3,206 reputation
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bio website math.wisc.edu/~janjigia
location Madison, WI
age
visits member for 2 years, 9 months
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I am a graduate student in the probability group of the mathematics department at the University of Wisconsin - Madison.


Apr
18
reviewed Close Proving that the set of irrational numbers is uncountable
Apr
18
reviewed Leave Open Equality of integrals: $ \int_{0}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x = 2 \cdot \int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x $
Apr
18
reviewed Close Mental Math Techniques
Apr
17
comment Is the martingale propertey preserved by taking weak$^*$-limits?
I think what you are calling weak* convergence is typically called strong convergence of measures in the literature. Unfortunately, I know nothing about that and my previous comment is not relevant, sorry. I was assuming your test functions were continuous. If they were continuous on say $\mathbb{R}^d$, then you could use a Skorokhod representation theorem to realize the sequence as an a.s. limit on a single probability space and uniform integrability would imply $L^1$ convergence.
Apr
17
comment Is the martingale propertey preserved by taking weak$^*$-limits?
I was implicitly assuming you were working on locally compact Polish space, since that is fairly standard in probability. I am confused, then. What does weak* convergence mean outside of a topological space?
Apr
17
comment Is the martingale propertey preserved by taking weak$^*$-limits?
It is not clear to me what you mean by weak*-convergence. Are the test functions bounded or vanishing at infinity? Probabilists refer to the former as convergence in distribution and the latter as vague convergence. At the absolute minimum, you are going to need uniform integrability of these measures. If you have that, then the last statement you wrote is easy to show. If not, then there are easy counterexamples with a trivial sigma-algebra (take anything positive which converges a.s. but not in L^1). The first question is similarly true with extra hypotheses but not without them.
Apr
16
comment Are “most” continuous functions also differentiable?
That question is in a finite dimensional setting, and the question plays nicely with countable operations. Probability on infinite dimensional spaces is extremely technical. For what it's worth, I would be shocked if you could find a reasonable interpretation in which the answer was not zero.
Apr
16
comment Are “most” continuous functions also differentiable?
This question is not well posed. You need to define a probability measure to ask the question and defining that measure will answer the question entirely. You can see this from the fact that the event that a function is not differentiable at a single point (much less any point) is going to depend on an uncountable collection of points. To make this question make sense, you push the question onto the sigma algebra or the sample space, rather than the measure. Further,there is no non-trivial translation invariant measure on $C(A)$ for any non-trivial $A$, so it isn't clear what measure to use
Apr
15
reviewed Leave Open Help regarding limit problem
Apr
9
comment Is convergence in probability to a uniformly continuous function a sufficient condition for stochastic equicontinuity?
If $\Theta$ is a compact Hausdorff space and you have uniform convergence in probability, this should follow from the Arzela-Ascoli theorem though.
Apr
9
comment Is convergence in probability to a uniformly continuous function a sufficient condition for stochastic equicontinuity?
I'm pretty sure you can reduce this to a functional analysis problem. I would have to write this out to be sure but I think it follows immediately from the fact that $X_n \stackrel{\tiny{p}}{\to} X$ if and only if every subsequence of $X_n$ has a further subsequence which converges to $X$ almost surely. With this, you can exchange uniform convergence in probability for almost sure uniform convergence. I cannot remember under what exact circumstances uniform convergence implies equicontinuity; I think that without compactness of $\Theta$, it may not be possible to say anything.
Apr
9
comment Is convergence in probability to a uniformly continuous function a sufficient condition for stochastic equicontinuity?
What are your thoughts on this? Have you considered what happens if you take $\Theta = [0,1]$ (which is compact) and $g_T$ to be continuous, non-random, uniformly bounded and pointwise convergent to a continuous function, but not uniformly convergent?
Apr
8
comment Coupled stochastic differential equations?
As written, these are random ODEs. Since Brownian motion (and most other processes built from it) is (Holder) continuous you can solve these pathwise the way you are used to. If you wanted to write this with a time derivative of $B$ there, then you would have a coupled family of stochastic differential equations. At this point, it becomes physically important to decide what type of SDE you mean.
Apr
8
reviewed Leave Open Find $\int \limits_0^\pi \sin(\sin(x))\sin(x)\mathrm dx$
Apr
7
reviewed Close Limits of series, proof of the convergence of two sequences
Apr
4
comment Fixed-time Jumps of a Lévy process
What exactly do you mean by "at fixed times"?
Apr
4
comment Use Ito's Lemma to show:
Can you add a statement of what your version of Ito's lemma is? What you have there is a slight rearrangement of what I would take to be the statement of Ito's lemma applied to f(t)B(t).
Apr
4
reviewed Close Properties of arithmetic functions
Apr
3
reviewed Leave Open $x\rightarrow \int_{0}^{x} \frac{\operatorname{sin}(t)}{t}$ is a bounded function
Apr
3
reviewed Close This is a probability mass function problem