# Tag Info

2

Intuitively, you are equating coefficients in the two different representations of the generator: $${1\over m(x)}\left[\left({f'(x)\over s(x)}\right)'-k(x)f(x)\right] = {1\over 2}a(x)^2f''(x)+b(x)f'(x)-c(x)f(x).$$ The left side expands out to $${1\over m(x)}\left[{f''(x)\over s(x)}-{s'(x)f'(x)\over s(x)^2}-k(x)f(x)\right].$$ Therefore $${a^2\over 2} = {... 2 I think so. The martingale X_t is a time-changed Brownian motion: X_t = B_{A_t}, where$$A_T = \int_0^T \sigma^2(t, X_t) dt$$and B_t is some Brownian motion. Now write$$ \int_0^T \mathbb 1_{[-\delta,\delta]}(X_t) dt = \int_0^{A_T} \mathbb 1_{[-\delta,\delta]}(B_s) dA^{-1}_s = \int_0^{A_T} \frac{\mathbb 1_{[-\delta,\delta]}(B_s)}{\sigma^2(A^{-1}_s, ...

2

My second advice for those who work with generators of Markov processes is: Use Dynkin's characteristic operator! [*] If $f$ is in the domain of the generator $L$ of $X$, then $$L f(x) = \lim_{B \to \{x\}} \frac{E_x f(X(\tau_B)) - f(x)}{E_x \tau_B} \, ,$$ where the expression under the limit means, for example, that $B = B(x, r)$ with $r \to 0^+$. The ...

2

The answer is yes, provided that you write your equation in Stratonovich form, rather than Itô form (and assuming that $\mu$ and $\sigma$ are sufficiently smooth in their arguments). The reason is that in one dimension the solution to the Stratonovich equation is a continuous map of $W$ in the sup-norm topology, as observed by Doss in 1977. This breaks in ...

1

As an approximation for a counterexample, consider $X_t=\sin(W_t+1)$. It has the stochastic differential $$dX_t=(-(1/2)\sin(W_t+1))dt+\cos(W_t+1)dW_t$$ with initial condition $X_0=\sin(1)$. Then $X|{\mathcal{S}_\varepsilon}$ converges to a constant function, which is not the solution of the diffusion-less equation $dX_t=(-(1/2)\sin(1))dt$. This equation is ...

1

Breaking the integral into two terms, the first term is simply $E_i\left[ \int_{t_i}^{t_{i+1}}\widehat{Z}_sds\right]$. The second term is $E_i \left[\int_{t_i}^{t_{i+1}} \overline{\widehat{Z}_i} ds\right]$. The term in the expectation is $\mathcal F_{t_i}$ measurable, and so the second term is just $\int_{t_i}^{t_{i+1}} \overline{\widehat{Z}_i} ds.$ The ...

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It seems no one checked this calculation prior to the publication of this paper. The factor $\frac{\lambda}{1-\lambda}$ in (1) seems mistaken, it should be $\frac{2-\lambda}{1-\lambda}$. This has no effect on the finiteness claimed in the lemma. The denominator in (1) indeed takes the given form because all the other summands in that denominator vanish due ...

1

Your statement holds if (and only if) every stopping time is predictable; also iff every optional process is predictable. Roughly, the filtration is such that there are no surprises as time passes. I don't know of a direct condition on $X_t$ ensuring this. Example: Let $T$ be a unit-rate exponential random variable, and define  X_t=\cases{0& \$0\le t<...

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