Find the number of paths for the simple random walk which go from (0, 0),
i.e. S0 = 0, to (n, v), i.e. Sn = v


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.

Find the number of paths for the simple random walk which go from (0, 0), i.e. S0 = 0, to (n, v), i.e. Sn = v.

Discover how to find the number of paths for a simple random walk from (0, 0) to (n, v). Learn key combinatorics formulas, such as using binomial coefficients, and understand the conditions for non-negative integer steps. Perfect for students studying undergraduate probability or combinatorics.

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