Taylor Martin
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 Apr 21 comment Solve differential equation $f'(x)+f^2(x)=4$ It can be seen in more detail from niemanski's answer. I also included an edit so you can see it from the analytical point of view. You don't need to worry about what Lipschitz means if you haven't studied it yet (it's a stronger form of continuity, and in the theory of ODE, it is a sufficient condition to ensure existence/uniqueness of solutions, at least locally). Jul 1 comment is intersection of a countable collection of dense, open subsets of a complete metric space also dense in X? Apr 10 comment $dx=\frac {dx}{dt}dt$. Why is this equality true and what does it mean? No precise meaning, and none is needed really. The point of my answer (the second part anyway) is to view the differentials in the expression as quantities $\delta x$ and $\delta t$ - when they are finite, you get the same form of the expression plus a small error; as you send the quantities closer to $0$, the error disappears, but so do the quantities - you fix this by dividing and examining the ratio (which in turn leads right back to the first interpretation). Thinking of $dx(\cdot)$ as a linear functional of $\delta t$ is likely beyond the scope of the OP's knowledge. Apr 10 comment $dx=\frac {dx}{dt}dt$. Why is this equality true and what does it mean? Correct, so you didn't read it apparently..... Apr 10 comment $dx=\frac {dx}{dt}dt$. Why is this equality true and what does it mean? I don't follow you - did you read the entire response? Mar 14 comment Let $M$ be a bounded subset of the space $C_{[a,b]}$. Prove that the set of all functions $F(x)=\int^{x}_{a}f(t)dt$ with $f\in{M}$ compact. I saw "$M$ compact" in the title and made that assumption, which makes the proof $\int M$ is compact trivial because of the uniform convergence norm. But I see now the assumption is that $M$ is merely bounded. Jan 17 comment Deriving the definition of stochastic integrals with respect to Ito processes from first principles @TheBridge - Regarding your second comment, the convergence is true for each $t\geq0$ in $L^{p}(\Omega)$ ($1\leq p\leq2$) no matter how you construct the sequence of partitions $\Pi_{n}$ as long as $||\Pi_{n}||\to0$. As is well known, you can construct a particular sequence $\Pi_{n}$ so that for a fixed $\omega$, $\text{QV}^{2}(W)(\omega)=\alpha$ for any $\alpha>0$ including $+\infty.$ However, you can avoid all of this by demanding that $||\Pi_{n}||\to0$ such that $\sum_{n=1}^{\infty}||\Pi||_{n}<\infty$; then by Borel-Cantelli the QV process does converge pathwise $\omega$-a.s. to $t$. Jan 17 comment Deriving the definition of stochastic integrals with respect to Ito processes from first principles @TheBridge - Regarding your first comment, please provide your definition if you disagree with mine; however, I am using the definition that is regularly encountered in mathematical finance texts. Also, there really isn't any need to be pedantic here with the conditions of the coefficient processes; they are more or less implied and in any case you can just assume the ones regularly imposed: adadpability, $L^{2}(\Omega\times[0,t])$ integrable, and $\omega$-a.s. continuous as a function of $t$. Dec 23 comment Is Hoffman-Kunze a good book to read next? "Covers quite a bit more" as in there are multiple topics and extensions of covered topics presented in H&K that are not available in Axler's presentation of the subject. The most noteworthy are systems of linear equations, determinants and multilinear forms. Dec 23 comment What is the explicit obstruction to almost sure convergence in stochastic integrals? I think what's actually amazing is the fact that $\int f\;dg$ exists for all continuous $f$ if and only if $g$ is of bounded variation (uniform boundedness principle). The fact that we have convergence in $d\mathbb{P}$-mean isn't really surprising because of Ito's isometry (which is the only technical part of the proof; once that's established the result is basically an application of well known $L^{p}$ approximation theorems). The question I'm asking is along the lines of what is the nature of the "bad" set $E_{n}$ corresponding to $\Pi_{n}$ that causes pathwise convergence to fail? Dec 21 comment What is the explicit obstruction to almost sure convergence in stochastic integrals? Here's another observation that leads me to believe the usual uniform $t/n$ partition is not sufficient to force pointwise convergence. If $E_{n}$ is the "bad" set on which $X_{n}$ is not close to the limit $X$, and if it marches around the sample space, then it will obstruct pointwise convergence if $P(E_{n})\to0$ sufficiently slow such that there is enough mass to always cover a set of points of positive mass. But Borel Cantelli says that if $\sum P(E_{n})<\infty$, then $E_{n}$ loses mass sufficiently fast that it can't do this, resulting in pointwise convergence; but $\sum n^{-1}=\infty$ Dec 21 comment What is the explicit obstruction to almost sure convergence in stochastic integrals? Since the question is somewhat imprecise, maybe answering this would be easier. Since convergence in probability implies a.s. convergence of a subsequence, if we take a rapidly decreasing sequence of partitions $\Pi_{n}$, we recover pathwise convergence. Is it possible to illustrate two explicit sequences of partitions whereby one results in convergence along both modes and the other just in probability? Furthermore, does the usual $t/n$ partition decrease rapidly enough to achieve a.s. convergence? Dec 3 comment Proving uniform convergence of an integral-defined function on compact sets Ah, then you will have to adjust the limits in the integral accordingly to take into account the domain. But note for instance that if $f$ is supported on $K\subset\mathbb{R}$, then $f_{\epsilon}$ is supported on a subset $K_{\epsilon}\subset K$. The specific range of $f$ is irrelevant, only that it allows for $\int_{\mathbb{R}} f=1$ and that $f\geq0$. Dec 2 comment Proving uniform convergence of an integral-defined function on compact sets Actually, since uniform convergence is usually harder to demonstrate than pointwise convergence, a good starting point would be to prove pointwise convergence first, even if restricted to compact sets. I elaborated on my answer a bit; hopefully it will get you started. Dec 1 comment Proving uniform convergence of an integral-defined function on compact sets In the statement of your problem, does $f$ have compact support? Dec 1 comment Orders of growth of typical sequences Are you using the $\ll$ symbol to mean $P(n)\ll Q(n)$ if there is an absolute constant $C$ such that $P(n)\leq CQ(n)$ or that $P(n)/Q(n)\leq C$ for $n\geq N$ for some $N$ sufficiently large; or is it notation for one quantity being "much smaller" than the other? Also, are the numbers $p_{j},q_{j}$ and $a_{j},b_{j}$ arbitrary pairs with no relation between the indices? Nov 20 comment Sum to infinity of the sum 1/n^2 Look to the right as you type this question and you'll see a ton of similar questions w/ answers. There's even one with literally dozens of solutions. Nov 4 comment Resource on Pathwise Computations Involving Brownian Motion Thanks - I edited my question to take some of this into account. Oct 26 comment Proving that if $f$ is Riemann integrable and $1/f$ is bounded then $1/f$ is Riemann integrable If you wanted to make it all workout to $\epsilon$ at the end, then just refine $P$ (if necessary) so that $|U(P,f)-L(P,f)|<(m_{f})^{2}\epsilon.$ But understand that this is not necessaty, since in general a quantity bounded by $C\epsilon$ for a constant that does not depend on $\epsilon$ is as good as being bounded by $\epsilon$ since both upper bounds can be made arbitrary small by sending $\epsilon\to0$. Oct 26 comment Proving that if $f$ is Riemann integrable and $1/f$ is bounded then $1/f$ is Riemann integrable No - consider the case where $(m_{f})^{-2}<1$.