User has2 - MathOverflow

Answer by has2 for Do convex and decreasing functions preserve the semimartingale property?

2010-10-30T18:37:55Z

What follows is too long to fit in a comment.

Some thoughts on the second problem. Let us consider the problem for $t$ taking values in the compact interval $[0,T]$. The general case perhaps can be approximated by this case. If the $t$ variable were discrete and finite the problem would be: we have a sequence of functions $f_1 \ge f_2 \ge f_3 \cdots \ge f_n$ (let us make the simplifying assumption that $f_i$ are all positive; otherwise, replace $f_i$ with $f_i +C$ where $C$ is $ C = -\min_{i,x} f_i(x)$), can one find convex and increasing ${h_i}$ and ${g_i}$ such that $f_i = h_i - g_i$? The answer to this question is obviously yes. For example, $h_i = \sum_{j\le i} f_i$ and $g_i = \sum_{j < i} f_i$ is one possible solution. The problem with this solution when $t$ is continuous is that the $h_i$ and $g_i$ would explode as the discretization of $t$ is refined. Thus, one needs to choose $h$ and $g$ in a way that they increase slowly. How slowly can this increase be? We can start with $h_1 = f_1$ and $g_1 = 0$. $h_2$ and $g_2$ will be of the following form: $$ h_2 = h_1 + S_2 = f_1 + S_2, $$ and $$ g_2 = g_1 + R_2 = R_2 $$ such that: 1) $S_2, R_2 \ge 0$, 2) $h_2 = f_1 + S_2$, $g_2 = R_2$ are convex and 3) $h_2 - g_2 = f_1 + S_2 - R_2 = f_2$. The last of these is equivalent to $R_2 = f_1 - f_2 + S_2$. Thus, what we are looking for is a function $S_2$ satisfying the above conditions. There will be many such $S_2$, the goal is to choose $S_2$ in a minimal way so that $h_i$ and $g_i$ grow slowly.The best would be: $S_2 = 0$, which is indeed a solution when $f_1 - f_i$ is convex. If $f_0 - f_t$ is convex for all $t \in [0,T]$ then this solution directly generalizes to the original continuous time problem (i.e., if $f_0-f_t$ is convex then $h(t,x) = f(0,x)$ and $g(t,x) = f(0,x) - f(t,x)$ is a solution).

Now, let us consider the case when $f(t,x)$ is such that $\frac{\partial^2}{\partial x^2}f(t,x)$ is continuous in $(t,x)$ and $x$ also takes values in a compact set $K$. The following type of argument quickly comes to mind. Define $$\tau \doteq \{t: \exists g, h:[0,t]\times K\rightarrow {\mathbb R}, f = h-g, h,g \text{ convex in } x \text{ and increasing in } t \}.$$ $0$ is clearly a member of this set. One can perhaps argue that $\sup \tau$ must be $T$ as follows. If $t_0 = \sup \tau < T$ then one can slightly modify $h(t_0,x)$ and $g(t_0,x)$ to obtain functions $h(t_0+\delta,x)$ and $g(t_0+\delta,x)$ whose difference will be $f(t_0+\delta,x)$ [the slight modification is made possible by the fact that the second derivative of $f$ with respect to $x$ barely changes between $t_0$ and $t_0 + \delta$].

Answer by has2 for Characteristic operator

2010-09-08T10:23:25Z

The wikipedia page cited in the question provides most of the answer: to get your operator compute \begin{equation} \lim_{\delta \rightarrow 0} \frac{ {\mathbb E}[f(X_\delta)] -f(x)}{\delta} \end{equation} The difference between your problem and the case covered in the wikipedia article is that $f$ in the above display is a function of $x$ only. However, your problem has an additional state variable (the binary variable that takes one of the values $1$ or $2$ depending on $\alpha$). So, the correct limit to study is: \begin{equation} \lim_{\delta \rightarrow 0} \frac{ {\mathbb E}[f(X_\delta,\alpha_t)] -f(x,1)}{\delta}. \end{equation} Thus, you don't have one function, but two functions $f(x,1)$ and $f(x,2)$ and two PDEs that these functions satisfy.

