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The problem you have uploaded involves a Wiener process \( \{ W(t) : t \geq 0 \} \), and it asks for the mean and variance of the random variable \( Y_n \), which is defined as:

\[
Y_n = \sum_{i=1}^{2^n} |W(i2^{-n}) - W((i-1)2^{-n})|
\]

### Key information from the problem:
1. **Wiener Process Properties**:
   - The increments of the Wiener process \( W(t) \) are normally distributed with mean 0 and variance proportional to the length of the time interval.
   - Specifically, for \( 0 \leq s < t \), the increment \( W(t) - W(s) \) follows a normal distribution with mean 0 and variance \( t - s \).

### Steps to solve:

#### 1. **Mean of \( Y_n \)**:
Each term inside the summation is of the form \( |W(i2^{-n}) - W((i-1)2^{-n})| \). From the properties of the Wiener process, the increment \( W(i2^{-n}) - W((i-1)2^{-n}) \) is normally distributed as \( \mathcal{N}(0, 2^{-n}) \).

The expected value of the absolute value of a normal random variable \( Z \sim \mathcal{N}(0, \sigma^2) \) is given by \( \mathbb{E}[|Z|] = \sigma \sqrt{2/\pi} \). Therefore, for each increment:

\[
\mathbb{E}\left[ |W(i2^{-n}) - W((i-1)2^{-n})| \right] = \sqrt{2^{-n}} \cdot \sqrt{\frac{2}{\pi}} = \sqrt{\frac{2^{-n}}{\pi}}
\]

Since there are \( 2^n \) such terms in the summation for \( Y_n \), the mean of \( Y_n \) is:

\[
\mathbb{E}[Y_n] = 2^n \cdot \sqrt{\frac{2^{-n}}{\pi}} = \sqrt{\frac{2^n}{\pi}}
\]

#### 2. **Variance of \( Y_n \)**:
To compute the variance of \( Y_n \), note that each increment \( |W(i2^{-n}) - W((i-1)2^{-n})| \) is independent, since the Wiener process has independent increments. The variance of the absolute value of a normal random variable \( Z \sim \mathcal{N}(0, \sigma^2) \) is given by:

\[
\text{Var}(|Z|) = \sigma^2 \left(1 - \frac{2}{\pi}\right)
\]

For each increment \( W(i2^{-n}) - W((i-1)2^{-n}) \), the variance is:

\[
\text{Var}\left( |W(i2^{-n}) - W((i-1)2^{-n})| \right) = 2^{-n} \left(1 - \frac{2}{\pi}\right)
\]

Since there are \( 2^n \) independent terms, the total variance of \( Y_n \) is:

\[
\text{Var}(Y_n) = 2^n \cdot 2^{-n} \left(1 - \frac{2}{\pi}\right) = 1 - \frac{2}{\pi}
\]

### Summary:
- The **mean** of \( Y_n \) is \( \sqrt{\frac{2^n}{\pi}} \).
- The **variance** of \( Y_n \) is \( 1 - \frac{2}{\pi} \).

Would you like further clarification or more details on any part of this solution?

Here are five related questions to deepen your understanding:
1. How would the result change if the increments of the Wiener process had a different variance?
2. What is the expected behavior of \( Y_n \) as \( n \to \infty \)?
3. Can you explain how the Wiener process properties affect the independence of increments?
4. How would the calculation change if we considered a more general Gaussian process?
5. What applications of Wiener processes are commonly seen in finance?

**Tip**: When dealing with Wiener processes, always remember that the increments are independent and normally distributed with mean 0 and variance proportional to the time difference. This independence simplifies variance calculations in many problems.

Let {W(t) : t ≥ 0} be a Wiener process, then for 0 ≤ s < t the increment W(t) - W(s) is normally distributed with mean 0 and variance t - s. Find the mean and variance of Y_n = Σ_{i=1}^{2^n} |W(i2^{-n}) - W((i-1)2^{-n})|.

Explore how to calculate the mean and variance of Y_n, which involves summing absolute differences in a Wiener process. The solution utilizes key concepts from stochastic processes and probability theory, including the properties of Wiener processes and normal distributions. Suitable for advanced students.

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