if x is uniform over (0 1) calculate e(xn) and var(xn)

Given that \( X \) is uniformly distributed over the interval \( (0, 1) \), we know that its probability density function (PDF) is:

\[
f_X(x) = 1 \quad \text{for} \quad 0 < x < 1
\]

We are asked to compute \( \mathbb{E}(X^n) \) and \( \text{Var}(X^n) \).

### 1. Calculation of \( \mathbb{E}(X^n) \)

The expected value \( \mathbb{E}(X^n) \) is given by:

\[
\mathbb{E}(X^n) = \int_0^1 x^n f_X(x) \, dx
\]

Since \( f_X(x) = 1 \), the integral simplifies to:

\[
\mathbb{E}(X^n) = \int_0^1 x^n \, dx
\]

The result of this integral is:

\[
\mathbb{E}(X^n) = \frac{1}{n+1}
\]

### 2. Calculation of \( \text{Var}(X^n) \)

The variance \( \text{Var}(X^n) \) is defined as:

\[
\text{Var}(X^n) = \mathbb{E}(X^{2n}) - \left( \mathbb{E}(X^n) \right)^2
\]

First, we need to calculate \( \mathbb{E}(X^{2n}) \), which follows the same process as above:

\[
\mathbb{E}(X^{2n}) = \int_0^1 x^{2n} \, dx = \frac{1}{2n+1}
\]

Now we can compute the variance:

\[
\text{Var}(X^n) = \frac{1}{2n+1} - \left( \frac{1}{n+1} \right)^2
\]

Simplifying this expression:

\[
\text{Var}(X^n) = \frac{1}{2n+1} - \frac{1}{(n+1)^2}
\]

Thus, the variance of \( X^n \) is:

\[
\text{Var}(X^n) = \frac{(n+1)^2 - (2n+1)}{(2n+1)(n+1)^2}
\]

After simplifying the numerator:

\[
\text{Var}(X^n) = \frac{n^2 + 2n + 1 - 2n - 1}{(2n+1)(n+1)^2} = \frac{n^2}{(2n+1)(n+1)^2}
\]

### Summary:

- \( \mathbb{E}(X^n) = \frac{1}{n+1} \)
- \( \text{Var}(X^n) = \frac{n^2}{(2n+1)(n+1)^2} \)

Would you like more details on any part of the solution?

Here are some related questions for deeper understanding:
1. What are the properties of the uniform distribution over any interval?
2. How does \( n \) influence \( \mathbb{E}(X^n) \) and \( \text{Var}(X^n) \)?
3. What is the variance formula for a general function of a random variable?
4. Can you derive the variance directly from the definition of variance without integration?
5. How would these calculations change if the uniform distribution had different bounds?

**Tip:** When working with moments of random variables, integrals involving powers of \( X \) can often be simplified using fundamental integration rules for polynomials.

Learn how to calculate the expected value and variance of X^n for a uniform distribution over the interval (0, 1). This solution involves key probability concepts such as expectation and variance, with detailed formulas for E(X^n) and Var(X^n). Suitable for undergraduate-level probability courses.

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