It is implicitly assumed that $\tau$ is independent of the dynamics of $X$ before $\tau$. Furthermore, before $\tau$ the dynamics of $X$ are governed by the first SDE given in the question. One can use these to write the above expectation in two pieces: one piece over the set $\{\delta < \tau\}$ the other over $\{\delta < \tau\}^c$. Once this is done, the usual use of Ito's formula gives: $$ L_1 f(x,1)=-\lambda f(x,2)~~~ (*) $$ and $$ L_2 f(x,2) = 0. $$ where $$ L_i = a(b_i - x) \frac{\partial}{\partial x} + \frac{1}{2} \sigma^2\frac{\partial^2}{\partial x^2} $$

Further details: \begin{align*} {\mathbb E}[ f(X_\delta,\alpha_\delta) ]&= {\mathbb E}[ f(X_\delta,\alpha_\delta) 1_{\{ \tau > \delta\}} ] + {\mathbb E}[ f(X_\delta,\alpha_\delta) 1_{\{ \tau \le \delta\} }]\\ &\approx (1-\lambda \delta){\mathbb E}[ f(X^1_\delta,\alpha_\delta)] + \lambda \delta f(x,2), \end{align*} where $X^1$ is a process that is independent of $\tau$ with dynamics determined by $L_1$.

Here you use several things: 1) $P( \tau < \delta) \approx \delta \lambda$ 2) if a jump occurs before $\delta$, you can ignore what happens between $\tau$ and $\delta$ (the contribution of this part is in the order of $\delta^2$ and when divided by $\delta$ and $\delta$ is let go to $0$, it disappears).

To get (*) from the previous display: use Ito's formula on the first expectation, subtract $f(x,1)$, divide by $\delta$ and let $\delta \rightarrow 0$. $f(x,2)$ is a function of what happens after $\tau$; after $\tau$ the stochastic process is a simple diffusion with generator $L_2$: this is why (**) holds.

Answer by has2 for Is the infimum of the Ky Fan metric achieved?

2010-09-12T14:47:22Z

Because what follows doesn't fit in a comment, I write it here as an answer; but they are merely comments. After computing the minimizer for several simple distributions, my impression is that the answer to this question is yes, and there will be many minimizers.

Intuitively, it seems possible to build an optimizer as follows: we are given the law $\mu$ and we would like to find a function $f$ such that 1) the distribution of $f$ is $\mu$ and 2) $\alpha(f,X)$ is minimum. Let $\epsilon > 0 $ be this minimum. Let $F(x) = \mu( (-\infty, x])$, i.e., $F$ is the distribution function associated with $\mu$. Let $G$ be the inverse function of $F$: $G(x) \doteq \inf\{y: F(y) \ge x \}$. By its definition $G$'s distribution is $\mu$. Draw the graphs of the functions $l(x) = x + \epsilon$ and $u(x) = x -\epsilon$ around the graph of the function $X(x) = x$. To get the minimizer, one cuts the graph of $G$ into $n$ small pieces with lines parallel to the $x$ axis and shifts around the pieces along these lines so that they lie between the graphs of $l$ and $u$ as much as possible. As the number of pieces increase and their size decreases you would expect this to converge to a function that is the desired minimizer. The result will depend on the particulars of this process.

As to non-uniqueness: suppose $f$ is a minimizer. Denote with $E$ the subset of $[0,1]$ over which $f$ differs from $X$ by at least $\epsilon$. The values that $f$ takes over $E$ can be freely permuted without affecting the distribution and the distance between $f$ and $X$. So there will be infinitely many minimizers, when there is one.

Answer by has2 for Question regarding divergence

2010-08-16T01:51:01Z

The inequality $D(P'|Q') \ge D(P^\star| Q^\star)$ does not need to hold.

Here is an example.

Let $A$ be the set $\{1,2,3,...,n\}$. Let $E$ be the set of measures $P$ on $A$ such that $P(\{1\}) = 0$. Projecting a measure $P$ on $E$ using $D$ is equivalent to conditioning $P$ on $ A- \{1\}$. Choose $P'$ and $Q'$ such that they both put equal and nonzero mass on $\{1\}$. By direct computation one sees: $D(P^\star| Q^\star) = \frac{1}{1-P'(\{1\})} D(P'|Q') > D(P' | Q')$.

The details of the above computation are as follows.

