Expected values when sampling is dependent

Question

Let $W$ be a standard Brownian motion. Let $\{t_{j,n}\}$ be the sequence of stopping times defined recursively as

$$ \begin{cases} t_{0,n} = 0\\ t_{j+1,n} = \inf\{t\geq t_{j,n}| W_t-W_{t_{j,n}}\geq n^{-1/2}\,a\textrm{ or }W_t-W_{t_{j,n}}\leq -n^{-1/2}\,b\} \end{cases} $$ with $a>0$ and $b>0$. Hence $W_{t_{j+1,n}}-W_{t_{j,n}}=n^{-1/2}\,Z_j$ with $Z_j$ iid sequence with mass in $a$ and $-b$ (with probability $p_a\doteq b/(a+b)$ and the complement $p_{-b}=1-p_a$). By using the Martingale property of $W_t$, $W_t^2-t$, $W_t^3-3\,t\,W_t$, etc... I can compute covariances of the type $$\mathbb{E}\left[(t_{j+1,n}-t_{j,n})\,(W_{t_{j+1,n}}-W_{t_{j,n}} )|\mathcal{F}_{t_{j,n}}\right]=\frac{1}{3}\,\mathbb{E}\left[(W_{t_{j+1,n}}-W_{t_{j,n}} )^3|\mathcal{F}_{t_{j,n}}\right]=\frac{1}{3}\left(a^3p_a-b^3\,p_{-b}\right),$$ which is a simple function of $a$ and $b$. My problem now is to extend the computation to covariances of the kind $$\mathbb{E}\left[(t_{j+1,n}-t_{j,n})^{-1/2}\,(W_{t_{j+1,n}}-W_{t_{j,n}} )^3|\mathcal{F}_{t_{j,n}}\right]$$ which cannot appears using only the martingale $t^{n/2}\,H_n\left(\frac{W_t}{\sqrt{t}}\right)$, where $H_n$ is the n-th Hermite polynomial. Any ideas?

SOME LATER THOUGHTS

From this post I can compute

$$\mathbb{E}\left[(t_{j+1,n}-t_{j,n})^{-1/2}|W_{t_{j+1,n}}-W_{t_{j,n}}=+\alpha\right]=\alpha\,\int_{0}^{\infty}t^{-1/2}\,\frac{\exp(-\alpha^2/(2\,t))}{\sqrt{2\,\pi\,t^{3}}}=\frac{1}{\alpha}\,\sqrt{\frac{2}{\pi}}$$ whence, using $\alpha=n^{-1/2}\,a$ and $\beta=n^{-1/2}\,b$ we get $$ \mathbb{E}\left[(t_{j+1,n}-t_{j,n})^{-1/2}\,(W_{t_{j+1,n}}-W_{t_{j,n}} )^3|\mathcal{F}_{t_{j,n}}\right] = \alpha^3\,\frac{1}{\alpha}\,\sqrt{\frac{2}{\pi}}-\beta^3\,\frac{1}{\beta}\,\sqrt{\frac{2}{\pi}} = n^{-1}\,\sqrt{\frac{2}{\pi}}\left(a^2-b^2\right)\quad(1) $$

can someone confirm?

SOME OTHER THOUGHTS

I think equation (1) is wrong. Using the fact that $W_t^2-t$ is a martingale we get

$$ \mathbb{E}_{t_{j,n}}\left[(t_{j+1,n}-t_{j,n})\right] = \mathbb{E}_{t_{j,n}}\left[(W_{t_{j+1,n}}-W_{t_{j,n}})^2\right] = \frac{a\,b}{n} $$ However, using the density of the first exit time of a Brownian motion we get

$$\mathbb{E}_{t_{j,n}}\left[(t_{j+1,n}-t_{j,n})\right]=+\infty$$

Hence, the correct density of $(t_{j+1,n}-t_{j,n})$ is not that of the first hitting time, but something more complicated (which I don't know).

Answer 1

An explicit closed form for $(\Delta T)^{-\frac{1}{2}} [\Delta W]^3,$ where $\Delta T = t_{j+1,n} - t_{j,n}$ is the random exit time of the Brownian motion from the interval $(-b n^{-1/2}, a n^{-1/2})$, and $\Delta W = W_{t_{j+1,n}} - W_{t_{j,n}}$ is the corresponding increment of the Brownian motion:

$$n^{-\frac{3}{2}} \frac{\sqrt{\pi}}{2(a+b)} \Big[ a^3 \tan\left(\frac{\pi b}{2(a+b)}\right) - b^3 \tan\left(\frac{\pi a}{2(a+b)}\right) \Big]$$