In the context of a simple random walk on the integer line, where at each step the walker moves either left or right with equal probability (1/2 for each), we are tasked with finding the probability $P(\text{path goes from } a \text{ to } v \text{ in time } n)$, where $a > 0$ and $v > 0$.

Key Variables:

• $a$ is the starting position.
• $v$ is the target position.
• $n$ is the number of steps.

Solution:

1. Step Count Analysis:

   • For the walker to move from position $a$ to position $v$ in $n$ steps, the walker must take a specific number of steps to the right ($r$) and to the left ($\ell$).
   • Let the total number of right steps be $r$, and the total number of left steps be $\ell$. Since each step is either left or right, we have the relation: $r + \ell = n.$
   • The net displacement after $n$ steps must be $v - a$, which means the difference between right and left steps should equal $v - a$: $r - \ell = v - a.$

2. Solving for $r$ and $\ell$: From these two equations, we can solve for $r$ and $\ell$: $r = \frac{n + (v - a)}{2}, \quad \ell = \frac{n - (v - a)}{2}.$

   • For this to be valid, both $r$ and $\ell$ must be non-negative integers. This means:
     • $n + (v - a)$ must be even, so $n$ and $v - a$ must have the same parity (i.e., both even or both odd).

3. Binomial Coefficient: Given the number of right steps $r$ and left steps $\ell$, the probability of any specific sequence of right and left steps is given by the binomial distribution. The total number of ways to arrange $r$ right steps and $\ell$ left steps in $n$ steps is: $\binom{n}{r} = \binom{n}{\frac{n + (v - a)}{2}}.$ Each such sequence occurs with probability $(1/2)^n$, since each step is independent and occurs with probability $1/2$.

4. Final Probability: The probability of the walker going from $a$ to $v$ in exactly $n$ steps is then: $P(\text{path from } a \text{ to } v \text{ in } n \text{ steps}) = \binom{n}{\frac{n + (v - a)}{2}} \cdot \left(\frac{1}{2}\right)^n,$ where $n$ and $v - a$ must have the same parity for this to be non-zero.

Summary:

The probability depends on the number of steps $n$ and the initial and target positions $a$ and $v$. It is computed using binomial coefficients, and the condition that $n$ and $v - a$ must have the same parity is critical for the solution to exist.

Do you have any questions or need more details? Here are some related questions to explore further:

1. How would the formula change if the random walk was biased, i.e., unequal probabilities for left and right steps?
2. What happens if the walker is allowed to stay in place (i.e., a step with probability $p$ of no movement)?
3. How would this change in higher dimensions, such as a 2D or 3D random walk?
4. What is the expected number of steps to reach $v$ starting from $a$?
5. Can we compute the probability that the path crosses a given position $x$ before reaching $v$?

Tip: Random walks are deeply connected to diffusion processes and Brownian motion, key concepts in both probability theory and physics.