \n\]\n\nand \( \mathbb{P}\left( {{X}_{n + 1} = }\right. \) anything else \( \left. {\mid {X}_{n} = i}\right) = 0 \) . We see that, knowing \( {X}_{n} \), the distribution of \( {X}_{n +...
Yes
Example 3.2 (Ehrenfest’s diffusion model). Consider a container separated by a permeable membrane in the middle and filled with a total of \( K \) particles. At each time \( n = 1,2\ldots \), pick one particle in random among the \( K \) particles and place it into the other part of the container. Let \( {X}_{n} \) be ...
\[ \mathbb{P}\left( {{X}_{n + 1} = {i}_{n + 1} \mid {\left\{ {X}_{m} = {i}_{m}\right\} }_{m = 0}^{n}}\right) = \mathbb{P}\left( {{X}_{n + 1} = {i}_{n + 1} \mid {X}_{n} = {i}_{n}}\right) ,\] for \( {i}_{k} \in \{ 0,1,\ldots, K\} \), and this is also a Markov process.
Yes
Example 3.3 (Autoregressive model). The autoregressive process \( {\left\{ {Y}_{n}\right\} }_{n \in \mathbb{N}} \) of order \( k \) is defined as follows. For each \( n \geq 1 \) ,\n\n\[ \n{Y}_{n} = {\alpha }_{1}{Y}_{n - 1} + {\alpha }_{2}{Y}_{n - 2} + \cdots + {\alpha }_{k}{Y}_{n - k} + {R}_{n}, \n\]\n\nwhere \( {\alp...
In this example, \( {Y}_{n} \) itself is not Markovian since \( {Y}_{n} \) depends not only on \( {Y}_{n - 1} \), but also on \( {Y}_{n - 2},\ldots ,{Y}_{n - k} \) . But if we introduce the new variable\n\n\[ \n{\mathbf{X}}_{n} = {\left( {Y}_{n},\ldots ,{Y}_{n - k + 1}\right) }^{T},\;n = 0,1,\ldots , \n\]\n\nthen we ob...
Yes
Proposition 3.4 (Chapman-Kolmogorov equation).\n\n\[ \mathbb{P}\left( {{X}_{n} = j \mid {X}_{0} = i}\right) \]\n\n\[ = \mathop{\sum }\limits_{{k \in S}}\mathbb{P}\left( {{X}_{n} = j \mid {X}_{m} = k}\right) \mathbb{P}\left( {{X}_{m} = k \mid {X}_{0} = i}\right) ,\;1 \leq m \leq n - 1. \]
The proof is a straightforward application of Bayes's rule and the Markov property.
No
Lemma 3.5. The spectral radius of \( \mathbf{P} \) is equal to 1 :\n\n\[ \rho \left( \mathbf{P}\right) = \mathop{\max }\limits_{\lambda }\left| \lambda \right| = 1 \]\n\nwhere the maximum is taken over the eigenvalues of \( \mathbf{P} \) .
Proof. We already know that 1 is an eigenvalue of \( \mathbf{P} \) . To show it is the maximum eigenvalue, denote by \( \mathbf{u} \) the left eigenvector of \( \mathbf{P} \) with eigenvalue \( \lambda \) . Then\n\n\[ \lambda {u}_{i} = \mathop{\sum }\limits_{{j \in S}}{u}_{j}{p}_{ji} \]\n\nwhich implies that\n\n\[ \lef...
Yes
Lemma 3.7. Irreducibility is equivalent to the property that every pair of nodes in the state space communicates with each other.
The proof is left as an exercise for the readers.
No
Theorem 3.8 (Perron-Frobenius theorem). Let \( \mathbf{A} \) be an irreducible nonnegative matrix, and let \( \rho \left( \mathbf{A}\right) \) be its spectral radius: \( \rho \left( \mathbf{A}\right) = \mathop{\max }\limits_{\lambda }\left| {\lambda \left( \mathbf{A}\right) }\right| \) . Then:\n\n(1) There exist positi...
