Jump Process - Random Walk 
A 1-D random walker strarting at time $t=0$ and location $x=0$, moves to the right ($x+1$) or the left ($x-1$) according to independent random variables $R_1,R_2,\ldots$ and $L_1,L_2,\ldots$, such that the $k^{\mathrm{th}}$ jump to the right occurs at the time $\sum_{i=1}^{k} R_i$ and the $k^{\mathrm{th}}$ jump to the left occurs at the time $\sum_{i=1}^{k} L_i$. Assume $R_i$s and $L_i$s are samples of the same probability density function $f(x)$. Show that the probability that the location of the random walker remains $x\leq M$ after the first $N$ steps to the right, tends to $1-\delta$, for all $\delta>0$, as $N, M \to\infty$, as long as $M=\mathcal{O}(\sqrt{N})$.

My Solution: If $f(x)=\lambda e^{-\lambda x}$, the memorylessness of exponential random variables makes this problem equivalent to a symmetric random walk, then we can find the survival probability of a random walk and use the Brownian motion limit to prove this (see Survival Probability in here). How about the general $f(x)$?
I think we make it equivalent to another Brownian motion, I don't know how to find the parameters of that Brownian motion.
 A: I agree with Shalop's comments, in particular, we need $M$ to grow faster than $\sqrt{n}$ to get a probability that converges to 1.   Here is an analysis of the general problem. 
Assume that $\{R_1, L_1, R_2, L_2, ...\}$ are i.i.d. positive random variables with mean $E[L]$ and variance $\sigma^2$. Assume $0<E[L]<\infty$, $0<\sigma^2<\infty$.  Let $X(t)$ be the (integer) location at time $t \geq 0$, with initial condition $X(0)=0$. Define
\begin{align}
t^R_n &= R_1 + R_2 + … + R_n\\
t^L_n &= L_1 + L_2 + … + L_n\\
N^R(t) &= \mbox{Number of right arrivals during $[0,t]$} \\
N^L(t) &= \mbox{Number of right arrivals during $[0,t]$} 
\end{align}
Then $X(t) = N^R(t) - N^L(t)$ for all $t \geq 0$ and 
$$ \boxed{X(t^R_n) = n - N^L(t^R_n)} \quad (Eq. *) $$
Now let $\{M_n\}_{n=1}^{\infty}$ be any sequence of non-negative integers that satisfies
\begin{align}
&n-M_n \geq 1 \quad \forall n \in \{1, 2, 3, ...\}\\
&\lim_{n\rightarrow\infty} \frac{M_n}{\sqrt{n}} = c \in [0, \infty)
\end{align}
We want to show that 
$$ \boxed{\lim_{n\rightarrow\infty} P[X(t^R_n) > M_n] = \int_{\frac{cE[L]}{\sqrt{2\sigma^2}}}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx}$$
We will use the fact that for all $t \geq 0$ and all positive integers $k$ we have
$$ \boxed{P[N^L(t)<k] = P[L_1 + … + L_k > t]}  \quad (Eq. **) $$

Define for each $i \in \{1, 2, 3, …\}$ 
\begin{align}
G_i &= \frac{1}{\sqrt{i}}\sum_{j=1}^i (L_i-E[L])\\
H_i &= \frac{1}{\sqrt{i}}\sum_{j=1}^i (R_i - E[L])
\end{align}
Then $\{G_i\}_{i=1}^{\infty}$ and $\{H_i\}_{i=1}^{\infty}$ are independent, and the central limit theorem tells us that $G_i$ and $H_i$ both converge in distribution to a Gaussian with 0 mean and variance $\sigma^2$. We have
\begin{align}
P[X(t^R_n)>M_n] &\overset{(a)}{=} P[n-N^L(t^R_n) > M_n] \\
&= P[N^L(t^R_n) < n-M_n] \\
&\overset{(b)}{=} P[L_1 + … +L_{n-M_n} >t^R_n]\\
&\overset{(c)}{=} P[L_1 + … + L_{n-M_n} > R_1 + … + R_n] \\
&\overset{(d)}{=}P\left[\sum_{j=1}^{n-M_n}(L_i-E[L]) > M_nE[L] + \sum_{j=1}^n (R_i-E[L])\right]\\
&\overset{(e)}{=} P\left[G_{n-M_n} > \frac{M_nE[L]}{\sqrt{n-M_n}} + \left(\frac{\sqrt{n}}{\sqrt{n-M_n}}\right)H_n  \right]
\end{align}
where (a) holds by  (Eq *); (b) holds by (Eq **) with $t=t^R_n$ and $k=n-M_n$; (c) holds by definition of $t^R_n$; (d) holds by subtracting $(n-M_n)E[L]$ from both sides of the inequality inside the $P[\cdot]$; (e) holds by dividing both sides of the inequality inside the $P[\cdot]$ by $\sqrt{n-M_n}$ and using definitions of $H_i$ and $G_i$.  Define 
$$ W_n = G_{n-M_n} - \left(\frac{\sqrt{n}}{\sqrt{n-M_n}}\right)H_n$$
Then
$$ P[X(t^R_n)>M_n] = P\left[W_n > \frac{M_nE[L]}{\sqrt{n-M_n}}\right]$$
Note that $\lim_{n\rightarrow\infty} \frac{M_n}{\sqrt{n}} = c \in [0, \infty)$ implies the following: 
$$ \lim_{n\rightarrow\infty} \frac{\sqrt{n}}{\sqrt{n-M_n}} = 1 \quad ,\quad  \lim_{n\rightarrow\infty} \frac{M_nE[L]}{\sqrt{n-M_n}} = cE[L] \quad, \quad \lim_{n\rightarrow\infty} (n-M_n) = \infty$$
Since $G_{n-M_n}$ and $H_n$ are independent and individually converge in distribution to a Gaussian with mean $0$ and variance $\sigma^2$,  we know that  $W_n$ converges in distribution to a Gaussian with mean $0$ and variance $2\sigma^2$. Hence
$$ \lim_{n\rightarrow\infty} P[X(t^R_n)>M_n] = P[W > cE[L]]$$
where $W$ is Gaussian with mean 0 and variance $2\sigma^2$. $\Box$
