# Criteria for Poisson Point Processes

If S is a finite random set in a standard Borel measurable space ${(E,\mathcal E)}$ satisfying the following two properties,

• if ${A,B\in\mathcal E}$ are disjoint, then the sizes of ${S\cap A}$ and ${S\cap B}$ are independent random variables,
• ${{\mathbb P}(x\in S)=0}$ for each ${x\in E}$,

then it is a Poisson point process. That is, the size of ${S\cap A}$ is a Poisson random variable for each ${A\in\mathcal E}$. This justifies the use of Poisson point processes in many different areas of probability and stochastic calculus, and provides a convenient method of showing that point processes are indeed Poisson. If the theorem applies, so that we have a Poisson point process, then we just need to compute the intensity measure to fully determine its distribution. The result above was mentioned in the previous post, but I give a precise statement and proof here. Continue reading “Criteria for Poisson Point Processes”

# Poisson Point Processes

The Poisson distribution models numbers of events that occur in a specific period of time given that, at each instant, whether an event occurs or not is independent of what happens at all other times. Examples which are sometimes cited as candidates for the Poisson distribution include the number of phone calls handled by a telephone exchange on a given day, the number of decays of a radio-active material, and the number of bombs landing in a given area during the London Blitz of 1940-41. The Poisson process counts events which occur according to such distributions.

More generally, the events under consideration need not just happen at specific times, but also at specific locations in a space E. Here, E can represent an actual geometric space in which the events occur, such as the spacial distribution of bombs dropped during the Blitz shown in figure 1, but can also represent other quantities associated with the events. In this example, E could represent the 2-dimensional map of London, or could include both space and time so that ${E=F\times{\mathbb R}}$ where, now, F represents the 2-dimensional map and E is used to record both time and location of the bombs. A Poisson point process is a random set of points in E, such that the number that lie within any measurable subset is Poisson distributed. The aim of this post is to introduce Poisson point processes together with the mathematical machinery to handle such random sets.

The choice of distribution is not arbitrary. Rather, it is a result of the independence of the number of events in each region of the space which leads to the Poisson measure, much like the central limit theorem leads to the ubiquity of the normal distribution for continuous random variables and of Brownian motion for continuous stochastic processes. A random finite subset S of a reasonably ‘nice’ (standard Borel) space E is a Poisson point process so long as it satisfies the properties,

• If ${A_1,\ldots,A_n}$ are pairwise-disjoint measurable subsets of E, then the sizes of ${S\cap A_1,\ldots,S\cap A_n}$ are independent.
• Individual points of the space each have zero probability of being in S. That is, ${{\mathbb P}(x\in S)=0}$ for each ${x\in E}$.

The proof of this important result will be given in a later post.

We have come across Poisson point processes previously in my stochastic calculus notes. Specifically, suppose that X is a cadlag ${{\mathbb R}^d}$-valued stochastic process with independent increments, and which is continuous in probability. Then, the set of points ${(t,\Delta X_t)}$ over times t for which the jump ${\Delta X}$ is nonzero gives a Poisson point process on ${{\mathbb R}_+\times{\mathbb R}^d}$. See lemma 4 of the post on processes with independent increments, which corresponds precisely to definition 5 given below. Continue reading “Poisson Point Processes”

# Quantum Coin Tossing

Let me ask the following very simple question. Suppose that I toss a pair of identical coins at the same time, then what is the probability of them both coming up heads? There is no catch here, both coins are fair. There are three possible outcomes, both tails, one head and one tail, and both heads. Assuming that it is completely random so that all outcomes are equally likely, then we could argue that each possibility has a one in three chance of occurring, so that the answer to the question is that the probability is 1/3.

