User kenneth - MathOverflow

limit of functionals on weak convergent random variables

2012-10-04T03:21:46Z

Suppose real value random variables satisfy $X_{n} \Rightarrow X$ (convergence in distribution) as $n\to \infty$ in the same probability space $(\Omega, \mathcal F, \mathbb P)$. It is well known that $\lim_{n\to \infty} \mathbb E f(X_{n}) = \mathbb E f(X)$ for all continuous bounded real functions $f:\mathbb R \to \mathbb R$.

[Q.] If $f$ is continuous and linear growth, i.e. $|f(x)| < K(1 + |x|)$ for some constant $K$, can you find counter-example for $\lim_{n\to \infty} \mathbb E f(X_{n}) = \mathbb E f(X)$? What additional conditions are needed to still have $\lim_{n\to \infty} \mathbb E f(X_{n}) = \mathbb E f(X)$?

Answer by kenneth for limit of functionals on weak convergent random variables

2012-10-04T09:03:10Z

Since $f(X_{n}) \Rightarrow f(X)$ as long as $f$ is continuous, by redefining $X_{n}:= f(X_{n})$, it is equivalent to ask following question:

[Q1.] If $X_{n}\Rightarrow X$, then what additional condition is needed to have $\lim_{n\to\infty}\mathbb E[X_{n}] = \mathbb E[X]$?

[Ex.] Let $X_{n}: [0,1] \mapsto \mathbb R$ be given by $X_{n} (\omega) = n I_{(0,1/n)}(\omega)$ and $\mathbb P$ be Lesbegue meaure. Then, $X_{n} \Rightarrow X:=0$, since it is indeed almost sure convergence. However, it violates $\lim_{n\to\infty}\mathbb E[X_{n}] = \mathbb E[X]$.

So one immediate sufficient condition needed to have $\lim_{n\to\infty}\mathbb E[X_{n}] = \mathbb E[X]$ is that,

$X_{n} \to X$ almost surely, and satisfies other conditions of Dominated (or Monotone) Convergence Theorem.

Generalized Ito's formula

2011-09-28T07:52:33Z

Consider classical statement of Ito's formula: Let $X$ be a continuous semimartingale and $F \in C^2(\mathbb{R}^d, \mathbb{R})$; then $F(X)$ is a continuous semimartingale and $$F(X_t) = F(X_0) + \sum_i \int_0^t {\partial_i F} dX_s^i + \frac 1 2 \sum_{i,j} {\partial^2_{ij} F} d \langle X^i, X^j \rangle_s.$$ In the above Ito formula, how much does function $F$ extendable in a Sobolev space? For example, is Ito formula true if $F\in W^{2,p}$ for some $p>1$? Note that, if we use Ito-Tanaka formula, then there exists some extra term from local time, and we wish to find Sobolev regularity to make sure this term being zero.

A wrong proof of Squared Bessel process

2012-08-03T06:09:21Z

The squared Bessel process with $\delta$-dimension for $\delta>0$, denoted by $BESQ^\delta(y)$, is given by $$d Y_t = \delta t + 2 \sqrt{Y_t} d B_t, \ Y_0 = y\ge 0$$ where $B_t$ is BM under $(\Omega, {\cal F}_t, P)$. Consider $\tau = \inf[ t>0: Y_t = 0].$

(Claim). $\tau = \infty$ almost surely.

(Proof). Let $X_t = Y_{\frac 2 \delta t}$. Then, $$d X_t = 2 t + 2 \sqrt{X_t} d W_t, \ X_0 = y,$$ where $W_t = B_{\frac 2 \delta t}$ is BM under $(\Omega, {\cal F}_{\frac 2 \delta t}, P)$. In other words, $X_t$ is $BESQ^2(y)$ w.r.t. time-scaled filtration under the same probability measure. Therefore, {0} is polar set of $X_t$, so is of $Y_t$. END.

However, it gives a contradiction to the fact that $\tau = 0$ for $BESQ^1(0)$ due to the properties of 1-D BM. Where is the gap of the above proof?

Is this process strictly positive?

