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Jun
14
comment Strange definition of Ergodicity
Take your sample space to be sequence space that you get for the joint law of all of the $X_i$ by using Kolmogorov's extension theorem and let your measure preserving transformation be the shift map taking you from coordinate $i$ to coordinate $i+1$. I think that this condition says that that shift map is ergodic in the usual definition. Proving that requires a bit of measure theory, since you need to show that the indicator of any invariant set is well approximated by a function of $k$ variables in the right sense. There may be some technical issues that make this a bit weaker though.
Jun
12
reviewed Close Convergence of infinite series involving $\frac{\sin(x)}x$.
Jun
12
reviewed Close Intuitively, why does $\int_{-\infty}^{\infty}\sin(x)dx$ diverge?
Jun
8
comment In Markov inequality proof, why is $\int_a^\infty xp(x) \, dx \ge \int_a^\infty ap(x) \, dx$
I added in bounds of integration to improve the formatting a bit.
Jun
8
revised In Markov inequality proof, why is $\int_a^\infty xp(x) \, dx \ge \int_a^\infty ap(x) \, dx$
improving the notation
Jun
6
reviewed Close Sum of odd numbers always gives a perfect square.
Jun
3
comment conditional expectation of brownian motion
This really is a lovely paper, thank you for posting it.
May
31
comment Expression for $B_1$
There is a similar formula which you can obtain through an enlargement of filtration argument. That formula is $B_1 = W_1 + \int_0^1 \frac{B_1 - B_t}{1-t}dt$, where $W_t$ is Brownian motion with respect to the filtration of $B_t$ enlarged by $\sigma(B_1)$. I do not know whether or not that is what you are looking for.
May
27
comment Proof of a theorem about Baire categories
@Demons94 sorry for the bad hint.
May
27
reviewed Leave Open Integrals in analysis and category theory
May
27
comment Proof of a theorem about Baire categories
@DaveL.Renfro Yes thanks, I definitely did not think that through carefully.
May
23
reviewed Leave Open How to prove $\theta$ which $\in(0,1)$ is unique & $\theta$ is related with $x$ &$\lim_{x\rightarrow+\infty}\theta=1$
May
19
comment Defining the scale function of a diffusion process
I suppose I should have said this in my previous comment. The scale function is not defined at zero for this process. If you want to define it consistently as a limit of $s(\epsilon)$ where $\epsilon \downarrow 0$ then the value is $\infty$, not $0$. Notice that if you formally assume $\frac{1}{\infty} = 0$ and $\frac{\infty}{\infty} = 1$, this still gives you the correct answer for this problem, which is that $P(\tau_0 < \tau_5) = 0$ and $P(\tau_5 < \tau_0) = 1$.
May
19
comment Frechet/Gateaux differentiability of an integral operator L^2 --> R
I don't think that this operator is well defined in general. Take a function which is just barely in $L^2$ and make it so that in any neighborhood of zero it is singular like $x^{-\frac{1}{2} + \epsilon}$ with terms that are both positive and negative. When you cube this function ($f(x) = x^3$), the positive part and the negative part will both be infinite, so the functional isn't well defined.
May
18
reviewed Close How to find the limit by Lebesgue Dominated Convergence Theorem
May
16
comment Defining the scale function of a diffusion process
There is a very simple reason for what you are seeing: this process almost surely cannot hit zero if you start from a non-negative number. I will leave this as as comment in case someone wants to write out the details, but the gist of it is that your drift is always a single positive number and your noise will vanish as $X$ gets close to zero. Therefore if the process ever gets close to zero, the drift pushes it away from zero.
May
15
reviewed Leave Open Interesting “real life” applications of serious theorems
May
10
comment Sample continuity of Brownian motion
@Did Indeed, one should also check that it is possible to define a process with such-and-such marginals. There are circumstances where the extension theorem fails, after all. Whether or not the definition makes sense is another question entirely. My point is that continuity is part of the definition and if you do not include continuity in the definition of Brownian motion, you cannot prove continuity from the other parts of the definition. The statement that all Gaussian processes with the same marginals as Brownian motion are continuous is false.
May
9
comment Sample continuity of Brownian motion
@Did I would say one shows that there is an equivalent process which is built by hand to be pathwise continuous, which is meaningfully different. Take a continuous Brownian Motion $B(t)(\omega)$ and consider the process $B(t)(\omega)1_{B(t)(\omega)\text{ is irrational} }.$ It is not hard to check that this satisfies the other conditions of the standard definitions of Brownian Motion, but is almost surely pathwise almost everywhere discontinuous. The event that a stochastic process is continuous (even at a point) is not measurable unless one works on a sample space where one assumes continuity
May
9
comment Sample continuity of Brownian motion
You have to take continuity as part of the definition of both. There are measurability issues which prevent one from asking about the probability that a function is continuous unless you build the function to be continuous by hand. For both regular and fractional Brownian motion, you can use Kolmogorov's Continuity Theorem to show that a continuous version exists and then we define that version to be the (fractional) Brownian motion.