For ease of notation set $n=3$. Let $E$ be the set of measures $P$ with $P(\{1\}) =\epsilon$; to obtain the example above, one sets $\epsilon = 0$. Let us parametrize the measures on $\{1,2,3\}$ as follows: $P(\{1\}) = p_1$, $P(\{2\}) =p_2$ and $P(\{3\}) = 1-p_1 -p_2$. Our problem is: $$ \inf_{ Q \in E}\left[ p_1 \log \frac{p_1}{q_1} + p_2 \log \frac{p_2}{q_2} + (1-p_1 -p_2) \log\frac{ 1- p_1 - p_2}{ 1- q_1 - q_2 } \right]. $$ Let $F$ denote the expression after the $\inf$. $F$ is strictly convex in $Q$ and therefore will have a unique optimizer. In the above coordinates, the normal to $E$ is the vector $(1,0)$. Then $$ \frac{\partial F} {\partial q_1} = -\frac{p_1}{q_1} + \frac{1-p_1-p_2}{1-q_1-q_2} = \lambda $$ and $$ \frac{\partial F} {\partial q_2} = -\frac{p_2}{q_2} + \frac{1-p_1-p_2}{1-q_1-q_2} = 0. $$ We have the constraint that $Q\in E$, i.e., $q_1 =\epsilon$. From the last two equalities one infers: $$ q_2 = \frac{(1-\epsilon) p_2}{ 1-p_1}. $$

Going back to the coordinates $(p_1,p_2,p_3)$ to denote a measure on $\{1,2,3\}$, projecting a measure on $E$ using $D$ corresponds to the following map: $$ (p_1,p_2,p_3) \rightarrow \left(\epsilon, (1-\epsilon)\frac{p_2}{p_2+p_3}, (1-\epsilon)\frac{p_3}{p_2 + p_3}\right). $$ For $\epsilon =0$, this is the same as conditioning $P$ on $\{2,3\}$.

One obtains the expression for the relative entropy given above by directly computing it using this formula for the projections.

Answer by has2 for Joint Law with 2 marginals and marginal of the spread

2010-02-08T23:19:39Z

Let us think about the discrete case; that is let us suppose we are interested in determining a probability distribution $P$ on the discrete set $\Omega ={\mathbb Z}_n^2$.

Such a probability distribution assigns a nonnegative weight to each $(i,j) \in \Omega$. $|\Omega| = n^2$, thus $P$ is determined by $n^2-1$ nonzero variables $p_{i,j}$ whose sum is less than $1$. To fix the marginals of $P$ means to put $2(n-1)$ constraints on ${p_{i,j}}$. In addition to these constraints, the present question also imposes a distribution on $X-Y$. These translate into $n-1$ further constraints on $p_{i,j}$. Thus, in general, $P$ will be a function of $n^2-1-3(n-1)$ free variables.

The special cases you mention (i.e, the clayton and gumbel copulas as well as the normal distribution), however, are determined by the marginal distributions and an additional real parameter. In general, if the given data makes sense, $\theta$ can be recovered by first writing an equation that it satisfies and then solving it.

Under any of the above mentioned copulas the joint distribution equals $\Phi(F(x),G(y),\theta)$ where $F$ is the $X$ marginal, $G$ is the $Y$ marginal and $\theta$ is a real number. The only unknown here is $\theta$. Knowing the distribution of $X-Y$, means in particular we know the probability that $X-Y=n-1$[again assuming that we are operating in the discrete setup]. There is only one way this can happen, i.e, if $Y=0$ and $X=n-1$. Thus, we know the weight $p_{(n-1,0)}$ of the point $(n-1,0)$. Then, $\theta$ is the solution of $$\Phi(F(n-1),G(0),\theta) -\Phi(F(n-2),G(0),\theta) = p_{(n-1,0)}.$$

In the case of ${\mathbb R}^2$, one can for example, write the following equation for $\theta$: $$ \int_{\mathbb R}^2 (x-y) d\Phi(F(x),G(y),\theta) = \int z dS(z) $$ where $S$ is the distribution of $X-Y$ given in the problem.

Edit: more importantly, it seems, one has to check if the given distribution of $X-Y$ is compatible with the copula in question. Because, there are many equations that one can write for $\theta$ and all must give the same answer for the solution to be meaningful.

