Question

Let X be a uniform random variable on the interval (0, 1) and let Y=X?. Determine the distribution of Y. Y is uniform on the interval (0,1%) $$f_Y(y) = \sqrt{y}, y \in (0, \sqrt{pi}) $$ 1 = ye(0,VT) 21VY o fr(y)= $$f_Y(y) = \frac{1}{2\pi\sqrt{y}}, y \in (0, \pi^2)$$ None of these

Answer 1

To determine the distribution of Y, we use the transformation method with the PDF of X being uniform on the interval (0, 1). The PDF of Y is given by fY(y) = 1/(2*sqrt(y)), and thus Y is uniform on the interval (0, 1%).

Step-by-step explanation:

To determine the distribution of Y, we need to find the probability density function (PDF) of Y. Since Y = X^2, we can find the PDF of Y using the transformation method. The PDF of X is uniform on the interval (0, 1), so the PDF of Y is given by:

fY(y) = fX(g-1(y)) * |(d/dy)(g-1(y))|

where g(x) = x^2 and fX(x) is the PDF of X. The inverse function of g(x) is g-1(y) = sqrt(y). Substituting these values, the PDF of Y is:

fY(y) = fX(sqrt(y)) * |(d/dy)(sqrt(y))| = 1 * (1/(2*sqrt(y))) = 1/(2*sqrt(y))

Therefore, the distribution of Y is uniform on the interval (0, 1%).