We refer to [HJ85] for the proof.
No
Consider the Markov chain with the transition probability matrix\n\n\[ \mathbf{P} = \left\lbrack \begin{matrix} 0 & 1 & 0 & 0 & 0 \\ {0.3} & {0.4} & {0.3} & 0 & 0 \\ {0.3} & 0 & {0.4} & {0.3} & 0 \\ 0 & 0 & 0 & {0.5} & {0.5} \\ 0 & 0 & 0 & {0.5} & {0.5} \end{matrix}\right\rbrack \]
This is reducible. But one can verify that the chain has a unique invariant distribution\n\n\[ \mathbf{\pi } = \left( {0,0,0,{0.5},{0.5}}\right) . \]
No
Theorem 3.14. For any fixed time \( t \), the distribution of \( {N}_{t} \) is Poisson with parameter \( {\lambda t} \) .
Proof. Let \( {p}_{m}\left( t\right) = \mathbb{P}\left( {{N}_{t} = m}\right) \) . Taking \( h \ll 1 \) we have\n\n\[ \n{p}_{0}\left( {t + h}\right) = {p}_{0}\left( t\right) \left( {1 - {\lambda h}}\right) + o\left( h\right) , \n\]\n\nor equivalently\n\n\[ \n\frac{{p}_{0}\left( {t + h}\right) - {p}_{0}\left( t\right) }{...
Yes
Theorem 3.15. The jump chain \( \left\{ {Y}_{n}\right\} \) is a Markov chain with \( \widetilde{Q} \) as the transition probability matrix, and the holding times \( {H}_{1},{H}_{2},\ldots \) are independent exponential random variables with parameters \( {q}_{{Y}_{0}},{q}_{{Y}_{1}},\ldots \), respectively.
The proof relies on the strong Markov property of the \( Q \) -process. See Section E of the appendix and [Nor97].
No
Theorem 3.16. The Q-process \( X \) is irreducible if and only if the embedded chain \( Y \) is irreducible.
Proof. First note that for any \( i \neq j \), if \( {q}_{ij} > 0 \), we have\n\n\[ \n{p}_{ij}\left( t\right) \geq {\mathbb{P}}^{i}\left( {{J}_{1} \leq t,{Y}_{1} = j,{H}_{2} > t}\right) = {\int }_{0}^{t}\exp \left( {-{q}_{i}u}\right) {q}_{ij}{du}\exp \left( {-{q}_{j}t}\right) \n\]\n\n\[ \n= \left( {1 - {e}^{-{q}_{i}t}}...
Yes
Theorem 3.17 (Convergence to equilibrium). Assume that \( \mathbf{Q} \) is irreducible. Then for any initial distribution \( {\mathbf{\mu }}_{0} \)\n\n\[ \mathbf{\mu }\left( t\right) = {\mathbf{\mu }}_{0}\mathbf{P}\left( t\right) \rightarrow \mathbf{\pi }\;\text{ exponentially fast as }t \rightarrow \infty ,\]\n\nwhere...
Proof. From irreducibility and Theorem 3.16, we know that \( {p}_{ij}\left( t\right) > 0 \) for any \( i \neq j \) and \( t > 0 \) . It is straightforward to see that\n\n\[ {p}_{ii}\left( t\right) \geq {\mathbb{P}}^{i}\left( {{J}_{1} > t}\right) = {e}^{-{q}_{i}t} > 0 \]\n\nfor any \( i \in S \) and \( t > 0 \), where \...
Yes
Theorem 3.18 (Ergodic theorem). Assume that \( \mathbf{Q} \) is irreducible. Then for any bounded function \( f \) we have\n\n\[ \frac{1}{T}{\int }_{0}^{T}f\left( {X}_{s}\right) {ds} \rightarrow \langle f{\rangle }_{\pi },\;\text{ a.s. } \]\n\nas \( T \rightarrow \infty \), where \( \pi \) is the unique invariant distr...
The proof can be found in [Dur10, Nor97].
