Question

Let \(X_1,X_2,X_3\) be a random sample of size \(3\) from \(Exp(1)\) distribution. Then, \(E\left(4X_{(2)}e^{X_{(2)}}\right)\) equals (in integer).

Hint:

For the \(r\)-th order statistic, use \[ f_{X_{(r)}}(x)=\frac{n!}{(r-1)!(n-r)!}[F(x)]^{r-1}[1-F(x)]^{n-r}f(x). \]

Answer 1

Correct Answer: 18

Step 1: Density of the middle order statistic.
For $Exp(1)$ with $f=e^{-x}$, $F=1-e^{-x}$, the second of three order statistics has $f_{X_{(2)}}(x)=6(1-e^{-x})e^{-x}\cdot e^{-x}=6(e^{-2x}-e^{-3x})$.

Step 2: Set up the expectation.
$E(4X_{(2)}e^{X_{(2)}})=\int_0^\infty 4xe^x\cdot6(e^{-2x}-e^{-3x})dx=24\int_0^\infty x(e^{-x}-e^{-2x})dx$.

Step 3: Use the standard integral.
$\int_0^\infty xe^{-ax}dx=\frac1{a^2}$, so the pieces are $1$ and $\frac14$.

Step 4: Combine.
$24\left(1-\frac14\right)=24\cdot\frac34=18$.

Step 5: Conclude.
\[ \boxed{18} \]