In a random walk with position $X_t$ at time $t$, starting at $X_0 = 0$, the probability of reaching a specific position after $n$ steps depends on the number of steps taken to the right and left. Here’s how we analyze each question:

Part 1: Probability $X_{10} = 10$

To reach $X_{10} = 10$, the random walk must have taken all 10 steps to the right, as any leftward steps would result in a lower position.

1. Since each step has probability $P$ of moving right and $1 - P$ of moving left, reaching $X_{10} = 10$ requires the probability of taking 10 steps to the right with probability $P$ each time.

   $\Pr(X_{10} = 10) = P^{10}$

Part 2: Probability $X_2 = 10$

To achieve $X_2 = 10$ after only 2 steps is impossible. Each step in the walk moves the position by $\pm 1$, meaning after 2 steps, the maximum possible position is $+2$ and the minimum possible position is $-2$. Thus:

$\Pr(X_2 = 10) = 0$

Part 3: Stationarity of the Random Walk

A random walk with biased probabilities $P$ and $1 - P$ is generally not stationary. A stationary process has a distribution that does not change over time, while a biased random walk typically has a drift (a tendency to move in one direction), leading to a non-stationary behavior. A symmetric random walk (where $P = 0.5$) could be stationary in a different sense, as it has no drift and is symmetric over time, but it lacks a fixed distribution.

Summary:

• $\Pr(X_{10} = 10) = P^{10}$
• $\Pr(X_2 = 10) = 0$
• The random walk is not stationary if $P \neq 0.5$.

Would you like more details or have any questions?

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Further Questions:

1. How would you calculate the probability of reaching $X_{10} = 4$ in a similar random walk?
2. What is the probability $X_{10} = -10$?
3. How does stationarity change if $P = 0.5$?
4. Can you explain how the central limit theorem applies to random walks?
5. How do we find the expected position after $n$ steps in this walk?

Tip: For large $n$, the position in a symmetric random walk approaches a normal distribution due to the central limit theorem.