Of course, this is wrong! A fair coin has a probability of 1/2 of showing heads and, by independence, standard probability theory says that we should multiply these together for each coin to get the correct answer of ${\frac12\times\frac12=\frac14}$, which can be verified by experiment. Alternatively, we can note that the outcome of one tail and one head, in reality, consists of two equally likely possibilities. Either the first coin can be a head and the second a tail, or vice-versa. So, there are actually four equally likely possible outcomes, only one of which has both coins showing heads, again giving a probability of 1/4. Continue reading “Quantum Coin Tossing”

# Local Time Continuity

The local time of a semimartingale at a level x is a continuous increasing process, giving a measure of the amount of time that the process spends at the given level. As the definition involves stochastic integrals, it was only defined up to probability one. This can cause issues if we want to simultaneously consider local times at all levels. As x can be any real number, it can take uncountably many values and, as a union of uncountably many zero probability sets can have positive measure or, even, be unmeasurable, this is not sufficient to determine the entire local time ‘surface’

 $\displaystyle (t,x)\mapsto L^x_t(\omega)$

for almost all ${\omega\in\Omega}$. This is the common issue of choosing good versions of processes. In this case, we already have a continuous version in the time index but, as yet, have not constructed a good version jointly in the time and level. This issue arose in the post on the Ito–Tanaka–Meyer formula, for which we needed to choose a version which is jointly measurable. Although that was sufficient there, joint measurability is still not enough to uniquely determine the full set of local times, up to probability one. The ideal situation is when a version exists which is jointly continuous in both time and level, in which case we should work with this choice. This is always possible for continuous local martingales.

Theorem 1 Let X be a continuous local martingale. Then, the local times

 $\displaystyle (t,x)\mapsto L^x_t$

have a modification which is jointly continuous in x and t. Furthermore, this is almost surely ${\gamma}$-Hölder continuous w.r.t. x, for all ${\gamma < 1/2}$ and over all bounded regions for t.

# The Kolmogorov Continuity Theorem

One of the common themes throughout the theory of continuous-time stochastic processes, is the importance of choosing good versions of processes. Specifying the finite distributions of a process is not sufficient to determine its sample paths so, if a continuous modification exists, then it makes sense to work with that. A relatively straightforward criterion ensuring the existence of a continuous version is provided by Kolmogorov’s continuity theorem.

For any positive real number ${\gamma}$, a map ${f\colon E\rightarrow F}$ between metric spaces E and F is said to be ${\gamma}$-Hölder continuous if there exists a positive constant C satisfying

 $\displaystyle d(f(x),f(y))\le Cd(x,y)^\gamma$

for all ${x,y\in E}$. The smallest value of C satisfying this inequality is known as the ${\gamma}$-Hölder coefficient of ${f}$. Hölder continuous functions are always continuous and, at least on bounded spaces, is a stronger property for larger values of the coefficient ${\gamma}$. So, if E is a bounded metric space and ${\alpha\le\beta}$, then every ${\beta}$-Hölder continuous map from E is also ${\alpha}$-Hölder continuous. In particular, 1-Hölder and Lipschitz continuity are equivalent.

Kolmogorov’s theorem gives simple conditions on the pairwise distributions of a process which guarantee the existence of a continuous modification but, also, states that the sample paths ${t\mapsto X_t}$ are almost surely locally Hölder continuous. That is, they are almost surely Hölder continuous on every bounded interval. To start with, we look at real-valued processes. Throughout this post, we work with repect to a probability space ${(\Omega,\mathcal F, {\mathbb P})}$. There is no need to assume the existence of any filtration, since they play no part in the results here

Theorem 1 (Kolmogorov) Let ${\{X_t\}_{t\ge0}}$ be a real-valued stochastic process such that there exists positive constants ${\alpha,\beta,C}$ satisfying

 $\displaystyle {\mathbb E}\left[\lvert X_t-X_s\rvert^\alpha\right]\le C\lvert t-s\vert^{1+\beta},$

for all ${s,t\ge0}$. Then, X has a continuous modification which, with probability one, is locally ${\gamma}$-Hölder continuous for all ${0 < \gamma < \beta/\alpha}$.

