# asymmetric random walk, hitting time probability

Let's consider an asymmetric Random Walk on $Z$, with transition probabilities $p_{i, i+1}=p$, $~~p_{i, i+1}=q$, $\forall i \in \mathcal{Z}$, $p+q=1$ and $p>q$.

I am interested in the probability of the first hitting time of a bareer in $i=n$, assuming that the walk started from $i=0$ at time $t=0$. Is there an explicit formula for it? How does it depend on $n$?

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This is an exercise. The reflection principle lets you count the walks which first arrive at $n$ at time $t$. These all have the same probability, since they all have $n$ more positive steps than negative steps. – Douglas Zare Mar 11 '13 at 17:28
This has been answered fully in math.stackexchange.com/questions/64919/… – Jeremy Voltz Mar 11 '13 at 17:56
The answer by Did suggests another solution by generating functions but the reflection method is much simpler. – Douglas Zare Mar 11 '13 at 20:14

The explicit formula is: $P[N_m=n]=(m/n)P[S_n=m]$, where $P[N_m=n]$ is the probability the position $m$ is hit after exactly $n$ steps, $S_n = X_1+X_2+\dots X_n$ and $P[S_n=m]$ is the probablity after $n$ steps the path to be at the position $m$. This last is well-known and is given by $P[S_n=m]=(n!/([(n+m)/2]![(n-m)/2]!))p^{(n+m)/2}q^{(n-m)/2}$. This is true when $n$ and $m$ have the same parity, else the probability is zero. Additionally $m$ is positive and $n$ is greater than or equal to $m$. A similar expression can be found for negative $m$.