Here's the question: on each round, Ann and Bob each simultaneously toss a fair coin. Let $X_n$ be the number of heads tossed in the $2n$ flips which occur during the first $n$ rounds. For each integer $m>0$, let $r_m$ denote the probability that there exists an $n$ such that $X_n = m$. Use a renewal theorem to find $\lim_{m\to\infty}r_m.$
My thought is, we can think of $X_n$ as summation of $n$ inter-arrival times of a renew process, each is independent and has mean 1; think of $m$ as a time point. Let $N(m)$ denote max number of hits (in this case each hit is 1 round of coin toss) before time $m$. We have $\lim_{m\to\infty} r_m = \lim_{m\to\infty} P(\exists \ n \ s.t. X_n = m) = \lim_{m\to\infty} P(\exists \ n \ s.t. X_n = m | N(m) = m-1)P(N(m) = m-1)+P(\exists \ n \ s.t. X_n = m | N(m) = m-2)P(N(m)=m-2) = \lim_{m\to\infty} \frac{1}{2} P(N(m) = m-1) + \frac{1}{4}P(N(m)=m-2).$
This is where I got stuck. Now by renew theorem, we have: $\lim_{m\to\infty} \frac{X_{N(m)}}{m} = \lim_{m\to\infty}\frac{X_{N(m)}}{N(m)}\frac{N(m)}{m} = 1 \ w.p.1$. Does this mean $\lim_{m\to\infty} P(\frac{X_{N(m)}}{m-1} = 1) = \lim_{m\to\infty} P(\frac{X_{N(m)}}{m-2} =1) = 1?$ This doesn't make intuitive sense to me since $X_{N(m)}$ can either be $m-1$ or $m-2$, and the two probabilities should sum up to 1, not each equal to 1. Any help on how to proceed is greatly appreciated!
I'm not sure how to apply the renewal theorem to solve this problem (I tried), but you can certainly solve it using a recurrence relation.
Take $A_m$ as the event that $X_1+\dots+X_n=m$ for some $n\geq 1$. Conditioning on $X_1$ yields $$\begin{eqnarray*}r_1 &=& \mathbb{P}(A_1) \\ &=& \frac{1}{4}\mathbb{P}(A_1|X_1=0)+\frac{1}{2}\mathbb{P}(A_1|X_1=1)+\frac{1}{4}\mathbb{P}(A_1|X_1=2) \\ &=& \frac{1}{4}\mathbb{P}(A_1)+ \frac{1}{2} \cdot 1 + \frac{1}{4}\cdot 0 \\&=& \frac{r_1}{4}+\frac{1}{2}\end{eqnarray*}$$ Solving for $r_1$ yields $r_1=\frac{2}{3}$. Similarly, $$\begin{eqnarray*}r_2 &=& \mathbb{P}(A_2) \\ &=& \frac{1}{4}\mathbb{P}(A_2|X_1=0)+\frac{1}{2}\mathbb{P}(A_2|X_1=1)+\frac{1}{4}\mathbb{P}(A_2|X_1=2) \\&=& \frac{1}{4}\mathbb{P}(A_2)+\frac{1}{2}\mathbb{P}(A_1)+\frac{1}{4}\cdot 1 \\ &=& \frac{r_2}{4}+\frac{r_1}{2}+\frac{1}{4}\end{eqnarray*}$$ Solve for $r_2$ and get $r_2=\frac{7}{9}$. Since $\mathbb{P}(A_m|X_1=j)=r_{m-j}$ for $m\geq 3$ and $j\in \{0,1,2\}$ we get $$\begin{eqnarray*}r_m&=& \mathbb{P}(A_m) \\ &=& \frac{1}{4}\mathbb{P}(A_m|X_1=0)+\frac{1}{2}\mathbb{P}(A_m|X_1=1)+\frac{1}{4}\mathbb{P}(A_m|X_1=2) \\ &=& \frac{r_m}{4}+\frac{r_{m-1}}{2}+\frac{r_{m-2}}{4}\end{eqnarray*}$$ This unveils the recurrence relation $$r_m=\frac{2}{3}r_{m-1}+\frac{1}{3}r_{m-2};r_1=\frac{2}{3},r_2=\frac{7}{9}$$ which has unique solution $$r_m=\frac{1}{4}\left(-\frac{1}{3}\right)^m+\frac{3}{4}$$ Taking $m\mapsto +\infty$ gives a limit of $\frac{3}{4}$. I would be very interested to see an approach using a renewal theorem.