# Almost Sure

## 2 January 19

### Proof of Measurable Section

I will give a proof of the measurable section theorem, also known as measurable selection. Given a complete probability space ${(\Omega,\mathcal F,{\mathbb P})}$, we denote the projection from ${\Omega\times{\mathbb R}}$ by

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\pi_\Omega\colon \Omega\times{\mathbb R}\rightarrow\Omega,\smallskip\\ &\displaystyle\pi_\Omega(\omega,t)=\omega. \end{array}$

By definition, if ${S\subseteq\Omega\times{\mathbb R}}$ then, for every ${\omega\in\pi_\Omega(S)}$, there exists a ${t\in{\mathbb R}}$ such that ${(\omega,t)\in S}$. The measurable section theorem says that this choice can be made in a measurable way. That is, using ${\mathcal B({\mathbb R})}$ to denote the Borel sigma-algebra, if S is in the product sigma-algebra ${\mathcal F\otimes\mathcal B({\mathbb R})}$ then ${\pi_\Omega(S)\in\mathcal F}$ and there is a measurable map

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle\tau\colon\pi_\Omega(S)\rightarrow{\mathbb R},\smallskip\\ &\displaystyle(\omega,\tau(\omega))\in S. \end{array}$

It is convenient to extend ${\tau}$ to the whole of ${\Omega}$ by setting ${\tau=\infty}$ outside of ${\pi_\Omega(S)}$.

Figure 1: A section of a measurable set

We consider measurable functions ${\tau\colon\Omega\rightarrow{\mathbb R}\cup\{\infty\}}$. The graph of ${\tau}$ is

$\displaystyle [\tau]=\left\{(\omega,\tau(\omega))\colon\tau(\omega)\in{\mathbb R}\right\}\subseteq\Omega\times{\mathbb R}.$

The condition that ${(\omega,\tau(\omega))\in S}$ whenever ${\tau < \infty}$ can then be expressed by stating that ${[\tau]\subseteq S}$. This also ensures that ${\{\tau < \infty\}}$ is a subset of ${\pi_\Omega(S)}$, and ${\tau}$ is a section of S on the whole of ${\pi_\Omega(S)}$ if and only if ${\{\tau < \infty\}=\pi_\Omega(S)}$.

The proof of the measurable section theorem will make use of the properties of analytic sets and of the Choquet capacitability theorem, as described in the previous two posts. [Note: I have since posted a more direct proof which does not involve such prerequisites.] Recall that a paving ${\mathcal E}$ on a set X denotes, simply, a collection of subsets of X. The pair ${(X,\mathcal E)}$ is then referred to as a paved space. Given a pair of paved spaces ${(X,\mathcal E)}$ and ${(Y,\mathcal F)}$, the product paving ${\mathcal E\times\mathcal F}$ denotes the collection of cartesian products ${A\times B}$ for ${A\in\mathcal E}$ and ${B\in\mathcal F}$, which is a paving on ${X\times Y}$. The notation ${\mathcal E_\delta}$ is used for the collection of countable intersections of a paving ${\mathcal E}$.

We start by showing that measurable section holds in a very simple case where, for the section of a set S, its debut will suffice. The debut is the map

$\displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle D(S)\colon\Omega\rightarrow{\mathbb R}\cup\{\pm\infty\},\smallskip\\ &\displaystyle \omega\mapsto\inf\left\{t\in{\mathbb R}\colon (\omega,t)\in S\right\}. \end{array}$

We use the convention that the infimum of the empty set is ${\infty}$. It is not clear that ${D(S)}$ is measurable, and we do not rely on this, although measurable projection can be used to show that it is measurable whenever S is in ${\mathcal F\otimes\mathcal B({\mathbb R})}$.

Lemma 1 Let ${(\Omega,\mathcal F)}$ be a measurable space, ${\mathcal K}$ be the collection of compact intervals in ${{\mathbb R}}$, and ${\mathcal E}$ be the closure of the paving ${\mathcal{F\times K}}$ under finite unions.

Then, the debut ${D(S)}$ of any ${S\in\mathcal E_\delta}$ is measurable and its graph ${[D(S)]}$ is contained in S.

