If $F (x) = \frac{1}{π} (\frac{π}{2} + {tan}^{- 1} x)$ , $- \infty < x < \infty$ is a distribution function of a continuous variable $X$ , then find $P (0 \leq x \leq 1)$

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Solution

Given, $F (x) = \frac{1}{π} (\frac{π}{2} + {tan}^{- 1} (x))$
Therefore, $P (0 \leq x \leq 1) = F (1) \sim F (0)$
$= \frac{1}{π} (\frac{π}{2} + {tan}^{- 1} (1)) - \frac{1}{π} (\frac{π}{2} + {tan}^{- 1} (0))$
$= \frac{1}{π} (\frac{π}{2} + \frac{π}{4}) - \frac{1}{π} (\frac{π}{2} + 0)$
$= \frac{1}{π} (\frac{π}{2} + \frac{π}{4} - \frac{π}{2}) = \frac{1}{4}$

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