Approximations of hitting times of biased random walks Let $X, X_i$ be iid with mean $\mu$ and variance $\sigma^2$ and $h>0$ be the stopping position. Let $S_n=X_1+...+X_n$ and $T$ be the number of steps it takes to walk beyond h. I need to find the distribution of $T$ in terms of the first two moments of $X$, at least approximately, since in my scenario $$h\gg\sigma\gg\mu.$$ We may assume $X_i$ is well behaved, i.e. having finite moments whenever you want. We have $T\leq n$ iff $max(S_1,...,S_n)\geq h$, but I did not find enough fact about the random variable $max(S_1,...,S_n)$. I only know $S_n$ by central limit theorem. If it is asking too much to find the distribution, I also conjecture that the variance of $T$ can be estimated, perhaps in the similar manner of proving $E(S_T)=E(T)E(X)$.
 A: Consider a standard Brownian motion $B$ and the Brownian motion with drift $Y$ defined by $$Y_t=\sigma B_t+\mu t.$$ Then, when $h$ is large when compared to $\sigma$ and $\mu$, $T$ can be approximated by $$\tau=\inf\{t\gt0\mid Y_t=h\}.
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The distribution of $\tau$ is characterized by its Laplace transform, such that, for every nonnegative $u$, $$E(\mathrm e^{-s\tau})=\exp\left(-\frac{h\mu}{\sigma^2}\left(\sqrt{1+2s\frac{\sigma^2}{\mu^2}}-1\right)\right).$$
Thus, the renormalized hitting time $$\theta=\frac{\mu^2}{\sigma^2}\tau,$$ is such that $$E(\mathrm e^{-s\theta})=\exp\left(-\left(\sqrt{1+2s}-1\right)h\mu/\sigma^2\right).$$ 
For every fixed nonzero $(h,\mu,\sigma^2)$, $$\tau\stackrel{d}{=}\frac{\sigma^2}{\mu^2}\Theta_\lambda,\qquad\lambda=\frac{h\mu}{\sigma^2},$$ where, for every positive $\lambda$, the distribution of the random variable $\Theta_\lambda$ is characterized by its Laplace transform $$E(\mathrm e^{-s\Theta_\lambda})=\exp\left(-\lambda\left(\sqrt{1+2s}-1\right)\right).$$
Thus, $$E(\Theta_\lambda)=\mathrm{var}(\Theta_\lambda)=\lambda.$$
Inverting the Laplace transform, one sees that $\Theta_\lambda$ has density $f_\lambda$, where, for every positive $t$, $$f_\lambda(t)=\frac{\lambda\exp(\lambda-t-\lambda^2/(4t))}{2\sqrt{\pi t^3}}.$$
The regime $h\mu\gg\sigma^2$ corresponds to the limit $\lambda\to\infty$, then one has the convergence in probability $$\frac{\tau}{h\sigma}\to1.$$ Finally, the regime $h\mu\ll\sigma^2$ corresponds to the limit $\lambda\to0$, then one has the convergence in distribution $$\frac{\sigma^2}{h^2}\tau\to\vartheta,$$ where the distribution of $\vartheta$ is characterized by $$E(\mathrm e^{-s\vartheta})=\mathrm e^{-\sqrt{2s}}.$$
Inverting the Laplace transform, one sees that $\vartheta$ has density $g$, where, for every positive $t$, $$g(t)=\frac{\exp(-1/(2t))}{\sqrt{2\pi t^3}},$$ hence $\vartheta=1/Z^2$ where $Z$ is standard normal, in particular $\vartheta$ is not integrable, and $$\tau\approx\frac{h^2}{\sigma^2Z^2}.$$
