Cumulative distribution functions (cdfs) question If I have two cumulative distribution functions $F(x)$ and $G(x)$, are $$F(x)G(x)\quad\text{ and }\quad [F(x)G(x)]^{0.5}$$ necessarily cdf? I feel a little weird, cause I am thinking to qualify to be a cdf, it has to qualify: 
a. $0≤F(x)≤1$;
b. $F(x)$ and $G(x)$ are non-decreasing;
c. $F(x)$ and $G(x)$ are continuous from left to the right; 
d. $\lim_{x\to-\infty} F(x)=0 $ and $ \lim_{x\to\infty}=1$
Right? 
I think $F(x)G(x)$ qualifies to be a cdf ( not sure about b though), but I don't think $[F(x)G(x)]^{0.5}$ is cdf. Can any one give a counterexample maybe? 
 A: Both of these are cumulative distribution functions.
The interval $[0,1]$ is closed under multiplication so $0\le F(x)G(x)\le 1$.  It is also closed under taking square roots, so $0\le\sqrt{F(x)G(x)}\le1$.
If $x_1<x_2$ then $$F(x_1)\le F(x_2) \tag 1$$ and $$G(x_1)\le G(x_2)\tag 2$$ and then one has
$$
F(x_1)G(x_1) \le F(x_2)G(x_1) \le F(x_2)G(x_2)
$$
where the first inequality follows from $(1)$ and the fact that $G(x_1)\ge0$, and the second from $(2)$ and the fact that $F(x_2)\ge 0$.  This shows $F\cdot G$ is nondecreasing.  Then since $a\mapsto\sqrt a$ is indecreasing on $[0,\infty)$, one gets $\sqrt{F(x_1)G(x_1)} \le \sqrt{F(x_2)G(x_2)}$, so $\sqrt{FG}$ is nondecreasing.
For left-to-right continuity and the two limits in (d), use the fact that the limit of the product equals the product of the two limits, if the two limits both exist and are finite (and in this case, they do and they are).
A: You could go further! If $F_{1}(x),F_{2}(x),\ldots,F_{n}(x)$ are all CDFs and $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\in\mathbb{R}$ and $\forall_i. \alpha_i>0$, then
$$F(x)=\prod_{i=1}^{n}F_{i}^{\alpha_{i}}(x)$$
is also a CDF and
$$f(x)=\left(\prod_{i=1}^{n}F_{i}^{\alpha_{i}}(x)\right)\sum_{j=1}^{n}\alpha_{j}F_{j}^{-1}(x)\frac{\mathrm{d}F_{j}(x)}{\mathrm{d}x} $$
the corresponding PDF.
