If $x$ is uniformly distributed over $(0, 1)$, we denote it as $x \sim \text{Uniform}(0, 1)$.

1. Expected Value $E(x)$

The expected value of a uniform random variable $x$ over $(0, 1)$ is calculated as:

$E(x) = \int_0^1 x \, dx$

This is a simple integration:

$E(x) = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}$

Thus, $E(x) = \frac{1}{2}$.

2. Variance of $x^n$

To calculate the variance of $x^n$, we first need $E(x^n)$ and $E(x^{2n})$.

a) $E(x^n)$

The expected value of $x^n$ is:

$E(x^n) = \int_0^1 x^n \, dx$

Solving this integral:

$E(x^n) = \left[ \frac{x^{n+1}}{n+1} \right]_0^1 = \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} = \frac{1}{n+1}$

Thus, $E(x^n) = \frac{1}{n+1}$.

b) $E(x^{2n})$

Similarly, for $x^{2n}$:

$E(x^{2n}) = \int_0^1 x^{2n} \, dx = \left[ \frac{x^{2n+1}}{2n+1} \right]_0^1 = \frac{1}{2n+1}$

Thus, $E(x^{2n}) = \frac{1}{2n+1}$.

c) Variance $\text{Var}(x^n)$

The variance of $x^n$ is given by:

$\text{Var}(x^n) = E(x^{2n}) - \left( E(x^n) \right)^2$

Substituting the values we found:

$\text{Var}(x^n) = \frac{1}{2n+1} - \left( \frac{1}{n+1} \right)^2$

Simplifying the expression:

$\text{Var}(x^n) = \frac{1}{2n+1} - \frac{1}{(n+1)^2}$

Final Result

• $E(x) = \frac{1}{2}$
• $\text{Var}(x^n) = \frac{1}{2n+1} - \frac{1}{(n+1)^2}$

Would you like further details on any part of this?

Here are five related questions to consider:

1. How does the uniform distribution affect other moments of $x$?
2. What happens to $E(x^n)$ as $n \to \infty$?
3. Can you derive the skewness for $x^n$?
4. How does the variance of $x^n$ behave for large $n$?
5. What is the median of $x$ when uniformly distributed over $(0, 1)$?

Tip: When working with uniform distributions, the expected value is always the midpoint of the interval, and higher moments follow simple polynomial patterns.