## 1 January 19

### Choquet’s Capacitability Theorem and Measurable Projection

In this post I will give a proof of the measurable projection theorem. Recall that this states that for a complete probability space ${(\Omega,\mathcal F,{\mathbb P})}$ and a set S in the product sigma-algebra ${\mathcal F\otimes\mathcal B({\mathbb R})}$, the projection, ${\pi_\Omega(S)}$, of S onto ${\Omega}$, is in ${\mathcal F}$. The previous post on analytic sets made some progress towards this result. Indeed, using the definitions and results given there, it follows quickly that ${\pi_\Omega(S)}$ is ${\mathcal F}$-analytic. To complete the proof of measurable projection, it is necessary to show that analytic sets are measurable. This is a consequence of Choquet’s capacitability theorem, which I will prove in this post. Measurable projection follows as a simple consequence.

The condition that the underlying probability space is complete is necessary and, if this condition was dropped, then the result would no longer hold. Recall that, if ${(\Omega,\mathcal F,{\mathbb P})}$ is a probability space, then the completion, ${\mathcal F_{\mathbb P}}$, of ${\mathcal F}$ with respect to ${{\mathbb P}}$ consists of the sets ${A\subseteq\Omega}$ such that there exists ${B,C\in\mathcal F}$ with ${B\subseteq A\subseteq C}$ and ${{\mathbb P}(B)={\mathbb P}(C)}$. The probability space is complete if ${\mathcal F_{\mathbb P}=\mathcal F}$. More generally, ${{\mathbb P}}$ can be uniquely extended to a measure ${\bar{\mathbb P}}$ on the sigma-algebra ${\mathcal F_{\mathbb P}}$ by setting ${\bar{\mathbb P}(A)={\mathbb P}(B)={\mathbb P}(C)}$, where B and C are as above. Then ${(\Omega,\mathcal F_{\mathbb P},\bar{\mathbb P})}$ is the completion of ${(\Omega,\mathcal F,{\mathbb P})}$.

In measurable projection, then, it needs to be shown that if ${A\subseteq\Omega}$ is the projection of a set in ${\mathcal F\otimes\mathcal B({\mathbb R})}$, then A is in the completion of ${\mathcal F}$. That is, we need to find sets ${B,C\in\mathcal F}$ with ${B\subseteq A\subseteq C}$ with ${{\mathbb P}(B)={\mathbb P}(C)}$. In fact, it is always possible to find a ${C\supseteq A}$ in ${\mathcal F}$ which minimises ${{\mathbb P}(C)}$, and its measure is referred to as the outer measure of A. For any probability measure ${{\mathbb P}}$, we can define an outer measure on the subsets of ${\Omega}$, ${{\mathbb P}^*\colon\mathcal P(\Omega)\rightarrow{\mathbb R}^+}$ by approximating ${A\subseteq\Omega}$ from above,

 $\displaystyle {\mathbb P}^*(A)\equiv\inf\left\{{\mathbb P}(B)\colon B\in\mathcal F, A\subseteq B\right\}.$ (1)

Similarly, we can define an inner measure by approximating A from below,

$\displaystyle {\mathbb P}_*(A)\equiv\sup\left\{{\mathbb P}(B)\colon B\in\mathcal F, B\subseteq A\right\}.$

It can be shown that A is ${\mathcal F}$-measurable if and only if ${{\mathbb P}_*(A)={\mathbb P}^*(A)}$. We will be concerned primarily with the outer measure ${{\mathbb P}^*}$, and will show that that if A is the projection of some ${S\in\mathcal F\otimes\mathcal B({\mathbb R})}$, then A can be approximated from below in the following sense: there exists ${B\subseteq A}$ in ${\mathcal F}$ for which ${{\mathbb P}^*(B)={\mathbb P}^*(A)}$. From this, it will follow that A is in the completion of ${\mathcal F}$.

It is convenient to prove the capacitability theorem in slightly greater generality than just for the outer measure ${{\mathbb P}^*}$. The only properties of ${{\mathbb P}^*}$ that are required is that it is a capacity, which we now define. Recall that a paving ${\mathcal E}$ on a set X is simply any collection of subsets of X, and we refer to the pair ${(X,\mathcal E)}$ as a paved space.

