Biased Random Walk and PDF of Time of First Return

I have a random walk process where each step the probability of $+1$ is $p$ and $-1$ is $q$, with $p+q=1$. $p$ may not equal $q$. The walker starts at zero. I want to know the probability that the first return to zero occurs at $t$. Obviously, the number of increases will equal the number of decreases and both will equal $t/2$. So obviously $t$ must be even.

So far, all I have seen is the unbiased random walk PDF for first return time and that when $p>q$ there is a probability that the walker may ever reach zero. It seems like that for any finite $t$, it should simply be a question of getting the right number of permutations. Or perhaps it is as easy as a rewrite of the unbiased PDF?

Additionally, if possible, I’d like to condition on the first step being to $1$.

Thank you!

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You are correct. It boils down to calculating the correct number of “permutations” or walks. I will leave you to figure out how to get the exact probability; here I only do the combinatorial problem.

We want to count the number of walks of length $t$ that start and end at $0$, such that the walk never visits $0$ in between. For convenience, we will restrict the first step to be $+1$; you will need to double the answer we get.
We can naturally represent any walk as a word over the alphabet $\{ +, – \}$, where the sign tells us whether the step is in the positive or negative direction.

In this notation, we are counting the number of words such that

1. the length of the word is $t$,
2. the starting symbol is $+$ (why?),
3. the number of ‘$+$’ and ‘$-$’ are equal (why?),
4. for any proper prefix of the word, the number of $+$ strictly exceeds the number of ‘$-$’ (why?).

Clearly, the last symbol must be a ‘$-$’. Let us delete the starting ‘$+$’ and ending ‘$-$’ symbol of the word, to get a subword of length $t-2$. Obviously, the original word is in one-to-one correspondence with the subword.

The final idea here is to recognize that the subwords themselves form the so-called Dyck words of length $t-2$. (A Dyck word is one with equal number of $+$ and $-$ such that every prefix of the word has at least as many ‘$+$’ as ‘$-$’.) It is a standard result that the number of Dyck words of length $t-2$ is equal to the Catalan number
$$C := \frac{1}{(t/2)}\binom{t-2}{\frac{t}{2}-1}.$$

Can you take it from here?

Added. Let $E$ be the event that the first return time is $t$. If you want to calculate the probability of $E$ conditioned on the first step being $+1$, you can use the usual definition:
$$\frac{\Pr[ E \wedge \text{ first step is +} ]}{\text{first step is +}} = \frac{C \cdot (pq)^{t/2}}{p}.$$

A standard approach for a probabilist here is to rely on generating functions. That is, fix $|s|<1$ and consider $u_x=\mathrm E_x(s^T)$, where the subscript $x$ means that the random walk starts from $x$ and where $T$ denotes the first return to $0$ of the random walk, that is, $T=\inf\{n\ge1\mid X_n=0\}$ and $(X_n)_n$ denotes the random walk starting from $X_0=x$ and making steps $+1$ with probability $p$ and $-1$ with probability $q=1-p$.

What you are asking amounts to computing $u_0$ and $u_1$, we shall explain how to compute $u_x$ for every integer $x$.

Taking into account the first step of the walk, one sees that $u_0=s(pu_1+qu_{-1})$, $u_1=s(pu_2+q)$ and $u_{-1}=s(p+qu_{-2})$. Furthermore, starting from $x=2$, to reach $0$ one must first hit $1$ and then starting from $1$ one must hit $0$. The times to hit $1$ starting from $2$ and to hit $0$ starting from $1$ to are i.i.d. hence $u_2=u_1^2$. Likewise, $u_{-2}=u_{-1}^2$ (and in fact $u_x=u_1^x$ and $u_{-x}=u_{-1}^x$ for every positive $x$).

This shows that $u_0=s(pu_1+qu_{-1})$ where $u_1=s(pu_1^2+q)$ and $u_{-1}=s(p+qu_{-1}^2)$. Solving this for $u_1$ and $u_{-1}$ yields
$u_1=\dfrac{1\pm\sqrt{1-4pqs^2}}{2sp}$ and a similar formula for $u_{-1}$ but since one wants that $u_1$ and $u_{-1}$ stay bounded when $s\to0$, the $\pm$ signs are in fact $-$ signs and
$$u_0=1-\sqrt{1-4pqs^2},\qquad u_1=\frac{u_0}{2sp},\qquad u_{-1}=\frac{u_0}{2sq}.$$
The PDF of $T$ starting from $0$, $1$ and $-1$ is fully encoded in $u_0$, $u_1$ and $u_{-1}$ respectively. For example, the probability to come back at $0$ starting from $0$ and to hit $0$ starting from $1$ and $-1$ respectively, are the limits of $u_0$, $u_1$ and $u_{-1}$ when $s\to1$, that is,
$$\mathrm P_0(T<\infty)=1-\sqrt{1-4pq}=1-|1-2p|,$$
and
$$\mathrm P_1(T<\infty)=\frac1{2p}\mathrm P_0(T<\infty),\quad \mathrm P_{-1}(T<\infty)=\frac1{2q}\mathrm P_0(T<\infty).$$
Let us assume from now on that $p\ge\frac12$ (hence $p\ge q$). Then,
$$\mathrm P_0(T<\infty)=2q,\quad \mathrm P_1(T<\infty)=q/p,\quad\mathrm P_{-1}(T<\infty)=1.$$
Likewise, to get the full PDF only requires to know the expansion along powers of $t$ of the square root involved, namely,
$$\sqrt{1-4t}=1-\sum\limits_{n=1}^{+\infty}c_nt^n,\quad c_n=\frac{(2n)!}{(2n-1)(n!)^2}.$$
One gets for example
$$u_1=\frac1{2ps}\sum\limits_{n=1}^{+\infty}c_n(pq)^ns^{2n},$$
hence, for every $n\ge1$,
$$\mathrm P_1(T=2n-1)=\frac1{2p}c_n(pq)^n.$$
Likewise, for every $n\ge1$,
$$\mathrm P_{-1}(T=2n-1)=\frac1{2q}c_n(pq)^n,$$
and finally,
$$\mathrm P_0(T=2n)=c_n(pq)^n.$$