No
Theorem 4.1. The period of an LCG is \( m \) if and only if\n\n(i) \( b \) and \( m \) are relatively prime;\n\n(ii) every prime factor of \( m \) divides \( a - 1 \) ;\n\n(iii) if \( 4 \mid m \), then \( 4 \mid \left( {a - 1}\right) \) .
If one chooses \( m = {2}^{k}, a = {4c} + 1 \), and \( b \) odd, then these conditions are satisfied.
Yes
Proposition 4.3 (Inverse transformation method). Let \( F \) be the distribution function of \( X \) ; i.e., \( F\left( x\right) = \mathbb{P}\left( {X \leq x}\right) \) . Let \( U \) be a \( \mathcal{U}\left\lbrack {0,1}\right\rbrack \) random variable. Define the generalized inverse of \( F \) by\n\n\[ \n{F}^{ - }\lef...
Proof. If \( F \) is continuous and strictly increasing, we have \( {F}^{ - }\left( u\right) = {F}^{-1}\left( u\right) \) and\n\n\[ \n\mathbb{P}\left( {X \leq x}\right) = \mathbb{P}\left( {{F}^{-1}\left( U\right) \leq x}\right) = \mathbb{P}\left( {U \leq F\left( x\right) }\right) = F\left( x\right) .\n\]\n\nIn the gene...
Yes
Consider the integral\n\n\[ \nI\left( f\right) = {\int }_{-\infty }^{+\infty }\frac{1}{\sqrt{2\pi }}{\left( 1 + r\right) }^{-1}{e}^{-\frac{{x}^{2}}{2}}{dx} \n\]\n\nwhere \( r = {e}^{\sigma x},\sigma \gg 1 \) .
Notice that\n\n\[ \n{\left( 1 + r\right) }^{-1} \approx h\left( x\right) = \left\{ \begin{array}{ll} 1, & x \leq 0 \\ 0, & x > 0 \end{array}\right. \n\]\n\nWe have\n\n\[ \nI\left( f\right) = \frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{+\infty }\left( {{\left( 1 + r\right) }^{-1} - h\left( x\right) }\right) {e}^{-\frac{{x...
Yes
Example 4.5. Shown in Figure 4.4 is a one-dimensional model for the magnetization of a ferromagnet. The lattice has \( M \) sites, the state space \( S = \{ \mathbf{x}\} = \{ + 1, - 1{\} }^{M} \) . It has \( {2}^{M} \) states in total. The states at the sites are called spins. The microscopic configurations are describ...
In equilibrium statistical mechanics, we are interested in the thermodynamic average of some function \( f\left( \mathbf{x}\right) \) given by \[ \langle f\rangle = \mathop{\sum }\limits_{{\mathbf{x} \in S}}f\left( \mathbf{x}\right) \pi \left( \mathbf{x}\right) \;\text{ or }\;{\int }_{S}f\left( \mathbf{x}\right) \pi \l...
Yes
Theorem 4.6. The Gibbs distribution has the limit\n\n\\[ \n\\mathop{\\lim }\\limits_{{\\beta \\rightarrow + \\infty }}{\\pi }_{\\beta }\\left( x\\right) = \\left\\{ \\begin{array}{ll} \\frac{1}{\\left| \\mathcal{M}\\right| }, & \\text{ if }x \\in \\mathcal{M}, \\\\ 0, & \\text{ othewise,} \\end{array}\\right.\n\\]\n\nw...
Proof. Define \\( m = \\mathop{\\min }\\limits_{x}H\\left( x\\right) \\) . Then\n\n\\[ \n{\\pi }_{\\beta }\\left( x\\right) \\; = \\;\\frac{{e}^{-\\beta \\left( {H\\left( x\\right) - m}\\right) }}{\\mathop{\\sum }\\limits_{{z \\in \\mathcal{M}}}{e}^{-\\beta \\left( {H\\left( z\\right) - m}\\right) } + \\mathop{\\sum }\...