# Quantum Entanglement States

In an earlier post, I described four simple thought experiments, involving some black boxes and two or more participants. As described there, the results of these experiments were inconsistent with any classical description, assuming that the boxes cannot communicate. However, I also stated that all of these experiments are consistent with quantum probability, and that I would give the mathematical details in a further post. I will do this now. Continue reading “Quantum Entanglement States”

# The Ito-Tanaka-Meyer Formula

Ito’s lemma is one of the most important and useful results in the theory of stochastic calculus. This is a stochastic generalization of the chain rule, or change of variables formula, and differs from the classical deterministic formulas by the presence of a quadratic variation term. One drawback which can limit the applicability of Ito’s lemma in some situations, is that it only applies for twice continuously differentiable functions. However, the quadratic variation term can alternatively be expressed using local times, which relaxes the differentiability requirement. This generalization of Ito’s lemma was derived by Tanaka and Meyer, and applies to one dimensional semimartingales.

The local time of a stochastic process X at a fixed level x can be written, very informally, as an integral of a Dirac delta function with respect to the continuous part of the quadratic variation ${[X]^{c}}$,

 $\displaystyle L^x_t=\int_0^t\delta(X-x)d[X]^c.$ (1)

This was explained in an earlier post. As the Dirac delta is only a distribution, and not a true function, equation (1) is not really a well-defined mathematical expression. However, as we saw, with some manipulation a valid expression can be obtained which defines the local time whenever X is a semimartingale.

Going in a slightly different direction, we can try multiplying (1) by a bounded measurable function ${f(x)}$ and integrating over x. Commuting the order of integration on the right hand side, and applying the defining property of the delta function, that ${\int f(X-x)\delta(x)dx}$ is equal to ${f(X)}$, gives

 $\displaystyle \int_{-\infty}^{\infty} L^x_t f(x)dx=\int_0^tf(X)d[X]^c.$ (2)

By eliminating the delta function, the right hand side has been transformed into a well-defined expression. In fact, it is now the left side of the identity that is a problem, since the local time was only defined up to probability one at each level x. Ignoring this issue for the moment, recall the version of Ito’s lemma for general non-continuous semimartingales,

 \displaystyle \begin{aligned} f(X_t)=& f(X_0)+\int_0^t f^{\prime}(X_-)dX+\frac12A_t\\ &\quad+\sum_{s\le t}\left(\Delta f(X_s)-f^\prime(X_{s-})\Delta X_s\right). \end{aligned} (3)

where ${A_t=\int_0^t f^{\prime\prime}(X)d[X]^c}$. Equation (2) allows us to express this quadratic variation term using local times,

 $\displaystyle A_t=\int_{-\infty}^{\infty} L^x_t f^{\prime\prime}(x)dx.$

The benefit of this form is that, even though it still uses the second derivative of ${f}$, it is only really necessary for this to exist in a weaker, measure theoretic, sense. Suppose that ${f}$ is convex, or a linear combination of convex functions. Then, its right-hand derivative ${f^\prime(x+)}$ exists, and is itself of locally finite variation. Hence, the Stieltjes integral ${\int L^xdf^\prime(x+)}$ exists. The infinitesimal ${df^\prime(x+)}$ is alternatively written ${f^{\prime\prime}(dx)}$ and, in the twice continuously differentiable case, equals ${f^{\prime\prime}(x)dx}$. Then,

 $\displaystyle A_t=\int _{-\infty}^{\infty} L^x_t f^{\prime\prime}(dx).$ (4)

Using this expression in (3) gives the Ito-Tanaka-Meyer formula. Continue reading “The Ito-Tanaka-Meyer Formula”

# The Stochastic Fubini Theorem

Fubini’s theorem states that, subject to precise conditions, it is possible to switch the order of integration when computing double integrals. In the theory of stochastic calculus, we also encounter double integrals and would like to be able to commute their order. However, since these can involve stochastic integration rather than the usual deterministic case, the classical results are not always applicable. To help with such cases, we could do with a new stochastic version of Fubini’s theorem. Here, I will consider the situation where one integral is of the standard kind with respect to a finite measure, and the other is stochastic. To start, recall the classical Fubini theorem.

