The number of paths for a simple random walk from $(0, 0)$ to $(n, v)$ is a classic combinatorial problem in probability theory. In a simple random walk, at each step, the walker either moves up by 1 unit or down by 1 unit. Given this, the total number of steps up and steps down must satisfy certain conditions for the walker to end up at the desired point $(n, v)$.

Conditions:

• The random walk starts at $(0, 0)$.
• After $n$ steps, the position must be $v$, meaning the total displacement is $v$.
• At each step, the walker can either move up (+1) or down (-1).

Solution:

Let the number of steps up be denoted as $U$ and the number of steps down as $D$. Then:

• The total number of steps is $U + D = n$.
• The net displacement after $n$ steps must be $v$, meaning $U - D = v$.

From these two equations, we can solve for $U$ and $D$:

• Adding the two equations: $2U = n + v$, so $U = \frac{n + v}{2}$.
• Subtracting the second equation from the first: $2D = n - v$, so $D = \frac{n - v}{2}$.

Thus, the number of steps up is $U = \frac{n + v}{2}$, and the number of steps down is $D = \frac{n - v}{2}$.

Feasibility Conditions:

For $U$ and $D$ to be non-negative integers:

• $n + v$ must be even (so that $U$ is an integer).
• $n - v$ must also be even (so that $D$ is an integer).
• $U \geq 0$ and $D \geq 0$, which implies $-n \leq v \leq n$.

Combinatorics:

Given that $U$ steps are up and $D$ steps are down, the number of distinct paths is given by the number of ways to choose $U$ upward steps from the $n$ total steps. This is a binomial coefficient: $\binom{n}{U} = \binom{n}{\frac{n+v}{2}}.$

Final Answer:

The number of paths for a simple random walk from $(0, 0)$ to $(n, v)$ is: $\binom{n}{\frac{n+v}{2}},$ subject to the conditions that $n + v$ is even and $-n \leq v \leq n$.

Would you like further details or explanations on any part of the solution?

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Here are 5 related questions that you may find interesting:

1. What is the expected value of the position after $n$ steps in a simple random walk?
2. How does the variance of the position in a simple random walk evolve with $n$?
3. How would the answer change if the walk was biased (i.e., probabilities for up and down are not equal)?
4. How many paths return to the origin $(0, 0)$ after $2n$ steps in a simple random walk?
5. Can you calculate the number of paths to reach $(n, v)$ using generating functions?

Tip: Symmetry plays a crucial role in many random walk problems. For unbiased random walks, paths leading up or down from a given point are equally likely.