Definition 1 Let ${(X,\mathcal E)}$ be a paved space. Then, an ${\mathcal E}$-capacity is a map ${I\colon\mathcal P(X)\rightarrow{\mathbb R}}$ which is increasing, continuous along increasing sequences, and continuous along decreasing sequences in ${\mathcal E}$. That is,

• if ${A\subseteq B}$ then ${I(A)\le I(B)}$.
• if ${A_n\subseteq X}$ is increasing in n then ${I(A_n)\rightarrow I(\bigcup_nA_n)}$ as ${n\rightarrow\infty}$.
• if ${A_n\in\mathcal E}$ is decreasing in n then ${I(A_n)\rightarrow I(\bigcap_nA_n)}$ as ${n\rightarrow\infty}$.

As was claimed above, the outer measure ${{\mathbb P}^*}$ defined by (1) is indeed a capacity.

Lemma 2 Let ${(\Omega,\mathcal F,{\mathbb P})}$ be a probability space. Then,

• ${{\mathbb P}^*(A)={\mathbb P}(A)}$ for all ${A\in\mathcal F}$.
• For all ${A\subseteq\Omega}$, there exists a ${B\in\mathcal F}$ with ${A\subseteq B}$ and ${{\mathbb P}^*(A)={\mathbb P}(B)}$.
• ${{\mathbb P}^*}$ is an ${\mathcal F}$-capacity.

## 24 December 18

### Analytic Sets

We will shortly give a proof of measurable projection and, also, of the section theorems. Starting with the projection theorem, recall that this states that if ${(\Omega,\mathcal F,{\mathbb P})}$ is a complete probability space, then the projection of any measurable subset of ${\Omega\times{\mathbb R}}$ onto ${\Omega}$ is measurable. To be precise, the condition is that S is in the product sigma-algebra ${\mathcal{F}\otimes\mathcal B({\mathbb R})}$, where ${\mathcal B({\mathbb R})}$ denotes the Borel sets in ${{\mathbb R}}$, and ${\pi\colon\Omega\times{\mathbb R}\rightarrow\Omega}$ is the projection ${\pi(\omega,t)=\omega}$. Then, ${\pi(S)\in\mathcal{F}}$. Although it looks like a very basic property of measurable sets, maybe even obvious, measurable projection is a surprisingly difficult result to prove. In fact, the requirement that the probability space is complete is necessary and, if it is dropped, then ${\pi(S)}$ need not be measurable. Counterexamples exist for commonly used measurable spaces such as ${\Omega= {\mathbb R}}$ and ${\mathcal F=\mathcal B({\mathbb R})}$. This suggests that there is something deeper going on here than basic manipulations of measurable sets.

The techniques which will be used to prove the projection theorem involve analytic sets, which will be introduced in this post, with the proof of measurable projection to follow in the next post. [Note: I have since posted a more direct proof of measurable projection and section, which does not make use of analytic sets.] These results can also be used to prove the optional and predictable section theorems which, at first appearances, seem to be quite basic statements. The section theorems are fundamental to the powerful and interesting theory of optional and predictable projection which is, consequently, generally considered to be a hard part of stochastic calculus. In fact, the projection and section theorems are really not that hard to prove, although the method given here does require stepping outside of the usual setup used in probability and involves something more like descriptive set theory. (more…)

## 12 October 16

### Do Convex and Decreasing Functions Preserve the Semimartingale Property — A Possible Counterexample

Figure 1: The function f, convex in x and decreasing in t

Here, I attempt to construct a counterexample to the hypotheses of the earlier post, Do convex and decreasing functions preserve the semimartingale property? There, it was asked, for any semimartingale X and function ${f\colon{\mathbb R}_+\times{\mathbb R}\rightarrow{\mathbb R}}$ such that ${f(t,x)}$ is convex in x and right-continuous and decreasing in t, is ${f(t,X_t)}$ necessarily a semimartingale? It was explained how this is equivalent to the hypothesis: for any function ${f\colon[0,1]^2\rightarrow{\mathbb R}}$ such that ${f(t,x)}$ is convex and Lipschitz continuous in x and decreasing in t, does it decompose as ${f=g-h}$ where ${g(t,x)}$ and ${h(t,x)}$ are convex in x and increasing in t. This is the form of the hypothesis which this post will be concerned with, so the example will only involve simple real analysis and no stochastic calculus. I will give some numerical calculations suggesting that the construction below is a counterexample, but do not have any proof of this. So, the hypothesis is still open.