Answer by has2 for Help with a double sum, please

2010-01-29T02:11:58Z

Let's define $$ \beta_n \doteq \sum_{i\le (n-1)/2 } \binom{n-(i+1)}{i} (-1)^i \frac{1}{ (2i+1) 2^{2i+1} }. $$ The following problem is equivalent to proving that $S=0$: prove that the sequence $\beta_n$ satisfies the recursion $$ \beta_{n+1} = \frac{2n+1}{2n+2} \beta_n +\frac{1}{(n+1) 2^{n+1}}. $$ Similar with $S=0$, numerical computations suggest that this statement is true. Unfortunately, I didn't see a straightforward way to prove it.

Below is one way to think about the problem, which led to the above reformulation.

The connection between the above problem and $S=0$.

Using the notation developed in the previous answer, let's define $$ F(m,n) = \sum_{k=0}^m (-1)^k \binom{m}{k} \binom{2(n+k)}{n+k} \frac{1}{2^{2(n+k)}} \sum_{l=1}^{k+n} \frac{2^l}{l \binom{2l}{l} }, $$ and $$ f(n)= F(0,n)= \binom{2n}{n} \frac{1}{2^{2n}} \sum_{l=1}^n \frac{2^l}{l \binom{2l}{l} }. $$ The statement $S=0$ is the same as $F(m,m)= 0$. Note that $F$ satisfies $$ F(m,n) = \frac{1}{2} F(m-1,n) - \frac{1}{2}F(m-1,n+1) ~~~~~~\text{(r1)} $$ Define the difference operator $D(x_1,x_2) = (x_1 - x_2)/2.$ (r1) in terms of $D$ is $$ F(m,n) = D( F(m-1,n), F(m-1,n+1) ). $$ Define $D^k$ by iterating $D$: $$ D^n(x_1,x_2,x_3,\ldots,x_{n+1}) = D( D^{n-1}(x_1,x_2,x_3,\ldots,x_{n}), D^{n-1}(x_2,x_3,\ldots,x_{n+1} )) $$ Iterating (r1) gives

$$ F(m,n) = D^m( f(n),f(n+1),f(n+2), f(n+3),\cdots,f(n+m)). $$

In particular: $$ F(m,m) = D^m( f(m),f(m+1),f(m+2), f(m+3),\cdots,f(m+m)). $$

Define ${\mathcal D}:{\mathbb R}^\infty\rightarrow {\mathbb R}^\infty$ as follows: the $i^{th}$ component of ${\mathcal D}(x_{1}^\infty)$ is $$D^n(x_n,x_{n+1},x_{n+2},\ldots,x_{2n}).$$

We can restate our original problem as follows: show that $(f(1),f(2),f(3),...,f(n),...)$ is in the kernel of ${\mathcal D}$.

Because we are looking for a zero of this operator, the $1/2$ in the definition of $D$ is not important; thus let us assume that $D(x_1,x_2)$ is simply $x_1 -x_2$.

Note that $D^{n}(f(n),f(n+1),...,f(2n)) =0$ is the same as $$ D^{n-1}(f(n),f(n+1),f(n+2),...,f(2n-1)) = D^{n-1}(f(n+1),f(n+2),f(n+3),...,f(2n)). $$ A numerical computation reveals that these discrete derivatives equal $\frac{1}{(2n-1)2^{2n-1}}$. One can go back from these values to an element of the kernel of ${\mathcal D}$ by inverting each $D$ in the above display. A bit of computation in this direction yields the vector $\beta$ in the first display. By its construction $\beta$ is in the kernel of ${\mathcal D}$. Thus if one can prove that $f$ equals $\beta$ then we are done.

Finally, using its definition, we see that $f$ satisfies: $$ f(n+1) = \frac{2n+1}{2n+2} f(n) + \frac{1}{(n+1)2^{n+1}}, ~~~ f(1) = 1/2. $$ These relations determine $f$ and thus we can take them as $f$'s definition. Thus to verify $f=\beta$ it is enough to show that $\beta$ satisfies this recursion.

Answer by has2 for limsup and liminf for a sequence of sets

2010-01-21T04:39:45Z

For a sequence of subsets $A_n$ of a set $X$, the $\limsup A_n$ $= \cap_{N=1}^\infty ( \cup_{n\ge N} A_n )$ and $\liminf A_n$ $= \cup_{N=1}^\infty (\cap_{n \ge N} A_n)$.

