Suppose a particle is moving in discrete steps. When it is at position $n\ge1$, it takes a step of size $X_n$, and moves to $n-X_n$, where $X_n$ is a random variable taking values in $\{i\in\Bbb{Z}\mid i\le n-1\}$. Suppose there is a non-decreasing function $g:\Bbb{R}^+\to\Bbb{R}^+$ (where $\Bbb{R}^+$ is the set of positive reals). All that we know about the distribution of $X_n$, is that $\Bbb{E}[X_n]\ge g(n)$.
Let $T_n$ be the number of steps the particle takes to reach $1$ (for the first time). How can I calculate an upper bound on $\Bbb{E}[T_n]$ in terms of $g(n)$?
I have solved the problem for a much simpler case, where $X_n$ could only take the values $1,2,\cdots,n-1$. This simpler version can be easily solved via induction, and it turns out
$$\Bbb{E}[T_n]\le\int_1^n\frac{dx}{g(x)}$$
This can be proved via induction. Suppose the statement is true for all integers $< n$. When the particle is at $n$, it takes the first step to go to $n-X_n$, so we have \begin{align*} \Bbb{E}[T_n]&=1+\Bbb{E}[T_{n-X_n}]\\ &\le 1+\Bbb{E}\left[\int_1^{n-X_n}\frac{dy}{g(y)}\right]\\ &\le 1+\Bbb{E}\left[\int_1^{n}\frac{dy}{g(y)}-\int_{n-X_n}^{n}\frac{dy}{g(y)}\right]\\ &\le 1+\int_1^{n}\frac{dy}{g(y)}-\frac{1}{g(n)}\Bbb{E}\left[\int_{n-X_n}^{n}dy\right]\ \ \ \ \ \ \ \ \ \ \ [\text{as}\ g(n)\ \text{is non-decreasing.}]\\ &\le1+\int_1^n\frac{dy}{g(y)}-\frac{\Bbb{E}[X_n]}{g(n)}\\ &\le \int_1^n\frac{dy}{g(y)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\text{as}\ \Bbb{E}[X_n]\ge g(n)] \end{align*}
How do I solve it for the more general case? Thanks in advance.