I am trying to understand how the following CDF expression is obtained.
$h_1 = \text{min}(f, g)$ ---(1)
where $f, g$ are independent and identically distributed random variables.
$F_{h_1}(h) = \text{Pr}(h_1 < h) = \text{Pr}(\text{min}(f, g) < h)$ ----(2)
$F_{h_1}(h) = 1-(1-\text{Pr}(f<h))(1-\text{Pr}(g<h))$ ----(3)
$F_{h_1}(h) = $F_{f}(h)+ F_{g}(h) - F_{f}(h)*F_{g}(h)$----(4)
where $F(.)$ denotes CDF.
I had understood eq. (1) and (2) clearly, but not getting how eq. (3) and (4) is obtained.
Any help in this regard will be highly appreciated.
Note that \begin{align} F_{h_1}(h) &= Pr(min(f,g) < h) \\ &= 1 - Pr(min(f,g) \geq h). \end{align} If the minimum of $f$ and $g$ is bigger than $h$, then this means that both are bigger than $h$. So, we get \begin{align} F_{h_1}(h) &= 1 - Pr(min(f,g) \geq h) \\ &= 1 - Pr(f \geq h \text{ and } g \geq h). \end{align} Since, $f$ and $g$ are independent we know that \begin{align} F_{h_1}(h) &= 1 - Pr(f \geq h \text{ and } g \geq h) \\ &= 1 -Pr(f \geq h) \cdot Pr(g \geq h)\\ &=1 -\big(1 - Pr(f < h) \big) \cdot \big(1 - Pr(g < h) \big)\\ &= 1 - \Big( \big(1-F_{f}(h)) \cdot (1-F_{g}(h) \big) \Big)\\ &= 1 - \big( 1 + F_{f}(h)F_{g}(h) - F_{f}(h) - F_{g}(h) \big)\\ &= F_{f}(h) + F_{g}(h) - F_{f}(h)F_{g}(h). \end{align} We have now established equation 3 and 4.