If $ x \in \limsup A_n$ then $x$ is in all of the $\cup_{n\ge N} A_n$, which means no matter how large you pick $N$ you will find an $A_n$ with $n>N$ of which $x$ is a member. Thus members of $\limsup A_n$ are those elements of $X$ that are members of infinitely many of the $A_n$'s. If $A_n$ are thought of as events (in the sense of probability) $\limsup A_n$ will be another event. It corresponds exactly to the occurance of infinitely many of the $A_n$'s. This is why $\limsup A_n$ is sometimes written $x \in A_n$ infinitely often.

Similarly, if $x\in \liminf A_n$ then $x$ is in one of $\cap_{n\ge N} A_n$, which means $x \in A_n$ for all $n > N$. Thus, for $x$ to be in the $\liminf$, it must be in all of the $A_n$, with finitely many exceptions. This is how the phrase "ultimately all of them" comes up.

Both of these operations, similar to their counterparts in metric spaces, concern the tail of the sequence ${A_n}$. I.e., neither changes if an initial portion of the sequence is truncated. As a previous response pointed out, often the sets $A_n$ are defined to track the deviation of a sequence of random variables from a candidate limit by setting $A_n = \{x: |Y_n(x) -Y(x)| \ge \epsilon\}$. The members of $\limsup A_n$ then represents those sequences that every now and then deviate $\epsilon$ away from $Y(x)$, which is solely determined by the tail of the sequence $Y_n$.

Here is a conceptual game that can be partially understood using these concepts: We have a deck of cards, on the face of each card an integer is printed; thus the cards are ${1,2,3...}.$ At the nth round of this game, the first $n^2$ cards are taken, they are shuffled. You pick one of them. If your pick is 1, you win that round. Let $A_n$ denote the event that you win the nth round. The complement $A_n^c$ of $A_n$ will represent that you lose the $n^{th}$ round. The event $\limsup A_n$ represents those scenarios in which you win infinitely many rounds. The complement of this event is $\liminf A_n^c$, and this represents those scenarios in which you ultimately lose all of the rounds. By the Borel Cantelli Lemma $P(\limsup A_n)$ $=0$ or equivalently $P(\liminf A_n^c)=1$. Thus, a player of this game will deterministically experience that there comes a time, after which he never wins.

Baire category theorem

2010-01-18T22:56:02Z

Let's call the following conditions (1): $X$ is a complete metric space with metric $d$, $X = \cup_{n=1}^\infty A_n$. Let $\bar{A}$ denote the closure of $A$.

Let's call the following statement (2): at least one of the $\bar{A}_n$ contains a ball.

Baire category theorem gives:

Fact1: (1) $\Rightarrow$ (2)

Now take any $x\in X$ and consider the closed ball $B = B(x,\delta)=\{y: d(x,y)\le \delta\}$. This is a complete metric space itself and it is covered by $ A_n \cap B$. Thus these sets satisfy (1). Fact1 gives us an $n$ such that the closure of $A_n \cap B$ contains a ball in $B$. Thus, we have strengthened Fact1 to

Fact1': (1) $\Rightarrow$ (2')

where (2') is: For every $x\in X$, every neighborhood of $x$ contains a ball that is contained in one of the $\bar{A}_n$.

Question: Can (2') be strengthened further? Here are some example statements, both of which are too strong:

For every $x$, there is a $\delta$ such that $B(x,\delta)$ is contained in one of the $\bar{A}_n$
For every $x$, there is an $A_n$ such that $\bar{A}_n$ contains an open set $G$ with $d(x,G)=0$.

Many thanks for the responses. The motivation for this question was as follows.

1) What does it mean for a set $A$ to have a closure with empty interior? Take an element $a \in \bar{A}$. This means that $a$ is either in $A$ or there is a sequence in $A$ that converges to $a$. This can be rephrased as `$a$ is a point that can be approximated with infinite precision by $A$.' If $\bar{A}$ has no interior, perturbing $a$ by a small amount will give an $a'$, such that $a'$ is a finite distance away from $\bar{A}$. Thus, $a'$ will be a point that can only be approximated by $A$ with finite precision. Then, one can think of $A$ as a multi resolution grid with discontinuous approximation capability: you perturb any point that can be approximated infinitely well by it and you get a point that can only be approximated with finite precision.

