Final answer:
To find the pdf, mean, and variance for Y, we need to determine the pdf of Y by finding the CDF and differentiating it. Then, we can find the mean and variance of Y by integrating the relevant expressions. P(Y < 1/2) can be found by integrating the pdf of Y over the appropriate range.
Step-by-step explanation:
To find the probability density function (PDF) of Y, we need to determine the cumulative distribution function (CDF) of Y and then differentiate it. Since Y = X², we can rewrite the equation as Y = X * X. The range of X is (-1,1), so the range of Y will be (0,1). The CDF of Y is given by P(Y ≤ y) = P(X² ≤ y) = P(-√y ≤ X ≤ √y). Since X is uniform on (-1,1), the probability of X lying between -√y and √y is 2√y. Therefore, the CDF of Y is F(y) = 2√y for 0 < y ≤ 1, and 0 for y ≤ 0 or y > 1. To find the PDF of Y, we differentiate the CDF with respect to y, which gives f(y) = (1/√y) for 0 < y ≤ 1, and 0 for y ≤ 0 or y > 1. The mean of Y can be found by integrating y * f(y) over the range of Y, which gives ∫(0,1) y * (1/√y) dy. Evaluating this integral gives the mean of Y as 2/3. The variance of Y can be found by integrating (y - mean)^2 * f(y) over the range of Y, which gives ∫(0,1) (y - 2/3)^2 * (1/√y) dy. Evaluating this integral gives the variance of Y as 1/18.
To find P(Y < 1/2), we need to integrate f(y) over the range (0, 1/2). The integral is given by ∫(0,1/2) (1/√y) dy. Evaluating this integral gives P(Y < 1/2) as 2 - 2√2/3.