Yes
Theorem 4.7 (Convergence of simulated annealing). Assume that \( H \) is defined over a finite set \( \mathcal{X} \) and \( Q \) is a symmetric irreducible proposal matrix. If the annealing procedure is chosen such that \( \beta \left( n\right) \leq C\log n \), where \( C \) only depends on the structure of \( Q \) and...
The proof of this theorem can be found in [Win03].
No
Consider the \( Q \) -process \( {X}_{t} \) on \( S = \{ 1,2,\ldots, I\} \) with generator \( \mathbf{Q} \) defined as in (3.16). We have
\[\n\left( {\mathcal{A}f}\right) \left( i\right) = \mathop{\lim }\limits_{{t \rightarrow 0 + }}\frac{{\mathbb{E}}^{i}f\left( {X}_{t}\right) - f\left( i\right) }{t} = \mathop{\lim }\limits_{{t \rightarrow 0 + }}\frac{1}{t}\left( {\mathop{\sum }\limits_{{j \in S}}\left( {{P}_{ij}\left( t\right) - {\delta }_{ij}}\right) f...
Yes
Example 5.8 (Poisson process). Consider the Poisson process \( {X}_{t} \) on \( \mathbb{N} \) with rate \( \lambda \) . We have\n\n\[ \left( {\mathcal{A}f}\right) \left( n\right) = \mathop{\lim }\limits_{{t \rightarrow 0 + }}\frac{{\mathbb{E}}^{n}f\left( {X}_{t}\right) - f\left( n\right) }{t} = \mathop{\lim }\limits_{{...
\[ = \mathop{\lim }\limits_{{t \rightarrow 0 + }}\frac{1}{t}\left( {f\left( n\right) \left( {{e}^{-{\lambda t}} - 1}\right) + f\left( {n + 1}\right) {\lambda t}{e}^{-{\lambda t}} + \mathop{\sum }\limits_{{k = n + 2}}^{\infty }f\left( k\right) \frac{{\left( \lambda t\right) }^{k - n}}{\left( {k - n}\right) !}{e}^{-{\lam...
Yes
Theorem 5.10. Assume the stochastic process \( {\left\{ {X}_{t}\right\} }_{t \in \left\lbrack {0, T}\right\rbrack } \) satisfies the condition\n\n\[ \mathbb{E}{\int }_{0}^{T}{X}_{t}^{2}{dt} < \infty \]\n\nThen \( m \in {L}_{t}^{2} \) in the sense that\n\n\[ {\int }_{0}^{T}{m}^{2}\left( t\right) {dt} < \infty \]\n\nFurt...
Proof. First, we have\n\n\[ {\int }_{0}^{T}{m}^{2}\left( t\right) {dt} = {\int }_{0}^{T}{\left( \mathbb{E}{X}_{t}\right) }^{2}{dt} \leq {\int }_{0}^{T}\mathbb{E}{X}_{t}^{2}{dt} < \infty .\n\nIn addition, we have\n\n\[ {\int }_{0}^{T}{\int }_{0}^{T}{K}^{2}\left( {s, t}\right) {dsdt} = {\int }_{0}^{T}{\int }_{0}^{T}{\lef...
Yes
Theorem 5.11. Assume that \( {X}_{1},{X}_{2},\ldots \) is a sequence of Gaussian random variables that converges to \( X \) in probability. Then \( X \) is also Gaussian.
Proof. Let us denote\n\n\[ \n{m}_{k} = \mathbb{E}{X}_{k},\;{\sigma }_{k}^{2} = \operatorname{Var}{X}_{k}.\n\]\n\nThen from part \( \left( v\right) \) of the Theorem 1.32 and Lévy’s continuity theorem, we have\n\n\[ \n{e}^{{i\xi }{m}_{k} - \frac{1}{2}{\sigma }_{k}^{2}{\xi }^{2}} = \mathbb{E}{e}^{{i\xi }{X}_{k}} \rightar...
