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Let the density of random variable $X$ be given as $$f_X(x)=\begin{cases} 12x^2 & -0.5 \le x \le 0.5 \\0 & \text{otherwise}\end{cases}$$

How can I determine $$f_X(x\mid-0.25 \le X \le 0.25)$$

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The usual way: apply conditional probability.

$$f_X(x\mid a\leq X\leq b) ~=~\dfrac{~ f_X(x)~\mathbf 1_{a\leq x\leq b}~}{~\displaystyle\int_a^b f_X(s)\operatorname d s~}$$


PS: $\mathbf 1_{a\leq x\leq b}$ is an indicator.

$$\mathbf 1_{a\leq x\leq b}=\begin{cases} 1 & : a\leq x\leq b \\[1ex] 0 & : \textsf{elsewhere}\end{cases}$$

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