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

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

1. Calculate $E(X^n)$:

The expected value $E(X^n)$ is given by the following integral:

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

Evaluating this integral:

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

So, the expected value is:

$E(X^n) = \frac{1}{n+1}$

2. Calculate $\text{Var}(X^n)$:

The variance of $X^n$ is calculated using the formula:

$\text{Var}(X^n) = E(X^{2n}) - (E(X^n))^2$

First, calculate $E(X^{2n})$:

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

Now, using the previously calculated $E(X^n)$:

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

Simplifying this expression:

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

Thus, the variance is:

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

Summary:

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

Would you like further details or explanations on any part of the process?

Here are 5 related questions you might find helpful:

1. How does the uniform distribution affect expected values?
2. Can you generalize this method for other distributions?
3. What is the relationship between $E(X^n)$ and $E(X^{2n})$ for uniform distributions?
4. How does the variance change as $n$ increases?
5. What is the role of the PDF in calculating expected values?

Tip: When dealing with uniform distributions, the calculations often simplify because the PDF is constant over the given interval.