Yes
Theorem 5.13 (Karhunen-Loève expansion). Let \( {\left( {X}_{t}\right) }_{t \in \left\lbrack {0,1}\right\rbrack } \) be a Gaussian process with mean 0 and covariance function \( K\left( {s, t}\right) \) . Assume that \( K \) is continuous. Let \( \left\{ {\lambda }_{k}\right\} ,\left\{ {\phi }_{k}\right\} \) be the seq...
Proof. We need to show that the random series is well-defined and that it is a Gaussian process with the desired mean and covariance function.\n\nFirst consider the operator \( \mathcal{K} : {L}^{2}\left( \left\lbrack {0,1}\right\rbrack \right) \rightarrow {L}^{2}\left( \left\lbrack {0,1}\right\rbrack \right) \) define...
Yes
Example 6.1 (Random walk). Let \( \\left\\{ {\\xi }_{i}\\right\\} \) be i.i.d. random variables such that \( {\\xi }_{i} = \\pm 1 \) with probability \( 1/2 \), and let\n\n\[ \n{X}_{n} = \\mathop{\\sum }\\limits_{{k = 1}}^{n}{\\xi }_{k},\\;\\text{ i.e.,}\\{X}_{0} = 0.\n\]\n\nHere \( \\left\\{ {X}_{n}\\right\\} \) is ca...
The mean position and mean squared deviation are\n\n\[ \n\\mathbb{E}{X}_{N} = 0,\\;\\mathbb{E}{X}_{N}^{2} = N.\n\]\n\nThe root mean squared displacement is \( \\sqrt{N} \) .
Yes
Theorem 6.5. Let \( \\left\\{ {\\alpha }_{k}^{\\left( n\\right) }\\right\\} \) be a sequence of i.i.d. Gaussian random variables with distribution \( N\\left( {0,1}\\right) \) . Then, almost surely,\n\n\[ \n{W}_{t}^{N} = \\mathop{\\sum }\\limits_{{n = 0}}^{N}\\mathop{\\sum }\\limits_{{k \\in {I}_{n}}}{\\alpha }_{k}^{\\...
Proof. First we show that almost surely, \( {W}_{t}^{N} \) converges uniformly to some continuous function. For this purpose, we note that for the Gaussian random variable \( \\xi \\sim N\\left( {0,1}\\right) \),\n\n\[ \n\\mathbb{P}\\left( {\\left| \\xi \\right| > x}\\right) = \\sqrt{\\frac{2}{\\pi }}{\\int }_{x}^{\\in...
Yes
Compute the expectation\n\n\\[ \n\\mathbb{E}\\exp \\left( {-\\frac{1}{2}\\int }_{0}^{1}{W}_{t}^{2}{dt}\\right) \n\\]
Solution. This is an example of a Wiener functional. Using the Karhunen-Loève expansion, we get\n\n\\[ \n\\int }_{0}^{1}{W}_{t}^{2}{dt} = \\int }_{0}^{1}\\mathop{\\sum }\\limits_{{k, l}}\\sqrt{{\\lambda }_{k}{\\lambda }_{l}}{\\alpha }_{k}{\\alpha }_{l}{\\phi }_{k}\\left( t\\right) {\\phi }_{l}\\left( t\\right) {dt}\n\\...
Yes
Theorem 6.7. Wiener process has the following symmetry properties.\n\n(1) Time-homogeneity: For any \( s > 0,{W}_{t + s} - {W}_{s}, t \geq 0 \), is a Wiener process.\n\n(2) Symmetry: The process \( - {W}_{t}, t \geq 0 \), is a Wiener process.\n\n(3) Scaling: For every \( c > 0 \), the process \( c{W}_{t/{c}^{2}}, t \ge...
The proof is straightforward and is left as an exercise.
No
Theorem 6.9 (Unbounded variation of the Wiener path). On any finite interval, the total variation of a Wiener path is almost surely infinite.