Although the construction given here will be self-contained, it is worth noting that it is connected to the example of a martingale which moves along a deterministic path. If ${\{M_t\}_{t\in[0,1]}}$ is the martingale constructed there, then

$\displaystyle C(t,x)={\mathbb E}[(M_t-x)_+]$

defines a function from ${[0,1]\times[-1,1]}$ to ${{\mathbb R}}$ which is convex in x and increasing in t. The question is then whether C can be expressed as the difference of functions which are convex in x and decreasing in t. The example constructed in this post will be the same as C with the time direction reversed, and with a linear function of x added so that it is zero at ${x=\pm1}$. (more…)

## 5 October 16

### A Martingale Which Moves Along a Deterministic Path

Figure 1: Sample paths

In this post I will construct a continuous and non-constant martingale M which only varies on the path of a deterministic function ${f\colon{\mathbb R}_+\rightarrow{\mathbb R}}$. That is, ${M_t=f(t)}$ at all times outside of the set of nontrivial intervals on which M is constant. Expressed in terms of the stochastic integral, ${dM_t=0}$ on the set ${\{t\colon M_t\not=f(t)\}}$ and,

 $\displaystyle M_t = \int_0^t 1_{\{M_s=f(s)\}}\,dM_s.$ (1)

In the example given here, f will be right-continuous. Examples with continuous f do exist, although the constructions I know of are considerably more complicated. At first sight, these properties appear to contradict what we know about continuous martingales. They vary unpredictably, behaving completely unlike any deterministic function. It is certainly the case that we cannot have ${M_t=f(t)}$ across any interval on which M is not constant.

By a stochastic time-change, any Brownian motion B can be transformed to have the same distribution as M. This means that there exists an increasing and right-continuous process A adapted to the same filtration as B and such that ${B_t=M_{A_t}}$ where M is a martingale as above. From this, we can infer that

$\displaystyle B_t=f(A_t),$

expressing Brownian motion as a function of an increasing process. (more…)

## 26 September 16

### Do Convex and Decreasing Functions Preserve the Semimartingale Property?

Some years ago, I spent considerable effort trying to prove the hypothesis below. After failing at this, I spent time trying to find a counterexample, but also with no success. I did post this as a question on mathoverflow, but it has so far received no conclusive answers. So, as far as I am aware, the following statement remains unproven either way.

Hypothesis H1 Let ${f\colon{\mathbb R}_+\times{\mathbb R}\rightarrow{\mathbb R}}$ be such that ${f(t,x)}$ is convex in x and right-continuous and decreasing in t. Then, for any semimartingale X, ${f(t,X_t)}$ is a semimartingale.

It is well known that convex functions of semimartingales are themselves semimartingales. See, for example, the Ito-Tanaka formula. More generally, if ${f(t,x)}$ was increasing in t rather than decreasing, then it can be shown without much difficulty that ${f(t,X_t)}$ is a semimartingale. Consider decomposing ${f(t,X_t)}$ as

 $\displaystyle f(t,X_t)=\int_0^tf_x(s,X_{s-})\,dX_s+V_t,$ (1)

for some process V. By convexity, the right hand derivative of ${f(t,x)}$ with respect to x always exists, and I am denoting this by ${f_x}$. In the case where f is twice continuously differentiable then the process V is given by Ito’s formula which, in particular, shows that it is a finite variation process. If ${f(t,x)}$ is convex in x and increasing in t, then the terms in Ito’s formula for V are all increasing and, so, it is an increasing process. By taking limits of smooth functions, it follows that V is increasing even when the differentiability constraints are dropped, so ${f(t,X_t)}$ is a semimartingale. Now, returning to the case where ${f(t,x)}$ is decreasing in t, Ito’s formula is only able to say that V is of finite variation, and is generally not monotonic. As limits of finite variation processes need not be of finite variation themselves, this does not say anything about the case when f is not assumed to be differentiable, and does not help us to determine whether or not ${f(t,X_t)}$ is a semimartingale.

Hypothesis H1 can be weakened by restricting to continuous functions of continuous martingales.