2) $X$ itself can be thought of as the perfect multi resolution grid for itself: every point $x\in X$ is well approximated by $X$, and perturbing $x$ will not change this. The way I wanted to think of $X= \cup_{n=1}^\infty A_n$ was this: for every $x \in X$, one of the $A_n$ provides a multiresolution grid around $x$ that has continuous approximation capability. I wanted to think, similar to Leonid's response, that the set of $X$ where this is not possible was to be in some sense to be exceptional.

Comment by has2

2010-10-31T07:50:36Z

Another thing that seem relevant is the following: the region where $x \rightarrow f(x,0) - f(x,t)$ is not convex must be small, because otherwise $f(x,0)$ could not dominate $f(x,t)$.

Comment by has2

2010-10-31T07:45:24Z

You are welcome; your questions above and your related results are very interesting, thanks for sharing. What makes the problem difficult seems to be that there are many things that influence how $g$ and $h$ are to evolve in time and there seems to be many choices. The optimization problem in the discrete formulation (the choice of $S_2$) referred to in my answer seems to be nontrivial (I think it can be formulated as a control problem).

Comment by has2

2010-09-13T12:08:40Z

You are welcome. I tried to explain how you get the equations in the answer. Now I will add some further details, hope they are useful.

Comment by has2

2010-08-20T12:30:44Z

You are welcome, thanks to you and Suresh for your input.

Comment by has2

2010-08-19T10:31:14Z

There is a limiting process involved in the case that reduces to conditioning. To get an example without infinite entropy, choose $1>\epsilon>0$ in the answer above. Then $D(P^\star|Q^\star)=\frac{1−\epsilon}{1−p'_1}D(P'|Q')$ and if one chooses $1>p'_1=q'_1>\epsilon$ and $(p'_2,p'_3)\neq(q'_2,q'_3)$, one gets $D(P^\star|Q^\star)> D(P'|Q')$

Comment by has2

2010-08-18T08:42:12Z

Relative entropy (KL-divergence) is strictly convex in both of its variables. Therefore, it will always have a unique minimizer over any convex set. I tried to provide the details of the calculations above; they seem to be correct.

Comment by has2

2010-08-17T20:49:49Z

Hi, one can slightly modify the above case to generate an example in which projection decreases relative entropy. The main idea appears to be this: conditioning can increase or decrease entropy; depends on what one conditions on.

Comment by has2

2010-03-27T22:08:40Z

An intuitive reasoning for why the axiom makes sense is as follows. A subset of a set is closed if it contains all of the points that its members approximate. The whole set must be closed because it contains everything, and in particular, everything that can be approximated by the members of the set. The empty set must be closed too because there are no points in it and no points are being approximated by the members of the empty set.

Comment by has2

2010-02-09T00:28:15Z

You are welcome AlGoRiS. Another thought, hopefully not totally nonesense is as follows: the density of $X-Y$ equals $f(x,\theta) = \int_{\mathbb R} \Phi_{x,y}(F(y+c),G(y),\theta)dy$. Perhaps one can try to choose $\theta$ to minimize the distance between $f(x,\theta)$ and $S'$.

Comment by has2

2010-02-02T23:17:43Z

If $f>0$ and $\lambda >0$ then $1-e^{\lambda f}$ is less than $0$ but $0.5 \lambda (f\wedge 1)$ is greater than $0$.

Comment by has2

2010-02-02T23:09:38Z

Thanks! Yes, I did see your post after I submitted mine and we seemed to have looked at different editions of the book. I am a bit confused about the argument you provided for the converse. You claim that $1-e^\lambda f > 0.5 \lambda (f\wedge 1)$. Suppose $f$ is nonnegative; I think you also implicitly assume that$t\lambda>0$. Then it appears as if this inequality is incorrect: its left side is negative and its right side is positive.

Comment by has2

2010-01-29T14:45:09Z

Now corrected, thanks!

Comment by has2

2010-01-19T08:26:09Z

Many thanks for the answers and the comments. The responses below has been equally enlightening to me. A coin flip decided which one of them to be the accepted answer.