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bio website math.wisc.edu/~janjigia
location Madison, WI
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visits member for 2 years, 8 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
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
Apr
2
comment For the function $Y = e^{-x}$ where $X$ is $N (0,1)$
What have you tried? Is there a specific point where you are getting stuck?
Mar
25
reviewed Close Determine whether the following series convergent?
Mar
24
comment Why is a brownian motion conditioned to stay positive a Bessel-3
Notice that $P(\min_{0 \leq s \leq t}B_s \geq 0) = 0$, so you have to say what you mean by saying "Brownian motion conditioned to stay positive." In the usual interpretation, this result is Pitman's 2M-B theorem. You can find a number of proofs of that result.
Mar
24
comment Why is $e^\pi - \pi$ so close to $20$?
Yeah, that was silly. I meant 19.9991, sorry. My point though is that it's probably just an accident of the fact that we're calling 20 special.
Mar
24
reviewed Close if the imaginary part of an entire function f is bounded, then f is constant.
Mar
23
reviewed Leave Open Seeking intuitive explanation of Clifford Algebra
Mar
22
reviewed Close Property of Homology and orientation