Question

Let \(X\) be a continuous random variable having distribution function \(F\). If
\[ Y=-3\ln F(X), \]
then \(E(Y)\) equals ________ (answer in integer).

Hint:

Hint:
For continuous X with CDF F, \(U=F(X)\) is Uniform(0,1). Rewrite Y in terms of U and use \(\int_0^1 \ln u\,du=-1\).

Answer 1

Correct Answer: 3

Step 1: Recall the probability integral transform.
For a continuous random variable X with distribution function F, the transformed variable \(U=F(X)\) always has a Uniform(0,1) distribution, regardless of what F actually is.

Step 2: Rewrite Y using U.
Since \(U=F(X)\), the given variable becomes
\[ Y=-3\ln U, \qquad U\sim \text{Uniform}(0,1). \]

Step 3: Set up E(Y) as an integral over U.
Because U is uniform on (0,1), its density is 1 on that interval, so
\[ E(Y)=-3E(\ln U)=-3\int_0^1 \ln u\,du. \]

Step 4: Evaluate the integral by parts.
Using \(\int \ln u\,du=u\ln u-u\),
\[ \int_0^1 \ln u\,du=\Big[u\ln u-u\Big]_0^1. \]
At \(u=1\): \(1\cdot\ln 1-1=0-1=-1\).
At \(u=0\): \(u\ln u\to 0\) as \(u\to 0^+\) (a standard limit), so the term is \(0-0=0\).
So
\[ \int_0^1 \ln u\,du=-1-0=-1. \]

Step 5: Substitute back.
\[ E(Y)=-3\times(-1)=3. \]

Final Answer:
The expected value works out to a clean positive integer. \[ \boxed{3} \]