To find the mean ($\mu$) and variance ($\sigma^2$) of $x$ for the probability density function

$f(x) = \frac{1}{\pi} \frac{4}{1 + x^2} \quad \text{for } 0 < x < 1,$

we need to calculate the expected values $E[X]$ and $E[X^2]$.

Step 1: Calculate $E[X]$ (the mean)

The mean is given by: $E[X] = \int_0^1 x f(x) \, dx = \int_0^1 x \cdot \frac{4}{\pi (1 + x^2)} \, dx.$

Step 2: Calculate $E[X^2]$

The second moment is: $E[X^2] = \int_0^1 x^2 f(x) \, dx = \int_0^1 x^2 \cdot \frac{4}{\pi (1 + x^2)} \, dx.$

Step 3: Find the Variance $\sigma^2$

The variance $\sigma^2$ is given by: [ \sigma^2 = E[X^2] - (E[X])^2. ]

Solution:

To proceed, we calculate these integrals. Both require specific techniques, such as substitution or integration by parts, due to the presence of $x$ and $x^2$ in the numerators.