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comment Smoothness requirement for Stratonovich Integral
I don't follow how you got $f'(X_t)\sigma(t) \circ dB_t = f'(X_t)\sigma(t) dB_t + \frac{1}{2}f''(X_t)\sigma^2(t) dt$. In particular, how did you show $df'(X_t) = f''(X_t)dX_t + BV$ without knowing $f' \in C^2$? The point of the question was to determine this detail.
Aug
23
comment If $E[(X \wedge b) \vee a] = E[(Y \wedge b) \vee a]$ all $a\leq b$ is $X =^d Y$?
So simple I don't know how I missed it!
Aug
23
accepted If $E[(X \wedge b) \vee a] = E[(Y \wedge b) \vee a]$ all $a\leq b$ is $X =^d Y$?
Aug
23
asked If $E[(X \wedge b) \vee a] = E[(Y \wedge b) \vee a]$ all $a\leq b$ is $X =^d Y$?
Aug
21
comment Smoothness requirement for Stratonovich Integral
I don't have either of those books, if you would like to type of the details and cite your reference, I would certainly reward you with the bounty.
Aug
21
awarded  Investor
Aug
18
comment Prove that $\int f(x)g(y) \,d(\mu \times \nu) = [\int f\, d\mu ][\int g \,d\nu]$
Ahh, the wiki en.wikipedia.org/wiki/Product_measure provides the answer in the affirmative.
Aug
18
comment Prove that $\int f(x)g(y) \,d(\mu \times \nu) = [\int f\, d\mu ][\int g \,d\nu]$
What do you mean by $\mu \times \nu$ if the spaces are not $\sigma$-finite?
Aug
16
answered Reference book for Brownian Motion
Aug
16
comment Random Walk Limit Behavior
That is correct. recall that $S_n(\omega)$ converges if and only if $\limsup S_n(\omega) = \liminf S_n(\omega)$. Since 4. occurs with probability $1$, this implies that with probability $1$, $S_n$ does not converge.
Aug
16
answered If $\lim\limits_{x\rightarrow 0}(f(x)\cdot y)=0\space\forall y\in\mathbb{R}^n$, show that $\lim\limits_{x\rightarrow 0}f(x)=0$
Aug
16
comment Random Walk Limit Behavior
That is correct, $P(S_n \to \infty) = 0$ because case 2 and case 4 are disjoint, and case 4 happens with probability $1$.
Aug
16
answered Random Walk Limit Behavior
Aug
16
comment Proving $\lim\limits_{x \to \infty} xf(x)=0$ if $\int_{0}^{\infty}f(x) dx$ converges.
$\int_0^b \sin(x)/x$ converges to $\pi/2$ but $x (\sin(x)/x) = \sin(x)$ does not converge to $0$. Can you see why limit comparison doesn't work?
Aug
16
comment Continuous-time Wiener Process probability question
The first equation is simply dividing both sides by $\sqrt{t}$. Since $X(t)$ is $N(0,t)$ distributed, $X(t)/\sqrt{t}$ is $N(0,1)$ distributed. If you want to take the limit as $t \to \infty$ just do so. From the right side of the equation we see the limit is $1 -( 0 - 0) = 1$. I.e. $P[|X(t)|>1] \to 1$ as $t \to \infty$.
Aug
16
answered Continuous-time Wiener Process probability question
Aug
15
accepted Weak $L^p$ implies strong $L^q$ for $q<p$
Aug
15
revised Weak $L^p$ implies strong $L^q$ for $q<p$
added 8 characters in body
Aug
15
asked Weak $L^p$ implies strong $L^q$ for $q<p$
Aug
15
answered Is “random variable” really random?