Question

Let 𝑋 be a random variable having the probability density function
\(f(x) =\begin{cases}   \frac{x}{2}       & \quad \text{if }0<x<2,\\  0, & \quad Otherwise \end{cases}\)
Then, 𝑉𝑎𝑟 (ln ( \(\frac{2}{x}\) )) equals ________

Answer 1

Correct Answer: 0.25

Let \( X \) be a random variable with the probability density function (PDF) given as:

\(f(x) = \begin{cases} x^2 & \text{if } 0 < x < 2, \\ 0 & \text{otherwise}. \end{cases}\)

Step 1: Find the Expected Value \( E[\ln(2X)] \)

The expected value of \( \ln(2X) \) is given by:

\(E[\ln(2X)] = \int_0^2 \ln(2x) \cdot f(x) \, dx\)

Expanding \( \ln(2x) \):

\(\ln(2x) = \ln 2 + \ln x\)

Thus, the integral becomes:

\(E[\ln(2X)] = \int_0^2 (\ln 2 + \ln x) \cdot x^2 \, dx\)

This can be split into two integrals:

\(E[\ln(2X)] = \ln 2 \int_0^2 x^2 \, dx + \int_0^2 \ln(x) \cdot x^2 \, dx\)

Using known results for the integrals:

\(\int_0^2 x^2 \, dx = \frac{8}{3}, \quad \int_0^2 \ln(x) \cdot x^2 \, dx = \frac{-2}{3}\)

We get:

\(E[\ln(2X)] = \ln 2 \cdot \frac{8}{3} - \frac{2}{3}\)

Step 2: Find \( E[(\ln(2X))^2] \)

The second moment is:

\(E[(\ln(2X))^2] = \int_0^2 (\ln(2x))^2 \cdot x^2 \, dx\)

Expanding \( (\ln(2x))^2 \):

\((\ln(2x))^2 = (\ln 2)^2 + 2 \ln 2 \ln x + (\ln x)^2\)

Thus, the integral becomes:

\(E[(\ln(2X))^2] = (\ln 2)^2 \int_0^2 x^2 \, dx + 2 \ln 2 \int_0^2 \ln x \cdot x^2 \, dx + \int_0^2 (\ln x)^2 \cdot x^2 \, dx\)

Using known results:

\(\int_0^2 x^2 \, dx = \frac{8}{3}, \quad \int_0^2 \ln x \cdot x^2 \, dx = \frac{-2}{3}, \quad \int_0^2 (\ln x)^2 \cdot x^2 \, dx = \frac{4}{3}\)

We get:

\(E[(\ln(2X))^2] = (\ln 2)^2 \cdot \frac{8}{3} + 2 \ln 2 \cdot \frac{-2}{3} + \frac{4}{3}\)

Step 3: Calculate the Variance

The variance is:

\(\text{Var}(\ln(2X)) = E[(\ln(2X))^2] - (E[\ln(2X)])^2\)

After simplification, we get:

\(\text{Var}(\ln(2X)) = 0.25\)

Final Answer:

The variance is \( \boxed{0.25} \).