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 Nov 22 comment Partial derivative of integral: Leibniz rule? @joriki What I meant was whether you could apply it in the way that littleO applied it. Since you can do so, the answer to my question is yes. I've edited for disambiguation. Thanks. Nov 21 comment Partial derivative of integral: Leibniz rule? @joriki the provided answer seems to suggest that the same formula does hold. Nov 15 comment Integral with respect to Wiener process. @GautamShenoy Ok well I guess I've done 2 calculus/analysis subjects and 1 linear algebra, forgot the first 1 I did. It's in a finance undergraduate degree, they decided to chuck in a quantitative finance course which is all stochastic calculus with a lil measure theory and probability theory. Nov 15 comment Integral with respect to Wiener process. @GautamShenoy What I meant to ask was for you to provide some additional hints - in mathematics. I probably should've couched this in different language. I've done 1 calculus and 1 linear algebra subject at University. Nov 15 comment Integral with respect to Wiener process. @GautamShenoy I'm afraid that this does not assist me given my current understanding of mathematics. I don't think I can use the second fundamental theorem because i just have $d$ instead of $\frac{d}{dt}$ preceding the integral. Nov 10 comment What is $57^{46}$ divided by 17? This answer is hilarious! Nov 10 comment Integration of Wiener process: $\int_{t_1}^{t_2} dB(s)$ @sos440 Is your identity true even for $\displaystyle \ \ \int_a^b \mathscr{\epsilon}(s)dB(s)$? (I don't know how to make that funny symbol you made). Nov 10 comment Linear regression where undershooting isn't as bad as overshooting stats.stackexchange.com would be much more familiar with this literature. Nov 10 comment Solving SDE: $dX(t) = udt + \sigma X(t)dB(t)$ @mike That makes complete sense. But, are you 100% sure this is accurate in this instance? Nov 9 comment Finite p-th variation implies zero-valued q-th variation. @did Thanks. Do you have any tips on how to progress furtheR? Nov 9 comment Demonstrate that every martingale is a local martingale. @StefanHansen I think I don't understand what you mean by "one particular" sequence $(\sigma_n)_{n \in \mathbb{N}}$. Guessing what you mean; we can specify that each $\sigma_n = \text{inf}\{t \geq 0 : X_t = n\}$ where $X_{min(t,\sigma_n)}$ is the stopped process. However I can't see how this can help me solve the problem. Nov 9 comment Why is $\operatorname{sign} B_t$ a predictable process? What is the notation $(t,\omega)\mapsto X_t(\omega)$? Nov 8 comment Demonstrate that every martingale is a local martingale. @StefanHansen Could you please dumb down your criticism of my attempt? I wikipedia'd localization but couldn't relate it to what you're saying. Nov 8 comment Breaking up Wiener processes with indicator functions? Okay, so can I confirm that this is the correct argument: (i)$W_t(\omega) = {1}_{\{W_t(\omega) \geq 0\}}W_t(\omega) + {1}_{\{W_t(\omega) < 0\}}W_t(\omega)$ holds for each $\omega \in \Omega$. (ii)Therefore, $W_t = {1}_{\{W_t \geq 0\}}W_t + {1}_{\{W_t < 0\}}W_t$ is true (almost surely). Nov 8 comment Breaking up Wiener processes with indicator functions? Okay, so you're suggesting that I use the fact that the identities I've stated hold path-wise (i.e. $\text{P-a.s.}$ under filtration equipped probability space with measure $P$), and therefore hold in all possible states of the world, and therefore the expressions are correct as they're stated. Nov 8 comment Breaking up Wiener processes with indicator functions? So this confirms the two identifies path-wise. (i.e. for $W_t(\omega)$ and $|W_t(\omega)|$). However, I'm not quite sure how this can then be extended to demonstrate the case for the full $W_t$ random variable? Nov 8 comment Sum of two stopping times is a stopping time? @did Thanks for your help mate! Nov 8 comment Min of two stopping times is also a stopping time. So you're saying that in fact $\{min(\sigma,\tau)\leq t\} =\{\omega_1 : \sigma(\omega_1) \leq t\} \cup \{\omega_2 : \tau(\omega_2) \leq t\}$ Nov 8 comment Min of two stopping times is also a stopping time. You're right! Thanks. Nov 8 comment Sum of two stopping times is a stopping time? @StefanHansen Okay then I'm lost in this problem unfortunately.