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!
 A: 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 


*

*the length of the word is $t$,

*the starting symbol is $+$ (why?),

*the number of '$+$' and '$-$' are equal (why?),

*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: 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,
$$
\mathrm P_0(T=2n)=c_n(pq)^n.
$$
