# Is it also a cumulative distribution function?

There is a statement which I am trying to evaluate to whether true or false:

Let $$F_X(x)$$ be a cumulative distribution function (cdf) of a r.v. $$X$$. Then the following $$G_X(x)$$ is also a cdf: $$G(x) = \frac{F_X(x) + 1 - F_X(-x)}{2}$$

It's very clear this is true for distributions which are symmetric around 0, because then $$P(X\leq - x) = P(X \geq x)$$. It's also simple to show that $$\lim_{x \to \infty} G(x) = 1$$ when $$F_X$$ is a distribution.

I tend to believe this is a true statement. What is bothering me is the continuity in the right it must have. So I tried to think of a counterexample. Let:

$$F_X(x) =\begin{cases} 0, \text{ if } x<0 \\ \frac{1}{15} + \frac{2t}{3} \text{ if } x \in [0,1) \\ 1 \text { if } x\geq 1 \end{cases}$$

which is a mixture of a Bernoulli r.v. with $$p=4/5$$ and an uniform in $$(0,1)$$ with weights $$1/3$$ and $$2/3$$. Now I believe that for $$x<0$$, $$G(x)$$ is not continuous in the right (particularly in $$x=-1$$).

Does it suffice as a counterexample? Or is the statement right?

Thanks!

You counterexample is correct. And any $$F$$ that is discontinuous in at least one point will be a counterexample. For example, take $$F(x) = \mathbb{I}_{x \geqslant 0}$$. Then $$G(0) = \frac{1 + 1 - 1}{2} = \frac{1}{2}$$, but if $$x > 0$$ we have $$G(x) = 1$$, thus $$\lim_{x \to 0+} G(x) = 1$$.
And in general, $$\lim_{x \to x_0+} G(x) = G(x_0) - \lim_{x \to x_0+} \frac{F(-x) - F(-x_0)}{2} = G(x_0) - \lim_{x \to -x_0+}\frac{F(x) - F(-x_0)}{2}$$, so $$G$$ is continuous in right at $$x_)$$ iff $$F$$ is continuous in left in $$-x_0$$.