2012-08-02T03:36:56Z

Let $W_t$ is standard Brownian motion under probability measure $P$. Consider 1-D stochastic differential equation $$ dY_t = dt + \sigma(Y_t) dW_t, \ Y_0 = y\ge 0.$$ We assume $\sigma(0) = 0$, and $\sigma(x)$ is locally Holder-1/2, i.e. for any bounded subinterval $I\subset [0,\infty)$, we assume there exists constant $K_I$ s.t. $$|\sigma(x_1)- \sigma(x_2)| \le K_I |x_1 - x_2|^{1/2}, \ \forall x_1, x_2 \in I.$$ Note that, the above SDE has strong non-negative solution by comparison with $d X_t = \sigma(X_t) dW_t$.

[Q1] Define $\tau = \inf[t>0: Y_t = 0]$. Is $\tau>0$ almost surely?

[Q2] Can one show that $Y_t > 0$ almost surely for arbitrary given $t>0$?

In fact, it's enough to show the above results with $y= 0$.

My guess is that, [Q2] is too strong to be true, but [Q1] is correct. It will be helpful to get a proof of [Q1] at least.

Does this series stopping times marching forward?

2012-07-27T06:30:57Z

Let $W_t$ is standard Brownian motion under probability measure $P$. Consider stochastic differential equation $$ dY_t = dt + Y_t dW_t, \ Y_0 = 0.$$ Note that, the above SDE has a strong non-negative solution.

Define stopping times $$\tau_0 = 0; \tau_{n+1} = \inf [ t>\tau_n: Y_t = 0 ].$$

[Q] Can one show that $\lim_{n\to \infty} \tau_n >1$ almost surely in $P$?

The above question is not true if the underlying SDE is $$ dY_t = dt + dW_t, \ Y_0 = 0,$$ since $Y_t$ is standard BM under some equivalent probability measure, and $\tau_n = 0$ for all $n\ge 1$.

Is a semicontinuous real function Borel measurable?

2012-03-10T04:16:09Z

Let $f(x,u): [0,1]^2 \mapsto \mathbb{R}$ be a continuous function.

[Q] Is $g(x) = \inf_{u\in [0,1]} f(x,u)$ always Borel measurable? If not, can one find a counter-example?

Note that, for any $c$, we have $$(x: g(x) < c) = \text{Proj}_x ((x,u): f(x,u) < c),$$ where $\text{Proj}_x$ is a projection operator to $x$-axis. In the context of measurable selection theorem, the projection of Borel set $((x,u): f(x,u) < c)$ of $\mathbb{R}^2$ is not necessarily a Borel set of $\mathbb{R}$. But, I can not find a counter-example.

If there exists a proper counter-example, then it also implies that a semicontinuous real function is not necessarily Borel measurable.

Thanks.

A non-degenerate martingale

2011-12-24T09:38:46Z

Let $(\Omega, \mathcal{F}, P)$ be a probability space, on which $\mathcal{F}_t$ is filtration satisfying general conditions. $W_t$ is a standard Brownian motion. Let $Y_t$ be a martingale given by $$Y_t = \int_0^t \sigma_r d W_r$$ where $\sigma_t$ is a bounded $\mathcal{F}_t$ measurable process.

The question is, assume $\sigma_t>0$ almost surely for all $t$, then can we prove $$P(Y_1 = c) = 0$$ for all constant $c$?

law of iterated logrithm

2011-06-28T07:28:22Z

Let $(\Omega, \mathcal{F}, P)$ be a probability space, on which $\mathcal{F}_t$ is filtration satisfying general conditions. $W_t$ is a standard Brownian motion. By law of iterated logarithm, one has $$ P\Big(\lim_{t\to \infty} \frac{W_t}{t} = 0\Big) = 1 $$ This implies $$ P\Big(\lim_{t\to \infty} \frac{t + W_t}{t} = 1\Big) = 1. \quad (1)$$ We have contradiction in this below. By Girsanov theorem, there exists a probability measure $Q$ equivalent to $P$, such that $t+ W_t$ is a Brownian motion w.r.t. $Q$. By law of iterated logarithm, $$ Q\Big(\lim_{t\to \infty} \frac{t + W_t}{t} = 0\Big) = 1, $$ which implies $$ P\Big(\lim_{t\to \infty} \frac{t + W_t}{t} = 0\Big) = 1, \quad (2)$$ since $P$ is equivalent to $Q$. Why does this argument leads to a contradiction between (1) and (2)?

