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.