consider the simple random walk (Sn) with S0 = 0 and P(Xi = +1) =
P(Xi = −1) = 1
2 . Find P(S10 > 0)

In a simple symmetric random walk, we have the following scenario:

- The process \((S_n)\) is defined such that \(S_0 = 0\), and each increment \(X_i\) is either \(+1\) or \(-1\) with equal probability (i.e., \(P(X_i = +1) = P(X_i = -1) = \frac{1}{2}\)).
- We are asked to find \(P(S_{10} > 0)\), the probability that after 10 steps, the sum \(S_{10}\) is greater than 0.

This is equivalent to asking for the probability that the random walk is positive after 10 steps. First, let's note that:

\[
S_{10} = X_1 + X_2 + \dots + X_{10}
\]

This is a sum of 10 independent, identically distributed (i.i.d.) random variables, each taking values \(+1\) or \(-1\) with probability \(\frac{1}{2}\).

### Approach:
Since \(S_{10}\) must be an even number (because 10 is even and the number of \(+1\) and \(-1\) steps must add to an even result), we only need to consider the possible values of \(S_{10}\).

The number of ways for \(S_{10} = k\) (where \(k\) is an even integer) is governed by the binomial distribution. Let \(Y\) represent the number of \(+1\) steps in 10 trials, so:

\[
S_{10} = 2Y - 10
\]

We are interested in finding \(P(S_{10} > 0)\), which is the same as \(P(Y > 5)\), because \(S_{10} > 0\) implies more than 5 steps are \(+1\). 

Now, \(Y \sim \text{Binomial}(10, \frac{1}{2})\), and we can calculate:

\[
P(S_{10} > 0) = P(Y > 5) = P(Y = 6) + P(Y = 7) + P(Y = 8) + P(Y = 9) + P(Y = 10)
\]

Each of these probabilities can be computed using the binomial probability formula:

\[
P(Y = k) = \binom{10}{k} \left( \frac{1}{2} \right)^{10}
\]

Let's calculate these values explicitly.

\[
P(S_{10} > 0) = \sum_{k=6}^{10} \binom{10}{k} \left( \frac{1}{2} \right)^{10}
\]

We compute the binomial coefficients:

\[
P(S_{10} > 0) = \frac{1}{2^{10}} \left( \binom{10}{6} + \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10} \right)
\]

Now we calculate the binomial coefficients:

\[
\binom{10}{6} = 210, \quad \binom{10}{7} = 120, \quad \binom{10}{8} = 45, \quad \binom{10}{9} = 10, \quad \binom{10}{10} = 1
\]

So, we get:

\[
P(S_{10} > 0) = \frac{1}{2^{10}} (210 + 120 + 45 + 10 + 1) = \frac{1}{1024} (386)
\]

Finally,

\[
P(S_{10} > 0) = \frac{386}{1024} \approx 0.377
\]

Thus, the probability that \(S_{10} > 0\) is approximately 0.377, or 37.7%.

Would you like further details or have any questions? Here are some related questions to explore:

1. What is the probability that \(S_{10} = 0\)?
2. How would the probability change for \(P(S_{n} > 0)\) for a general \(n\)?
3. How can we extend this method to calculate \(P(S_{n} \geq 0)\)?
4. How does the distribution of \(S_n\) behave as \(n\) becomes large (Central Limit Theorem)?
5. What is the expected value and variance of \(S_n\) in this random walk?

**Tip:** In binomial problems, symmetry often allows us to relate probabilities for large \(n\) to normal distribution approximations for faster calculations.

Consider the simple random walk (Sn) with S0 = 0 and P(Xi = +1) = P(Xi = −1) = 1/2. Find P(S10 > 0)

Explore the probability that S10 > 0 in a simple random walk where S0 = 0 and each step Xi is +1 or -1 with equal probability. This detailed solution uses the binomial distribution to calculate the probability P(S10 > 0) with clear steps and key formulas. Ideal for undergraduate students studying probability theory.

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