Answer by kenneth for Theorems that are 'obvious' but hard to prove

2011-02-28T02:46:04Z

Dynamic programing principle (DPP) is one of the 'obvious' and also intuitive one, in the control problem. Many papers proves its validity in various setup, and all proofs are very complicated. But, there is rarely a counter example of DPP. I wonder, if there is general framework on it. See, http://mathoverflow.net/questions/22689/dynamic-programming-principle-dpp

infimum of a set of stopping times

2011-02-26T15:12:21Z

Let $(Y^a: a\in \Lambda)$ be a set of random processes given by $$Y^a(s) = \int_0^s \sigma^a(r) dW(r)$$ where $W$ is Brownian motion w.r.t. filtered probability space $(\Omega, \mathcal{F}, P, \mathcal{F}_t)$, $\sigma^a(\cdot)$ is uniformly bounded predictable process, s.t. $|\sigma^a(r)|<1$ for all $r$ and $a$. Let $\theta^a = \inf(s>0, Y^a(s) \ge 1)$ be a stopping time.

[Q.] Is the following true, $$\inf_{a\in \Lambda} (\theta^a) >0, \quad a.s.-P$$

When $\Lambda$ is finite set, it is clearly true. But, I am not sure otherwise. Thanks.

infimum of a set of positive r.v. with the same distribution

2011-02-26T15:30:32Z

Let $Y$ be real valued random variable on probability space $(\Omega, \mathcal{F}, P)$, such that $Y>0$ almost surely. Suppose $(X^a: a\in \Lambda)$ be a set of random variables in the same probability space with the same distribution as $Y$.

[Q.] Is the following true? $$\inf_a (X^a) >0, \quad a.s.-P$$

Distribution of running maximum of a local martingale

2010-07-30T17:08:22Z

Let $(\Omega, \mathcal{F}, \mathbb{P}, \mathcal{F}_t)$ be a given probability space with usual conditions, on which $W$ is a standard Brownian motion. For $x \ge 0$, consider $$X(t) = x + \int_0^t \sigma (X(s)) dW(s)$$ Assume $\sigma \in C^{0,1/2}_{loc}$, $\sigma(0) = 0$, $\sigma>0$ on $(0,\infty)$. By [Karatzas and Shereve 98], there exists a unique strong solution with absorbing state at zero. Denote the running maximum by $X^*(T) = \sup_{s\in [0,T]} X(s)$.

Question: For a fixed $T$, is this possible to show that $\mathbb{P} ( X^*(T) \ge \beta) = o(\beta^{-1})$ as $\beta \to \infty$?

I am trying to use time-changed Brownian motion, i.e. $X(t) = x + B([X]_t)$, where $B$ is BM, and $[X]$ is quadratic variation. There is also density function available for running maximum $B^* (T)$, i.e. $\mathbb{P}(B^*(T) \ge \beta) = 2 - 2 \Phi(\beta/\sqrt{T}) = o(\beta^{-1})$, where $\Phi(\cdot)$ is c.d.f of standard normal distribution. But, I could not succeed using those facts to prove it.

Thank you for your time.

Is the truncated Brownian motion of the class DL?

2010-06-04T12:26:24Z

Let $W$ be a standard Brownian motion under given probability space. For a given constant $a$, $W^a$ is a truncated Brownian motion by stopping time $T^a = \inf(t>0:W(t) = a)$. That is, $W^a(t) = W(t \wedge T^a)$.
We want to consider the following question: Is the process $W^1$ of the class DL?

(Solution1): Yes. Indeed, for any fixed $t>0$, we can prove the collection of random variables $( W(s), 0< s< t)$ is uniformly integrable by definition, since $E [|W^1(t)|] < \infty$.

We provide completely different answer using the following proposition from the Problem 1.5.19 (i) of Book [Karazas and Shereve 98].

[Proposition] A local martingale of class DL is martingale.

(Solution2): No. $W^1$ is strict local martingale, since $E [W^1(T^1)] = 1> E [W(0)]$. By [Proposition], $W^1$ is not of class DL.

In the above, we obtained completely two different solutions. Where is wrong?

Continuity in intial state of Brownian Motion

2010-07-31T15:33:48Z

$ B = (B_t, \mathcal{F}_t; t\ge 0 ) $ is a 1-d Brownian family on a measurable space $(\Omega, \mathcal{F})$ with a family of probability measures ${\mathbb{P}^x}$, i.e. $\mathbb{P}^x(B_0 = x) = 1$, and $B$ is 1-d BM starting from $x$ under $\mathbb{P}^x$.