Theorem 1 (Fubini) Let ${(E,\mathcal E,\mu)}$ and ${(F,\mathcal F,\nu)}$ be finite measure spaces, and ${f\colon E\times F\rightarrow{\mathbb R}}$ be a bounded ${\mathcal E\otimes\mathcal F}$-measurable function. Then,

 $\displaystyle y\mapsto\int f(x,y)d\mu(x)$

is ${\mathcal F}$-measurable,

 $\displaystyle x\mapsto\int f(x,y)d\nu(y)$

is ${\mathcal E}$-measurable, and,

 $\displaystyle \int\int f(x,y)d\mu(x)d\nu(y)=\int\int f(x,y)d\nu(x)d\mu(y).$ (1)

# WordPress LaTeX Issues

This is just a brief notice. There appears to be a new problem with WordPress.com displaying LaTeX maths equations. The issue appears to be sporadic, with some of the formulas on both old and new posts now showing “Formula does not parse”. It is affecting other blogs hosted on WordPress too. Thanks to Timothy Gowers on twitter for pointing this out.
The WordPress team are looking into it. Hopefully this will be fixed soon. Update: This issue has now been fixed.

For now, the following method can be used to display the equations using MathJax instead. Click here, then drag the displayed link to the address bar of your browser. Then click on it while viewing any page on this blog, and all formulas will be displayed with MathJax. Magic!

Some tests follow. At the time of writing, some of the following formulas show an error:
Test: ${X^n\xrightarrow{\rm ucp}X}$.

Another test: ${X^n\xrightarrow{\rm ucp}X}$

Yet another test: ${X^n\rightarrow X}$

And another: ${X^n X}$

And another: $X^n X$

And another: ${[S]_\infty}$

yet more: ${[S]_\infty}$

and more: $[S]_\infty$

and more: $[S]$

more: $S_\infty$

# An Unexpected Quartic Solution

Many years ago, while in high school, I tried my hand at solving cubic and quartic formulas. Although there are entirely systematic approaches, using Galois theory, this was not something that I was familiar with at the time. I had just heard that it is possible. Here, ‘solving’ means to find an expression for the roots of the polynomial in terms of its coefficients, involving the standard arithmetical operations of addition, subtraction, multiplication and division, as well as extracting square roots, cube roots, etc.

The solution for cubics went very well. In class one day, the teacher wrote a specific example of a quartic on the blackboard, and proceeded to solve it by reducing to two easy quadratics. The reason that his example worked so easily is because the coefficients formed a palindrome. That is, they were the same when written in reverse order. As an example, consider the equation,

 $\displaystyle x^4+2x^3-x^2+2x+1=0.$

If we divide through by ${x^2}$ then, with a little rearranging, this gives,

 $\displaystyle (x+1/x)^2+2(x+1/x)-3=0.$

As a quadratic in ${x+1/x}$, this is easily solved. One solution is ${x+1/x=-3}$. Multiplying by x and rearranging gives a new quadratic,

 $\displaystyle x^2+3x+1=0.$

By the standard formula for quadratics, we obtain

 $\displaystyle x=(-3\pm\sqrt{5})/2.$

It can be checked that this does give two real solutions to the original quartic.

Now, the approach that I attempted for the general quartic was to apply a substitution in order to simplify it, so that a similar method can be applied. Unfortunately, this resulted in a very messy equation, which seemed to be giving a sextic. That is, I went from the original fourth order polynomial, to what was looking like a sixth order one. This was complicating the problem, and getting further away from the goal than where I had started. I am not sure why I did not give up at that point, but I continued. Then, something amazing happened. Computing the coefficients of the sixth, fifth and fourth powers in this sextic, they all vanished! In fact, I had succeeded in reducing the quartic to a cubic, which can be solved. This still seems surprising, that such a messy looking expression should cancel out like this, in just the way that was needed. See equation (2) below for what I am talking about. As this was such a surprise at the time, and is still so now, I have decided to write it up in this post. It just demonstrates that, even if something seems hopeless, if you continue regardless then everything might just fall into place. Continue reading “An Unexpected Quartic Solution”