Question

If \(X\) is uniform over \((0,1)\), calculate \(E\left[X^{n}\right]\) and \(\operatorname{Var}\left(X^{n}\right)\).

Short answer

The expected value of \(X^n\) is \(E(X^n) = \frac{1}{n+1}\), and the variance of \(X^n\) is \(\operatorname{Var}(X^n) = \frac{1}{2n+1} - \frac{1}{(n+1)^2}\).

Step-by-step solution

Calculate the Expected Value

To calculate the expected value \(E(X^n)\), we need to integrate the product of \(x^n\) and the probability density function \(f_X(x)\) over the support of the random variable \(X\) i.e., from \(0\) to \(1\). Using the definition of expectation for a continuous random variable, we have: \[E(X^n) = \int_0^1 x^n f_X(x) dx\] Since \(f_X(x) = 1\) for \(0 \le x \le 1\), the integral becomes: \[E(X^n) = \int_0^1 x^n dx\]

Solve the Integral

Now, let's solve the integral: \[\int_0^1 x^n dx = \frac{x^{n+1}}{n+1} \Big|_0^1 = \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1}\] \[E(X^n) = \frac{1}{n+1}\] So, the expected value \(E(X^n) = \frac{1}{n+1}\).

Calculate the Second Moment

To calculate the variance, we also need the second moment of the random variable, that is \(E\left[(X^n)^2\right] = E(X^{2n})\). Following the same steps as before: \[E(X^{2n}) = \int_0^1 x^{2n} f_X(x) dx\] \[E(X^{2n}) = \int_0^1 x^{2n} dx = \frac{x^{2n+1}}{2n+1} \Big|_0^1 = \frac{1}{2n+1}\] So, the second moment of \(X^n\) is \(E(X^{2n}) = \frac{1}{2n+1}\).

Calculate the Variance

Using the formula for variance of a continuous random variable, we have: \[\operatorname{Var}(X^n) = E\left[(X^n)^2\right] - \left[E(X^n)\right]^2\] Plugging the values of \(E(X^n)\) and \(E(X^{2n})\): \[\operatorname{Var}(X^n) = \frac{1}{2n+1} - \left(\frac{1}{n+1}\right)^2\] So, the variance of \(X^n\) is \(\operatorname{Var}(X^n) = \frac{1}{2n+1} - \frac{1}{(n+1)^2}\).