Hypothesis H2 Let ${f\colon{\mathbb R}_+\times{\mathbb R}\rightarrow{\mathbb R}}$ be such that ${f(t,x)}$ is convex in x and continuous and decreasing in t. Then, for any continuous martingale X, ${f(t,X_t)}$ is a semimartingale.

As continuous martingales are special cases of semimartingales, hypothesis H1 implies H2. In fact, the reverse implication also holds so that hypotheses H1 and H2 are equivalent.

Hypotheses H1 and H2 can also be recast as a simple real analysis statement which makes no reference to stochastic processes.

Hypothesis H3 Let ${f\colon{\mathbb R}_+\times{\mathbb R}\rightarrow{\mathbb R}}$ be such that ${f(t,x)}$ is convex in x and decreasing in t. Then, ${f=g-h}$ where ${g(t,x)}$ and ${h(t,x)}$ are convex in x and increasing in t.

## 14 September 16

### Failure of the Martingale Property For Stochastic Integration

If X is a cadlag martingale and ${\xi}$ is a uniformly bounded predictable process, then is the integral

 $\displaystyle Y=\int\xi\,dX$ (1)

a martingale? If ${\xi}$ is elementary this is one of most basic properties of martingales. If X is a square integrable martingale, then so is Y. More generally, if X is an ${L^p}$-integrable martingale, any ${p > 1}$, then so is Y. Furthermore, integrability of the maximum ${\sup_{s\le t}\lvert X_s\rvert}$ is enough to guarantee that Y is a martingale. Also, it is a fundamental result of stochastic integration that Y is at least a local martingale and, for this to be true, it is only necessary for X to be a local martingale and ${\xi}$ to be locally bounded. In the general situation for cadlag martingales X and bounded predictable ${\xi}$, it need not be the case that Y is a martingale. In this post I will construct an example showing that Y can fail to be a martingale. (more…)

## 12 September 16

### Martingales with Non-Integrable Maximum

Filed under: Examples and Counterexamples — George Lowther @ 12:01 PM
Tags: , ,

It is a consequence of Doob’s maximal inequality that any ${L^p}$-integrable martingale has a maximum, up to a finite time, which is also ${L^p}$-integrable for any ${p > 1}$. Using ${X^*_t\equiv\sup_{s\le t}\lvert X_s\rvert}$ to denote the running absolute maximum of a cadlag martingale X, then ${X^*}$ is ${L^p}$-integrable whenever ${X}$ is. It is natural to ask whether this also holds for ${p=1}$. As martingales are integrable by definition, this is just asking whether cadlag martingales necessarily have an integrable maximum. Integrability of the maximum process does have some important consequences in the theory of martingales. By the Burkholder-Davis-Gundy inequality, it is equivalent to the square-root of the quadratic variation, ${[X]^{1/2}}$, being integrable. Stochastic integration over bounded integrands preserves the martingale property, so long as the martingale has integrable maximal process. The continuous and purely discontinuous parts of a martingale X are themselves local martingales, but are not guaranteed to be proper martingales unless X has integrable maximum process.

The aim of this post is to show, by means of some examples, that a cadlag martingale need not have an integrable maximum. (more…)

## 11 September 16

### The Optimality of Doob’s Maximal Inequality

One of the most fundamental and useful results in the theory of martingales is Doob’s maximal inequality. Use ${X^*_t\equiv\sup_{s\le t}\lvert X_s\rvert}$ to denote the running (absolute) maximum of a process X. Then, Doob’s ${L^p}$ maximal inequality states that, for any cadlag martingale or nonnegative submartingale X and real ${p > 1}$,

 $\displaystyle \lVert X^*_t\rVert_p\le c_p \lVert X_t\rVert_p$ (1)

with ${c_p=p/(p-1)}$. Here, ${\lVert\cdot\rVert_p}$ denotes the standard Lp-norm, ${\lVert U\rVert_p\equiv{\mathbb E}[U^p]^{1/p}}$.

An obvious question to ask is whether it is possible to do any better. That is, can the constant ${c_p}$ in (1) be replaced by a smaller number. This is especially pertinent in the case of small p, since ${c_p}$ diverges to infinity as p approaches 1. The purpose of this post is to show, by means of an example, that the answer is no. The constant ${c_p}$ in Doob’s inequality is optimal. We will construct an example as follows.

