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Assuming that $\mu$ and $S_0$ are positive, the process stays almost surely non-negative. This is easily seen as when $S$ hits zero, it has a deterministic drift upwards. However, the process does not …

Yes, it is explicitly known, but a bit tedious to write it down. You can find it in section 5.1 (compare also section 4) of Paavo Salminen. On the First Hitting Time and the Last Exit Time for a Brow …

Ignoring first the issue of boundary conditions, we note that by adding and subtracting an artificial $\frac{\theta^2}{2}t$ term in the exponential, $\mathbb{E}\bigl[e^{i \theta X_t - \theta Y_t} \bi …

The proof is not correct, as without additional integrability condition you will not be able to conclude that $g_n \in L^2$ for $n$ large enough, and therefore the $L^2$ convergence argument fails. As …

You can solve this by reducing it to a problem of Brownian motion: Define the scale function $\varsigma(x) = \int_{X_0}^x e^{-2\int_{X_0}^y \frac{b(z)}{\sigma^2(z)} dz} dy$ the process $M_t = \var …

A counterexample should be just the deterministic $$Z_t = \frac{1}{\sqrt{T-t}}$$ with $$\tau := \inf{\biggl\{s>0 : \int_0^s Z_u \, dW_u =1\biggr\}}$$. You have $\tau < T$ a.s.and thus $$\mathbb{E}\big …