Uniform Distribution
- The given random variable X has a uniform distribution on the interval (0,1).
- This means that X can take any value between 0 and 1, with equal probability.
- The probability density function (PDF) of X is given by f(x) = 1, for 0 < x="" />< 1,="" and="" f(x)="0" />

Transformation of Random Variable
- We are interested in finding the expected value of the random variable Y, which is defined as Y = -2 log X.
- To find the expected value of Y, we need to determine the probability density function (PDF) of Y.
- Let's start by finding the cumulative distribution function (CDF) of Y, denoted as F(y), which gives the probability that Y is less than or equal to y.
- Since Y = -2 log X, we can express X in terms of Y as X = e^(-y/2).
- The probability that X is less than or equal to e^(-y/2) can be found using the CDF of X, which is F(x) = x for 0 < x="" />< />
- Therefore, F(y) = P(Y ≤ y) = P(-2 log X ≤ y) = P(X ≥ e^(-y/2)) = 1 - P(X < e^(-y/2))="1" -="" />

Deriving the PDF of Y
- To find the PDF of Y, we differentiate the CDF with respect to y.
- The PDF of Y, denoted as f(y), is given by f(y) = dF(y)/dy.
- Differentiating the CDF F(y) = 1 - F(e^(-y/2)), we get f(y) = dF(y)/dy = 0 - dF(e^(-y/2))/dy.
- Using the chain rule, we have f(y) = -dF(e^(-y/2))/d(e^(-y/2)) * d(e^(-y/2))/dy.
- The first term, -dF(e^(-y/2))/d(e^(-y/2)), is the PDF of X evaluated at e^(-y/2), which is 1.
- The second term, d(e^(-y/2))/dy, is -1/2 * e^(-y/2).
- Therefore, the PDF of Y is f(y) = 1 * (-1/2 * e^(-y/2)) = -1/2 * e^(-y/2), for y > 0, and f(y) = 0 otherwise.

Finding the Expected Value of Y
- The expected value of Y, denoted as E(Y), is given by the integral of y times the PDF of Y, i.e., E(Y) = ∫ y * f(y) dy.
- Substituting the PDF of Y, we have E(Y) = ∫ y * (-1/2 * e^(-y/2)) dy.
- Integrating by parts, let u = y and dv = -1/2 * e^(-y/2) dy, we get du = dy and v = e^(-y/