Final answer:
1. The distribution of Y is given by Y = 1-2X for 0 ≤ X ≤ 0.5 and Y = 2X-1 for 0.5 < X ≤ 1. 2. The covariance cov(X,Y) is -1/12. 3. X and Y are not independent.
Step-by-step explanation:
1. Distribution of Y:
The random variable Y is defined as the absolute value of 2X-1. To determine the distribution of Y, we need to look at the probability density function (PDF) of X. Since X is uniformly distributed over the interval [0,1], its PDF is a constant value of 1 over this interval. We can find the distribution of Y by finding the PDF of 2X-1 and taking the absolute value. For x between 0 and 0.5, Y is equal to 1-2x. For x between 0.5 and 1, Y is equal to 2x-1. Therefore, the distribution of Y is given by:
Y = 1-2X, for 0 ≤ X ≤ 0.5
Y = 2X-1, for 0.5 < X ≤ 1
2. Covariance of X and Y:
To compute the covariance of X and Y, we need to know the means and variances of X and Y. Since X is uniformly distributed, its mean is (0+1)/2 = 0.5 and its variance is (1-0)^2/12 = 1/12. The mean of Y can be found by taking the expected value of Y, which is:
E(Y) = ∫[0,1] |2x-1| dx = 1/2
We can now calculate the covariance using the formula:
cov(X,Y) = E(XY) - E(X)E(Y)
Substituting the values we have:
cov(X,Y) = E(X(1-2X)) - (0.5)(0.5)
By integrating, we find:
cov(X,Y) = -1/12
3. Independence of X and Y:
To determine if X and Y are independent, we can check if the joint distribution of X and Y factors into the product of their individual distributions. The joint distribution of X and Y can be found by finding the PDF of Y and X. If the joint PDF factors into the product of their individual PDFs, then X and Y are independent.
In this case, we found that the distribution of Y depends on the value of X, so X and Y are not independent.