Question

Let X be a continuous uniform random variable U[0, 1], whose cumulative distribution func- tion is F(x).

(a) Find the expectation of F(X), that is, E[F(X)].
(b) Suppose the variance of F(X) is defined as E([F(X)]²)-(E[F(X)])². Find the variance of F(X).

Answer 1

The expectation of F(X), E[F(X)], for a uniform random variable X ~ U(0, 1) is 1/2. The variance of F(X) is calculated as 1/12.

Step-by-step explanation:

Let X be a continuous uniform random variable X ~ U(0, 1). The expectation of the cumulative distribution function of X, E[F(X)], can be calculated by integrating the product of the random variable's value and its probability density function. Since X is uniformly distributed, the density function is 1 over the range (0,1). The mean, μ, is therefore the integral from 0 to 1 of x dx, which is 1/2. Therefore, E[F(X)] = 1/2.

For the variance of F(X), which is defined as
`E([F(X)]^2) - (E[F(X)])^2`, we first find
`E([F(X)]^2)` by integrating the square of the random variable's value multiplied by its probability density function from 0 to 1. We calculate this as the integral from 0 to 1 of x2 dx, resulting in 1/3. Then, we use the earlier found E[F(X)] = 1/2 and calculate the variance:

`E([F(X)]^2) - (E[F(X)])^2` =
`1/3 - (1/2)^2`, which simplifies to 1/12.