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Feb
1
comment Prove that the limit in probability of normally distributed random variables is normally distributed, too
@G.Sassatelli A sequence converges in probability to a random variable if and only if every subsequence has a further subsequence which converges to that random variable almost surely, so the questions are really the same. The answer in that question also only deals with limits in distribution, which is implied by convergence in probability.
Jan
31
comment Prove that the limit in probability of normally distributed random variables is normally distributed, too
math.stackexchange.com/questions/232540/…
Jan
28
comment Prove that the stochastic process can not have continuous paths.
It looks like you are missing a hypothesis. The constant function $W_t = 0$ is a solution with your current conditions. @Ant you do not need any assumptions more than the first and an assumption that the process is not constant.
Jan
22
comment What is the use of moments in statistics
@kjetilbhalvorsen This post is meant to be heuristic, so I was not terribly careful about assumptions. Nevertheless, under the assumption in paragraph 3 (finiteness of the moment generating function on a neighborhood of the origin), the distribution is determined by its moments. I am happy to defer to your better judgment about the quality of sample moments.
Jan
15
comment Reference request: correlation and spectral analysis of stochastic processes
You may want to look at Ash and Gardner's "Topics in Stochastic Processes." These topics are covered rigorously, though perhaps not in the level of generality you might want for all applications.
Nov
4
comment Tightness of a vector valued sequence of stochastic processes
Tight in what space? If you mean the product of Skorokhod spaces (with the usual J1 topology), $D_{\mathbb{R}}[0,\infty)^2$, then yes. If you mean the Skorokhod space of the product (again, J1 topology), $D_{\mathbb{R}^2}[0,\infty)$, then no. Having one component be constant does not help if you mean the Skorokhod space of the product, but having one component be continuous does.
Jan
12
comment Showing that lim sup of sum of iid binary variables $X_i$ with $P[X_i = 1] = P[X_i = -1] = 1/2$ is a.s. infinite
If you just want a hint, try to show the stronger claim that $P(\limsup \frac{S_n}{\sqrt{n}} = \infty) = 1$ using the 0-1 law and the CLT. If you want a full solution using this hint, I wrote it up here: math.stackexchange.com/questions/210131/…
Jan
7
comment Distribution of integral of exponential Gaussian process?
As @Did said, this question is not currently well posed. Even if you mean that $X(t)$ is Brownian motion, this question is non-trivial. The distribution of $\int_0^T \text{exp}(u X(t))dt$ ($u$ real) has been studied by Matsumoto and Yor: arxiv.org/pdf/math/0511517.pdf I do not recall seeing anything for the complex case in any of their papers, but that is where I would start looking.
Nov
19
comment The measure generated by the Cantor staircase and the intersection of the Cantor set with its translate
Sorry, yes I was not reading carefully enough.
Nov
19
comment The measure generated by the Cantor staircase and the intersection of the Cantor set with its translate
Think of the Cantor set as $\sum_{k=1}^\infty \frac{2x_k}{3^k}$ where $x_k \in \{0,1\}$. This identifies the set with the compact group $\mathbb{Z}_2^{\mathbb{N}}$. The Cantor measure is a Haar measure with respect to that group structure.
Nov
19
comment The measure generated by the Cantor staircase and the intersection of the Cantor set with its translate
I think that this is going to depend on $\alpha$. You can work out a partial result from the fact that the Cantor measure is Haar with respect to its natural group structure, though.
Oct
29
comment The ito integral is gaussian
The distributional limit normally distributed random variables is (possibly degenerate) normal. See math.stackexchange.com/questions/232540/…
Oct
23
comment In frequentism, does every event have a probability?
Not measurable with respect to what? Probability is not analysis. When you specify a sigma algebra on your underlying space in probability, you are saying something about what events are accessible to your experiment.
Oct
23
comment how to related a weakly convergent random variable with its k-th moment
Whenever you have convergence in distribution of real random variables, you can switch probability spaces and make it almost sure convergence. This is called the Skorokhod representation theorem. Your question is then whether or not convergence a.s. implies convergence of integrals, which is not true in general. A necessary and sufficient condition for $L^k$ convergence (which implies what you want) is uniform integrability. This is slightly stronger than just having convergence of the expectations though.
Oct
23
comment how to related a weakly convergent random variable with its k-th moment
This result holds if the family $\{S_n^k\}_{n=1}^\infty$ is uniformly integrable.
Oct
17
comment Spotting mistake: unnecessary given condition
@PatrickDaSilva This is pretty fundamentally different from your example. Preference relations are the first topic covered in a first course in graduate microeconomics using the standard textbook (MWG). I've never heard of any physics program that starts with elliptic curves.
Oct
2
comment Can $\|f\|_p\to\infty$ arbitrarily slowly? (Looking for hints.)
This is a great classical analysis question. Hints: You may assume without loss of generality $\Phi(p)$ is as regular as you like (continuous for example). Consider a step function on a disjoint partition of $(0,1)$ and choose your coefficients and the measure of the sets in the partition carefully.
Sep
5
comment CLT - infinite variance
I haven't thought much about this, but how do you know that the histogram is normal? Is it just the general shape or have you tried testing your samples with normality tests?
Aug
31
comment If $f(\mathbb{C})\subset \mathbb{C}-[0,1]$ then $f$ is constant
Hint: Think about the codomain of the entire function $\frac{1}{f(z)}$.
Aug
12
comment Show $\lim\limits_{n\to\infty}\mathbf E(f(X_n)g(Y))=\mathbf E(f(X)g(Y))$
I do not understand the second paragraph. Why is it clear that the limit as $R \to \infty$ of the supremum is zero? At some point, you need to use the fact that the hypotheses give you tightness, otherwise all you have from using compactly supported functions as your test functions is vague convergence.