Given that $X$ is uniformly distributed over $(0,1)$, we can calculate the expected value $E(X^n)$ and variance $\text{Var}(X^n)$.

1. Expected Value $E(X^n)$

For a uniform random variable $X \sim U(0,1)$, its probability density function (PDF) is:

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

The expected value $E(X^n)$ is defined as:

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

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

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

Thus:

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

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

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

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

• From the first part, we know $E(X^n) = \frac{1}{n+1}$.
• Next, we need to compute $E(X^{2n})$, which is:

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

Now, we can compute the variance:

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

Simplifying:

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

To combine these terms, find a common denominator:

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

Simplify the numerator:

$(n+1)^2 - (2n+1) = n^2 + 2n + 1 - 2n - 1 = n^2$

Thus:

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

Final Results:

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

Would you like further clarification or a detailed explanation on a specific step?

Here are 5 related questions to consider:

1. How does the uniform distribution impact the expected value formula for different functions of $X$?
2. How would you compute $E(X^n)$ for a different distribution, such as exponential?
3. What happens to $\text{Var}(X^n)$ as $n \to \infty$?
4. How would you use integration by parts to solve similar expected value problems?
5. How does the variance change if $X$ is uniformly distributed over a different interval, say $(a,b)$?

Tip: When dealing with expectations involving powers of random variables, leveraging known integral results can save time in calculations.