Final answer:
The density function of Y, when X is uniformly distributed over (0,1), is the derivative of the cumulative distribution function of Y, resulting in fY(y) = 1/y for y in the interval (1,e).
Step-by-step explanation:
If X is uniformly distributed over the interval (0,1), we can denote this as X ~ U(0,1). The density function of X, fX(x), is constant in this interval and the value of this constant is 1 since the total area under the curve of a probability density function must be equal to 1. Considering Y = eX, we can find the probability density function of Y by applying the transformation technique.
Firstly, we find the cumulative distribution function (CDF) of Y, FY(y), which gives the probability that Y is less than or equal to y:
FY(y) = P(Y ≤ y) = P(eX ≤ y) = P(X ≤ ln(y)). Since X follows U(0,1), it implies that P(X ≤ ln(y)) = ln(y).
The probability density function of Y, denoted as fY(y), is the derivative of FY(y). Hence, fY(y) = d/dy[ln(y)] = 1/y, for y in the interval (1, e).