jpv

Hyderabad, India

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 process
I will go with this extra assumption for now. Thanks for your effort and time.
Oct
11
comment Integral of a Gaussian process
Thanks. 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 process
I 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 process
saz: 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 process
saz: 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 process
saz: 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 variance
As $W_s$ 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 variance
If $X$ 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 Integrals
No, I am not aware of that. Is that useful for the proof? I shall try to read about it.
Oct
4
comment Stopped Ito Integrals
saz: 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 Integrals
saz: Thanks for the reference.
Oct
3
asked Stopped Ito Integrals
Sep
29
awarded Teacher
Sep
29
answered Tightness and Inner Regularity
Sep
29
comment Tightness and Inner Regularity
saz:: Thanks. I will try to fill in the details in an answer for record.
Sep
29
accepted Tightness and Inner Regularity
Sep
24
comment Tightness and Inner Regularity
saz: It would be very helpful if you can do that. Thanks.
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