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| bio | website | math.wisc.edu/~janjigia |
|---|---|---|
| location | Madison, WI | |
| age | ||
| visits | member for | 1 year, 5 months |
| seen | 2 hours ago | |
| stats | profile views | 380 |
I am a graduate student in the probability group of the mathematics department at the University of Wisconsin - Madison.
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Mar 1 |
answered | Distribution of $\int_0^t e^s dB(s)$ |
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Feb 27 |
comment |
Are vague convergence and weak convergence of measures both weak* convergence? Take a compactly supported function on that set plus a little $\epsilon$ ball which agrees with $f$ and smoothly tapers to $0$ on that $\epsilon$ ball without increasing the sup norm then do a little analysis to show that you're only off by a multiple of $\epsilon$ in the integrals. |
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Feb 27 |
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Are vague convergence and weak convergence of measures both weak* convergence? It's a short jump from that result: let's suppose that relative compactness implies tightness. Here's a sketch of the proof of equivalence. Weak convergence of a sequence of probability measures ${\mu_n}$ clearly implies both relative compactness and vague convergence, which implies tightness. Conversely, if we have vague convergence all that remains to be checked is that the limit measure has mass 1, which follows from tightness: To see that you get convergence for any bounded continuous functions, fix a bounded continuous function $f$ and a compact set with mass $1-\epsilon$ |
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Feb 27 |
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Are vague convergence and weak convergence of measures both weak* convergence? The only issue with vague convergence is that the limit measure need not conserve mass. If you add in the requirement that the limit variable has mass 1, then vague convergence is weak convergence and you can prove tightness (assume you're working in a sufficiently nice space). in (v) you just know that a limit measure exists. You have to prove that it has mass 1 and that there exists a random variable with that distribution. |
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Feb 27 |
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Are vague convergence and weak convergence of measures both weak* convergence? (4) Yes, since the supremum norm closure of $C_K$ is $C_0$. |
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Feb 27 |
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Are vague convergence and weak convergence of measures both weak* convergence? (1) actually I assumed that we are working on a Polish space when I said that. That is the usual assumption for the equivalence of weak convergence and vague convergence + tightness. You can find a proof of that result in Billingsley's Convergence of Probability Measures as theorem 5.2. (2) The condition of tightness is fairly specific to finite measures and you can always assume a finite measure with positive mass is a probability measure by rescaling, so I am not sure what you mean. (3) The same counterexample as above shows that you need tightness. |
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Feb 25 |
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Are vague convergence and weak convergence of measures both weak* convergence? There are a lot of questions here and I don't think I can answer them all, so I'll just leave this part as a comment. A critical fact for probability is that vague convergence of subprobability measures is what you get with $C_0$ or $C_K$ test functions (that is, compactly supported or vanishing at infinity) while weak convergence, which is what you get with $C_b$ test functions, is equivalent to vague convergence plus tightness. Intuitively, compactly supported functions can't tell if mass escapes to infinity, while bounded functions can. Example: $\delta_n \to 0$ vaguely, but not weakly. |
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Feb 23 |
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If $f\in C^1([0,1])$ and $\mu (E)=0$, is it true that $\mu(f(E))=0$, or that $\mu(f'(E))=0$? Also think about the definition of absolute continuity, since this type of question is intimately connected to that definition. |
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Feb 20 |
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If $A$ is measurable does $m^*(B)=m^*(A)$ imply that $B$ is measurable? What exactly is $m*$ here? How are you defining measurable? This is clearly false for the Borel sigma algebra, but is true for Lebesgue and m* the usual outer measure. |
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Feb 4 |
accepted | Fixed-Time Brownian Motion Exit Probabilities |
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Jan 28 |
awarded | Good Question |
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Jan 21 |
answered | Proving $k e^{x}$ is the ONLY solution to the differential equation $f(x) = f'(x)$ |
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Jan 20 |
answered | How do you manage your “pedanticism”? |
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Jan 20 |
awarded | Informed |
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Jan 19 |
asked | Fixed-Time Brownian Motion Exit Probabilities |
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Jan 19 |
comment |
Showing that $\lim_{\delta \to 0^+} \frac{1}{\delta} \int_x^{x + \delta} f(t) \ \mathrm{d}t = f(x)$ This is not true as written. Consider the function $f(0) = 0, f(x) = 1$ otherwise and look what happens if you evaluate the above expression at $0$. You want $f$ continuous. In that case, think about the mean value theorem. |
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Jan 17 |
awarded | Yearling |
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Jan 14 |
answered | $f$ converges then $f'$ converges to $0$? |
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Jan 11 |
answered | Prove that $f(x)=x$ is Riemann-integrable on $[0,1]$ with $\int_0^1xdx=\frac{1}{2}$ |
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Jan 3 |
comment |
Is this a valid method for time-integrating a stochastic process? If you assume that $X_t$ is positive or that the integral of the first absolute moments is finite, then yes. |