Example 1 For any ${p > 1}$ and constant ${1 \le c < c_p}$ there exists a strictly positive cadlag ${L^p}$-integrable martingale ${\{X_t\}_{t\in[0,1]}}$ with ${X^*_1=cX_1}$.

For X as in the example, we have ${\lVert X^*_1\rVert_p=c\lVert X_1\rVert_p}$. So, supposing that (1) holds with any other constant ${\tilde c_p}$ in place of ${c_p}$, we must have ${\tilde c_p\ge c}$. By choosing ${c}$ as close to ${c_p}$ as we like, this means that ${\tilde c_p\ge c_p}$ and ${c_p}$ is indeed optimal in (1). (more…)

## 6 September 16

### The Maximum Maximum of Martingales with Known Terminal Distribution

In this post I will be concerned with the following problem — given a martingale X for which we know the distribution at a fixed time, and we are given nothing else, what is the best bound we can obtain for the maximum of X up until that time? This is a question with a long history, starting with Doob’s inequalities which bound the maximum in the ${L^p}$ norms and in probability. Later, Blackwell and Dubins (3), Dubins and Gilat (5) and Azema and Yor (1,2) showed that the maximum is bounded above, in stochastic order, by the Hardy-Littlewood transform of the terminal distribution. Furthermore, this bound is the best possible in the sense that there do exists martingales for which it can be attained, for any permissible terminal distribution. Hobson (7,8) considered the case where the starting law is also known, and this was further generalized to the case with a specified distribution at an intermediate time by Brown, Hobson and Rogers (4). Finally, Henry-Labordère, Obłój, Spoida and Touzi (6) considered the case where the distribution of the martingale is specified at an arbitrary set of times. In this post, I will look at the case where only the terminal distribution is specified. This leads to interesting constructions of martingales and, in particular, of continuous martingales with specified terminal distributions, with close connections to the Skorokhod embedding problem.

I will be concerned with the maximum process of a cadlag martingale X,

$\displaystyle X^*_t=\sup_{s\le t}X_s,$

which is increasing and adapted. We can state and prove the bound on ${X^*}$ relatively easily, although showing that it is optimal is more difficult. As the result holds more generally for submartingales, I state it in this case, although I am more concerned with martingales here.

Theorem 1 If X is a cadlag submartingale then, for each ${t\ge0}$ and ${x\in{\mathbb R}}$,

 $\displaystyle {\mathbb P}\left(X^*_t\ge x\right)\le\inf_{y < x}\frac{{\mathbb E}\left[(X_t-y)_+\right]}{x-y}.$ (1)

Proof: We just need to show that the inequality holds for each ${y < x}$, and then it immediately follows for the infimum. Choosing ${y < x^\prime < x}$, consider the stopping time

$\displaystyle \tau=\inf\{s\ge0\colon X_s\ge x^\prime\}.$

Then, ${\tau \le t}$ and ${X_\tau\ge x^\prime}$ whenever ${X^*_t \ge x}$. As ${f(z)\equiv(z-y)_+}$ is nonnegative and increasing in z, this means that ${1_{\{X^*_t\ge x\}}}$ is bounded above by ${f(X_{\tau\wedge t})/f(x^\prime)}$. Taking expectations,

$\displaystyle {\mathbb P}\left(X^*_t\ge x\right)\le{\mathbb E}\left[f(X_{\tau\wedge t})\right]/f(x^\prime).$

Since f is convex and increasing, ${f(X)}$ is a submartingale so, using optional sampling,

$\displaystyle {\mathbb P}\left(X^*_t\ge x\right)\le{\mathbb E}\left[f(X_t)\right]/f(x^\prime).$

Letting ${x^\prime}$ increase to ${x}$ gives the result. ⬜

The bound stated in Theorem 1 is also optimal, and can be achieved by a continuous martingale. In this post, all measures on ${{\mathbb R}}$ are defined with respect to the Borel sigma-algebra.

Theorem 2 If ${\mu}$ is a probability measure on ${{\mathbb R}}$ with ${\int\lvert x\rvert\,d\mu(x) < \infty}$ and ${t > 0}$ then there exists a continuous martingale X (defined on some filtered probability space) such that ${X_t}$ has distribution ${\mu}$ and (1) is an equality for all ${x\in{\mathbb R}}$.

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