Let $\tau$ be a given stopping time w.r.t. underlying filtration, $f$ be a given continuous bounded real function. Consider $V(x) = \mathbb{E}^x [f(B_\tau)]$, where $\mathbb{E}^x$ is the expectation under $\mathbb{P}^x$.

[Question] Is $V(\cdot)$ continuous for any given stopping time $\tau<\infty$? If not, is there any counter example? Or does continuity hold with further conditions?

If $\tau$ is deterministic, then $V$ has no doubt to be continuous. I am not sure, even if the problem is well formulated with the extension to stopping time $\tau$. Thanks for any of your comments.

Dynamic programming principle (DPP)

2010-04-27T06:26:39Z

In stochastic control problem, one shall use the measurable selection theorem to prove DPP. It was discussed in discrete time case in [Bertsekas and Shreve 1978]. Is there unified framework in continuous time problem, to derive sufficient condition under which DPP is valid? Is there an easy way to explain why measurable selection theorem is important in DPP?

For ex., here is one control problem: Let $(\Omega, \mathcal{F}, \mathbb{P}, \mathbb{F})$ be a filtered probability space. Let $\mathcal{A}$ be admissible control space, and $X^{t,x,\alpha}(s)$ be a controlled Markov process with initial $(t,x)$ and control $\alpha \in \mathcal{A}$. The value function is $V(t,x) = \inf \mathbb{E}[ g(X^{t,x, \alpha}(\tau)) ]$, where $\inf$ is over $\alpha \in \mathcal{A}$, and $\tau$ is a given stopping time. The question is, what is the sufficient condition for $V(\cdot)$ to have $V(t,x) = \inf \mathbb{E}[V(\theta, X^{t,x,\alpha}(\theta))]$ for all stopping time $\theta \in [t,\tau]$?

Comment by kenneth

2012-10-06T06:56:48Z

thanks, Alexander

Comment by kenneth

2012-10-04T14:45:10Z

thanks for your answer

Comment by kenneth

2012-09-26T05:32:52Z

Thank you. I also actually found that from that book.

Comment by kenneth

2012-08-03T12:04:51Z

@Mateusz, Thanks for your answer

Comment by kenneth

2012-08-03T03:31:10Z

@Steve, It's helpful to have the paper. I guess you are saying Cor 3.6 of the paper. However, the probability is about the probability of $Y_u$ hits zero during $0\le u \le t$. In particular, if $y = 0$, then this probability is 1 always. In fact, we need the same hitting probability during $0< u \le t$.

Comment by kenneth

2012-08-03T02:11:07Z

@George, I apologize for [Q2], which seems trivial. I meant to ask, instead [Q2'] If $\tau = \infty$ almost surely?

Comment by kenneth

2012-08-02T13:02:02Z

@Mike, could you be more specific? I have difficulty use scale function, since $\sigma^{-2}$ may not be locally integrable at $0$. I am assuming initial $y = 0$.

Comment by kenneth

2012-07-27T13:30:41Z

@George, oops, you are right.

Comment by kenneth

2012-03-11T04:32:30Z

GH. thanks for your answer.

Comment by kenneth

2011-12-28T06:21:58Z

@James, Thanks for the proof. It seems correct to me. I originally confused why we only need to show uniform upper bound for some $0<a<1$, not for all $a$. Levy Zero-one law plays a role here.

Comment by kenneth

2011-12-28T06:20:09Z

@George, Thanks for the explanations, it does help.

Comment by kenneth

2011-12-27T06:07:56Z

@George, I appreciate if you can post a short hint in the case of $\sigma$ is bounded away from zero. Thanks.

Comment by kenneth

2011-12-26T14:56:29Z

Hi James, Thanks for the hint. Unfortunately, I could not follow your idea completely. Could you make it a bit more detail?

Comment by kenneth

2011-12-26T07:10:17Z

Dear James, Elegant example. In the above,$\sigma_t$ in the first half shall be $\sigma_t = \sqrt{\alpha_n}$? Also, if we restrict $\alpha_t >K>0$ for some constant $K$, then is it possible to prove the above claim $P(Y_1 = c) = 0$?

Comment by kenneth

2011-12-26T03:37:31Z

Dear Yuri, $\sigma_t$ shall be given bounded as condition, but your example has to make $\sigma_t$ blow up near 1, as you mentioned.