Let \(X\) be a continuous random variable having distribution function \(F\). If \[ Y=-3\ln F(X), \] then \(E(Y)\) equals ________ (answer in integer).
Show Hint
Hint: For continuous X with CDF F, \(U=F(X)\) is Uniform(0,1). Rewrite Y in terms of U and use \(\int_0^1 \ln u\,du=-1\).
Step 1: Apply the probability integral transform.
Whenever $X$ is continuous with CDF $F$, the quantity $U=F(X)$ is always Uniform(0,1), no matter what $F$ looks like. This is a standard fact used to simulate any continuous distribution from uniform random numbers.
Step 2: Recognize the transform of a uniform variable.
A well known result is that if $U\sim\text{Uniform}(0,1)$, then $Z=-\ln U$ has the standard exponential distribution with rate 1, that is $Z\sim\text{Exp}(1)$. This is because the survival function of $Z$ is
\[ P(Z>z)=P(-\ln U>z)=P(U<e^{-z})=e^{-z}, \]
which is exactly the Exp(1) survival function.
Step 3: Express Y in terms of Z.
Since $Y=-3\ln U=3(-\ln U)=3Z$, and we just showed $Z\sim\text{Exp}(1)$, the variable $Y$ is simply 3 times a standard exponential variable.
Step 4: Use the known mean of an exponential variable.
For $Z\sim\text{Exp}(1)$, $E(Z)=1$. Scaling a random variable by a constant scales its mean by the same constant, so
\[ E(Y)=3E(Z)=3\times 1=3. \]
Final Answer:
Recognizing $-\ln U$ as a standard exponential variable gives the mean directly, without doing the integral by hand.
\[ \boxed{3} \]