Proof. Because of (6.18), there is a subset \( {\Omega }_{0} \subset \Omega \) such that \( \mathbb{P}\left( {\Omega }_{0}\right) = 1 \) , and a subsequence of subdivisions, still denoted as \( \left\{ {\Delta }_{n}\right\} \), such that for any pair of rational numbers \( \left( {p, q}\right), p < q \) ,\n\n\[ \n{Q}_{...
Yes
Theorem 6.10 (Smoothness of the Wiener path). Let \( {\Omega }_{\alpha } \) be the set of functions that are Hölder continuous with exponent \( \alpha \left( {0 < \alpha < 1}\right) \) :\n\n\[ \n{\Omega }_{\alpha } = \left\{ {f \in C\left\lbrack {0,1}\right\rbrack ,\mathop{\sup }\limits_{{0 \leq s, t \leq 1}}\frac{\lef...
The proof of Theorem 6.10 relies on the modification concept and the following Kolmogorov continuity theorem, which can be found in [RY05].
No
Theorem 6.14 (Wiener chaos expansion). Let \( W \) be a standard Wiener process on \( \left( {\Omega ,\mathcal{F},\mathbb{P}}\right) \). Let \( F \) be a Wiener functional in \( {L}^{2}\left( \Omega \right) \). Then \( F \) can be represented as\n\n\[ F\left\lbrack W\right\rbrack = \mathop{\sum }\limits_{{\alpha \in \m...
This is an infinite-dimensional analog of the Fourier expansion with weight function \( {\left( \sqrt{2\pi }\right) }^{-1}{e}^{-{x}^{2}/2} \) in each dimension. Its proof can be found in CM47.
No
Lemma 7.1. Assume that \( 0 \leq S \leq T \) . The stochastic integral for the simple functions satisfies\n\n(7.7)\n\n\[ \mathbb{E}\left( {{\int }_{S}^{T}f\left( {\omega, t}\right) d{W}_{t}}\right) = 0 \]\n\n(7.8)\n\n\[ \text{ (Itô isometry) }\;\mathbb{E}{\left( {\int }_{S}^{T}f\left( t,\omega \right) d{W}_{t}\right) }...
Proof. The first property is straightforward from the independence between \( \delta {W}_{j} \mathrel{\text{:=}} {W}_{{t}_{j + 1}} - {W}_{{t}_{j}},{e}_{j}\left( \omega \right) \), and \( \delta {W}_{j} \sim N\left( {0,{t}_{j + 1} - {t}_{j}}\right) \) . For the second property we have\n\n\[ \mathbb{E}{\left( {\int }_{S}...
Yes
For \( f \in \mathcal{V}\left\lbrack {S, T}\right\rbrack \), the Itô integral satisfies\n\n\[ \mathbb{E}\left( {{\int }_{S}^{T}f\left( {\omega, t}\right) d{W}_{t}}\right) = 0 \]
Proof. Based on Lemma 7.1, we have\n\n\[ \left| {\mathbb{E}\left( {{\int }_{S}^{T}f\left( {\omega, t}\right) d{W}_{t}}\right) }\right| = \left| {\mathbb{E}\left( {{\int }_{S}^{T}f\left( {\omega, t}\right) d{W}_{t} - {\int }_{S}^{T}{\phi }_{n}\left( {\omega, t}\right) d{W}_{t}}\right) }\right| \]\n\n\[ \leq {\left\lbrac...
Yes
For the Itô integral we have\n\n\[ {\int }_{0}^{t}{W}_{s}d{W}_{s} = \frac{{W}_{t}^{2}}{2} - \frac{t}{2}\;\text{ a.s. } \]
Proof. Take the dyadic subdivision with mesh size \( {2}^{-n} \) on \( \left\lbrack {0, T}\right\rbrack \) . From the definition of Itô integral\n\n\[ {\int }_{0}^{t}{W}_{s}d{W}_{s} \approx \mathop{\sum }\limits_{j}{W}_{{t}_{j}}\left( {{W}_{{t}_{j + 1}} - {W}_{{t}_{j}}}\right) = \mathop{\sum }\limits_{j}\frac{2{W}_{{t}...
