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.