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: Identify the distribution of \(X_{(2)}\).
For \(X_i\sim Exp(1)\), the density and distribution functions are
\[ f(x)=e^{-x},\qquad x>0 \] and
\[ F(x)=1-e^{-x},\qquad x>0 \] For a sample of size \(3\), the density of the second order statistic \(X_{(2)}\) is
\[ f_{X_{(2)}}(x) = \frac{3!}{1!1!}F(x)\{1-F(x)\}f(x) \] \[ =6(1-e^{-x})(e^{-x})e^{-x} \] \[ =6(e^{-2x}-e^{-3x}),\qquad x>0 \]

Step 2: Write the required expectation.
\[ E\left(4X_{(2)}e^{X_{(2)}}\right) = \int_0^\infty 4xe^x f_{X_{(2)}}(x)\,dx \] Substitute the density:
\[ = \int_0^\infty 4xe^x\cdot 6(e^{-2x}-e^{-3x})\,dx \] \[ = 24\int_0^\infty x(e^{-x}-e^{-2x})\,dx \]

Step 3: Use the standard integral.
We know that
\[ \int_0^\infty xe^{-ax}\,dx=\frac{1}{a^2},\qquad a>0 \] Therefore,
\[ \int_0^\infty xe^{-x}\,dx=1 \] and
\[ \int_0^\infty xe^{-2x}\,dx=\frac14 \]

Step 4: Calculate the value.
\[ E\left(4X_{(2)}e^{X_{(2)}}\right) = 24\left(1-\frac14\right) \] \[ = 24\cdot \frac34 \] \[ = 18 \]

Step 5: Final conclusion.
Hence,
\[ \boxed{18} \]