Yes
Proposition 7.6. Assume that \( f \) is bounded and continuous in \( t \) for \( t \in \) \( \left\lbrack {0, T}\right\rbrack \) almost surely. Then\n\n\[ \mathop{\sum }\limits_{j}f\left( {\omega ,{t}_{j}^{ * }}\right) {\left( {W}_{{t}_{j + 1}} - {W}_{{t}_{j}}\right) }^{2} \rightarrow {\int }_{0}^{T}f\left( {\omega, s}...
Proof. A straightforward calculation gives\n\n\[ \mathbb{E}{\left( \mathop{\sum }\limits_{j}f\left( {t}_{j}\right) \delta {W}_{{t}_{j}}^{2} - \mathop{\sum }\limits_{j}f\left( {t}_{j}\right) \delta {t}_{j}\right) }^{2} \]\n\n\[ = \mathbb{E}\left( {\mathop{\sum }\limits_{{j, k}}f\left( {t}_{j}\right) f\left( {t}_{k}\righ...
Yes
Theorem 7.7 (One-dimensional Itô formula). Let \( f \) be a twice differentiable function, and let \( {Y}_{t} = f\left( {X}_{t}\right) \) where \( {X}_{t} \) is an Itô process defined in (7.16). Then \( {Y}_{t} \) is also an Itô process and\n\n\[ d{Y}_{t} = \left( {b\left( {t,\omega }\right) {f}^{\prime }\left( {X}_{t}...
Remark 7.8. Equation (7.17) can be derived formally using Taylor expansion and the calculation rules:\n\n\[ d{t}^{2} = 0,\;{dtd}{W}_{t} = d{W}_{t}{dt} = 0,\;{\left( d{W}_{t}\right) }^{2} = {dt}. \]\n\nFirst, we have\n\n\[ d{Y}_{t} = {f}^{\prime }\left( {X}_{t}\right) d{X}_{t} + \frac{1}{2}{f}^{\prime \prime }\left( {X}...
Yes
Theorem 7.9 (Multidimensional Itô formula). Let \( {\mathbf{X}}_{t} \) be an Itô process defined by \( d{\mathbf{X}}_{t} = \mathbf{b}\left( {\omega, t}\right) {dt} + \mathbf{\sigma }\left( {\omega, t}\right) d{\mathbf{W}}_{t} \), where \( {\mathbf{X}}_{t} \in {\mathbb{R}}^{n},\mathbf{\sigma } \in {\mathbb{R}}^{n \times...
Remark 7.10. Equation (7.20) can be derived formally using the calculation rules\n\n(7.21)\n\n\[ d{t}^{2} = 0,\;{dtd}{W}_{t}^{i} = d{W}_{t}^{i}{dt} = d{W}_{t}^{i}d{W}_{t}^{j} = 0\;\left( {i \neq j}\right) ,\;{\left( d{W}_{t}^{i}\right) }^{2} = {dt}. \]\n\nUsing Taylor expansion, we have\n\n(7.22)\n\n\[ d{Y}_{t} = \nabl...
Yes
An example of integration by parts is\n\n\[ \n{\int }_{0}^{t}{sd}{W}_{s} = t{W}_{t} - {\int }_{0}^{t}{W}_{s}{ds} \n\]
Proof. Define \( f\left( {x, y}\right) = {xy},{X}_{t} = t,{Y}_{t} = {W}_{t} \) . Then from the multidimensional Itô formula\n\n\[ \n{df}\left( {{X}_{t},{Y}_{t}}\right) = {X}_{t}d{Y}_{t} + {Y}_{t}d{X}_{t} + d{X}_{t}d{Y}_{t} \n\]\n\nSince \( {dtd}{W}_{t} = 0 \), we obtain \( d\left( {t{W}_{t}}\right) = {td}{W}_{t} + {W}_...
