# jpv

 Mar 26 comment When does compact convergence imply convergence of integrals on the whole space?The example was nice anyway. I do not assume that the functions are Lebesgue Integrable on $[0,\infty)$. Mar 26 comment When does compact convergence imply convergence of integrals on the whole space?Thanks. I realized that this is in general not possible and therefore I have edited the question to "when"? Mar 26 asked When does compact convergence imply convergence of integrals on the whole space? Nov 2 comment Strong Markov property of Ito Diffusion - why must the stopping time be a.s. finite ? (Oksendal 6th edition p117 )I think the reason is that the Ito integrals will not be defined otherwise. For example $\tau+h$ can equal infinity and then there is no definition of Ito integral in an infinite interval. Oct 11 comment Integral of a Gaussian processI will go with this extra assumption for now. Thanks for your effort and time. Oct 11 comment Integral of a Gaussian processThanks. I completely understand the part where the sums are Gaussian and hence their a.s. limit is Gaussian. I am unable to get my head around this: "If $f : [0,T] \to \mathbb{R}$ is Lebesgue measurable and $\int_0^T|f(s)|ds < \infty$, then $f$ is Reimann integrable and both the integrals coincide". If this is false, then we atleast need to show that the Lebesgue integral can be approximated by Riemann sums. Lebesgue integral is defined using partitioning the range of the function so I am unable to see a direct connection. It may be that I am missing some very important knowledge. Oct 10 comment Integral of a Gaussian processI still have some troubles with the proof. For simplicity assume $f \geq 0$, then the simple functions which approximate $f$ are $f_n (t):= \sum_{k=0}^{n2^n-1}(k/2^n)\chi_{[k/2^n \leq f(t) <(k+1)/2^n)}$. Then, $\int_0^T f_n(s)ds = \sum_{k=0}^{n2^n-1}(k/2^n)\lambda[k/2^n \leq f <(k+1)/2^n)$ and it is not directly related to the Riemann sums. Oct 10 accepted Integral of a Gaussian process Oct 10 comment Integral of a Gaussian processsaz: I am assuming that $Y_t(\omega) = \int_0^t X_s(\omega) ds$ exists for all $\omega,t$ in the Lebesgue sense. I think that the Lebesgue integral cannot be approximated by Riemann sums. The way to approximate the Lebesgue integral is by using $Y_t^+,Y_t^-$ and using the fact that these non-negative random variables can be approximated by step functions. However, I cannot see how this entire process of approximation will preserve Gaussianity. Oct 9 comment Integral of a Gaussian processsaz: When the process is not continuous, the Riemann sums might not converge to the integral. Isn't that so? Oct 9 comment Integral of a Gaussian processsaz: Your answer to the linked question uses continuity of Brownian paths to prove this. I am not sure if I am missing something. Oct 9 asked Integral of a Gaussian process Oct 8 comment Prove that integral is a Gaussian random variable, compute its mean and varianceAs $W_s$ is a Gaussian process, all finite linear combinations are Gaussian. The integral in question is a Riemann integral due to continuity of $W_s$ and so $\sum_i W_{t_i} (t_{i+1}-t_i) \to \int_0^t W_s ds$ a.s. Now, the almost sure limit of Gaussian random variables is Gaussain and hence the result follows. Oct 8 comment Prove that integral is a Gaussian random variable, compute its mean and varianceIf $X$ and $Y$ are Gaussian, is it true that $aX+bY$ is also Gaussian? I think $X,Y$ have to be independent for this to be true. Oct 5 comment Stopped Ito IntegralsNo, I am not aware of that. Is that useful for the proof? I shall try to read about it. Oct 4 comment Stopped Ito Integralssaz: Thanks. It would be great if you can give a proof. The book you suggested seems to be very nice. Oct 4 comment Stopped Ito Integralssaz: Thanks for the reference. Oct 3 asked Stopped Ito Integrals Sep 29 awarded Teacher Sep 29 answered Tightness and Inner Regularity