Question

Let X be uniform on [0,1] and compute the expectation and variance of Xⁿ where n is a positive integer.

Let X be uniform on [0,1] and compute the probability density function of Y=X⁵.

Answer 1

The expectation of X^n for a uniformly distributed variable X on [0,1] is 1/(n+1), and its variance can be derived similarly. The probability density function of Y=X^5 is 1/5y^(1/5-1) for 0 ≤ y ≤ 1.

Step-by-step explanation:

To compute the expectation and variance of Xn where X is uniformly distributed on [0,1], we can make use of the definitions for the expectation and variance of a continuous random variable.

The expectation of Xn is calculated using the integral:

E(Xn) = ∫01 xn dx = ¼ xn+1 | 01 = 1/(n+1)

The variance of Xn is found by calculating E(X2n) and using the formula Var(Xn) = E(X2n) - [E(Xn)]2.

Similarly, to derive the probability density function (PDF) of Y = X5, we apply the transformation technique. The PDF of X is 1 for 0 ≤ x ≤ 1, and for Y we have:

fY(y) = fX(x) |dx/dy| = 1/5y1/5 - 1 for 0 ≤ y ≤ 1, and 0 elsewhere.