Final answer:
The probability density function (PDF) of Y, where Y=X² and X has a uniform distribution on (-1, 1), is found using the transformation of variables. The PDF of Y turns out to be f_Y(y) = 1/√(4y) for 0 < y < 1.
Step-by-step explanation:
The student is asking about the probability density function (PDF) of a random variable which is a transformation of another random variable with a known distribution. In this case, the original variable X has a uniform distribution on the interval (-1, 1), and we are seeking the density of Y = X².
The density function of X is f(x) = 1/2 for -1 < x < 1, since it's uniform. To find the density of Y, consider that Y can only take values from 0 to 1 because squaring any value in the range of X will result in a number between 0 and 1.
To find the PDF of Y, we need to use the method of transformation of variables. The formula for a change of variables from X to Y, where Y=g(X), is given by f_Y(y) = f_X(g⁻¹(y)) | d/dy g⁻¹(y) | for any y in the range of Y, where f_Y and f_X are the probability density functions of Y and X respectively.
For Y = X², the inverse transformation is X = ±√y. As squaring a positive or negative value of X gives the same Y, we have to account for both when transforming the variable. The derivative of the inverse transformation with respect to y is ½y⁻½.
Hence, the PDF of Y is calculated by summing the contributions from both positive and negative values of X:
f_Y(y) = f_X(√y)*½y⁻½ + f_X(-√y)*½y⁻½, which simplifies to f_Y(y) = 1/√(4y) for 0 < y < 1.