Given $F$ is a CDF, what can we say about $H(x) = 1 - (1-F(x))^2$?

Question

Let's say we know that a function $F$ is a CDF. What about the function $H(x) = 1 - (1-F(x))^2$? I know there's a theorem that says a function is a CDF if

• $\lim_{x \to -\infty}F(x) = 0$ and $\lim_{x \to \infty}F(x) = 1$
• $F$ is non-decreasing
• $F$ is right continuous

I try to walk through these three conditions, but I get stuck applying them to the function, $H$. Any help would be appreciated.

Answer 1

Starting from the properties of F, it is very easy to verify that also $H=1-(1-F)^2$ satisfies the conditions to be a nice CDF.

For example:

1. $H(-\infty)=1-[1-F(-\infty)]^2=1-(1-0)^2=0$

2. ...

3. ...

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$H(x)$ it is a very well known CDF. It's the CDF of the minimum of 2 iid rv's with CDF F

Let's suppose to have two iid rv's with CDF F, say $X,Y$.

Let's set $U=min(X,Y)$

$$\mathbb{P}(U > u)=\mathbb{P}(X >x,Y>y)=\text{using independence}=$$

$$=\mathbb{P}(X >x)\mathbb{P}Y>y)=[1-F_X(u)][1-F_Y(u)]=[1-F_X(u)]^2$$

Thus

$$H_U(u)=\mathbb{P}[U \leq u]=1-\mathbb{P}[U>u]=1-(1-F)^2$$