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.