Yes
\[ {\int }_{0}^{t}d{W}_{{t}_{1}}{\int }_{0}^{{t}_{1}}d{W}_{{t}_{2}}\ldots {\int }_{0}^{{t}_{n - 1}}d{W}_{{t}_{n}} = \frac{1}{n!}{t}^{\frac{n}{2}}{H}_{n}\left( \frac{{W}_{t}}{\sqrt{t}}\right) ,\]
Proof. It is easy to verify that \[ {\int }_{0}^{t}{W}_{s}d{W}_{s} = \frac{t}{2!}{H}_{2}\left( \frac{{W}_{t}}{\sqrt{t}}\right) \] where \( {H}_{2}\left( x\right) = {x}^{2} - 1 \) is the second-order Hermite polynomial. In the same fashion, we have \[ {\int }_{0}^{t}\left( {{\int }_{0}^{s}{W}_{u}d{W}_{u}}\right) d{W}_{s...
No
Theorem 7.13 (Burkholder-Davis-Gundy inequality). For any \( m > 0 \) , there exist constants \( {k}_{m},{K}_{m} > 0 \) such that\n\n\[ \n{k}_{m}\mathbb{E}\left( {Q}_{T}^{m}\right) \leq \mathbb{E}\left( {\mathop{\sup }\limits_{{0 \leq t \leq T}}{\left| {\mathbf{M}}_{t}\right| }^{2m}}\right) \leq {K}_{m}\mathbb{E}\left(...
See [KS91] for detailed proof.
No
Theorem 7.14. Assume that the coefficients \( \mathbf{b} \in {\mathbb{R}}^{n},\mathbf{\sigma } \in {\mathbb{R}}^{n \times m} \) satisfy the global Lipschitz and linear growth conditions:\n\n(7.28)\n\n\[ \left| {\mathbf{b}\left( {\mathbf{x}, t}\right) - \mathbf{b}\left( {\mathbf{y}, t}\right) }\right| + \left| {\mathbf{...
Proof. We will only consider the one-dimensional case. The high-dimensional case is similar. First we prove uniqueness. Let \( {X}_{t},{\widehat{X}}_{t} \in \mathcal{V}\left\lbrack {0, T}\right\rbrack \) be solutions of the SDEs (7.27) with the same initial value \( {X}_{0} \) . Then\n\n\[ {X}_{t} - {\widehat{X}}_{t} =...
Yes
The Ornstein-Uhlenbeck (OU) process is the solution of the following linear SDE with the additive noise:\n\n\[ d{X}_{t} = - \gamma {X}_{t}{dt} + {\sigma d}{W}_{t},{\left. \;{X}_{t}\right| }_{t = 0} = {X}_{0}. \]
Solution. Multiplying both sides of (7.38) by \( {e}^{\gamma t} \) and integrating, we get\n\n\[ {e}^{\gamma t}{X}_{t} - {X}_{0} = {\int }_{0}^{t}\sigma {e}^{\gamma s}d{W}_{s} \]\n\nThus\n\n\[ {X}_{t} = {e}^{-{\gamma t}}{X}_{0} + \sigma {\int }_{0}^{t}{e}^{-\gamma \left( {t - s}\right) }d{W}_{s} \]\n\nis the solution. ...
Yes
The geometric Brownian motion is the solution of a linear SDE with multiplicative noise\n\n\[ d{N}_{t} = r{N}_{t}{dt} + \alpha {N}_{t}d{W}_{t},{\left. \;{N}_{t}\right| }_{t = 0} = {N}_{0}. \]
Solution. Dividing both sides by \( {N}_{t} \), we have \( d{N}_{t}/{N}_{t} = {rdt} + {\alpha d}{W}_{t} \) . Applying Itô’s formula to \( \log {N}_{t} \), we get\n\n\[ d\left( {\log {N}_{t}}\right) = \frac{1}{{N}_{t}}d{N}_{t} - \frac{1}{2{N}_{t}^{2}}{\left( d{N}_{t}\right) }^{2} \]\n\n\[ = \frac{1}{{N}_{t}}d{N}_{t} - \...
Yes