I have been scratching my head about this question for a long time. I found one other question on here that included a minimum function for probability functions but unfortunately there wasn't enough information to make me understand this enough to proceed.
The question states:
Let X is a continuous random variable with probability density function
$$f(x) = c \min \left( 1, \dfrac 1 {x^4} \right)$$
(a) Find c and the (cumulative) distribution function of X.
(b) Find EX, VarX, and the median of X.
Based on the example I saw, this gave me a density function:
$$f(x) = \begin{cases} c & \mbox{for } -1 ≤ x ≤ 1\\ \tfrac{c}{x^4} & \mbox{for } |x| > 1\end{cases}$$
Is this correct? If so then by setting the summation of the integrals of these f(x) cases gives me a value of 3/8 for c.
And if so would I be correct to get a distribution function as:
$$F(x) = \begin{cases} \int_{1}^{\infty} \tfrac{3}{8x^4} dx & \text{if } x>1 \\ \int_{-1}^{1} \tfrac{3}{8} dx & \text{if } -1 ≤ x ≤ 1 \\ \int_{-\infty}^{-1} \tfrac{3}{8x^4} & \text{if } x < -1\end{cases}$$
I'm not sure about this though and even if this is correct, I'm not sure how to use this for part b.
By definition,
$$F(x) = \int_{-\infty}^x f(y) \, dy, $$
so we obtain:
$$F(x) = \begin{cases} \int_{-\infty}^{x} \tfrac{3}{8y^4} dy = \frac{1}{8} \, |x|^{-3}& \text{if } x < -1\\ F(-1) + \int_{-1}^{x} \tfrac{3}{8} dy = \frac{1}{8} (4 + 3x) & \text{if } -1 ≤ x ≤ 1 \\ F(1) + \int_{1}^{x} \tfrac{3}{8y^4}dx = 1 - \frac{1}{8} |x|^{-3} & \text{if } x>1 \end{cases} $$