Is the definition of symmetric distribution using cdf correct? Based on wikipedia (https://en.wikipedia.org/wiki/Symmetric_probability_distribution), a distribution is symmetric about $x_0$ if and only if it is a distribution whose pdf(or pmf) $f(\cdot)$ satisfies $f(x_0+x)=f(x_0-x)$ for any $x\in \mathbb{R}$.    Can we also define symmetry about $x_0$ using cdf?  That is, we say a distribution is symmetric about $x_0$ if and only if $F(x_0+x)=1-F(x_0-x)$ for any $x\in\mathbb{R}$. It seems that this definition is also okay, right?
 A: You can characterize symmetry in terms of the CDF -- but it's a little tricky.
Recall that a random variable $X$ has a symmetric distribution when there exists a number $a$ for which $X$ and $2a-X$ have the same distribution function.  Since this is the same thing as saying that $Y=X-a$ and $-Y=a-X$ have the same distribution, we lose no generality by analyzing the variable $Y.$  Let $F$ be its distribution function, which by definition is
$$F(y) = \Pr(Y\le y) = \Pr(X-a\le y) = \Pr(x \le a+y)$$
for all numbers $y.$  The symmetry of $Y$ implies
$$\begin{aligned}
F(y) &= \Pr(Y\le y)\\
& = \Pr(-Y\le y) \\
& = \Pr(Y \ge -y) \\
& = \Pr(Y \gt -y) + \Pr(Y=-y)\\
& = (1-\Pr(Y \le -y))+\Pr(Y=-y)\\
& = 1 - F(-y) + \Pr(Y=-y).
\end{aligned}$$
That extra term $\Pr(Y=-y)$ is a nuisance unless the distribution is continuous (which means that $\Pr(Y=-y)=0$ for all $y$).  Our challenge is to express the probabilities $\Pr(Y=-y)$ in terms of the distribution function $F.$
To accomplish this, equate the set $\{-y\}$ with the intersection of the countable family of intervals $(-y-1/n,-y],$ $n=1,2,3,\ldots,$ and exploit the axioms of probability to deduce
$$\Pr(Y=-y) = \lim_{n\to\infty}\Pr\left(-y-\frac{1}{n} \lt Y \le -y\right) =  F(-y) -\lim_{\epsilon\downarrow 0} F(-y-\epsilon).$$
Substituting this into the foregoing equation for $F(y)$ leads to the general characterization

$F$ is symmetric around $0$ if and only if $$F(y) = 1 - \lim_{\epsilon\downarrow 0}F(-y-\epsilon).$$

As an example, consider this graph of $F$ when $X$ has a Bernoulli$(1/2)$ distribution.  Here, the center of symmetry is $a=1/2,$ so $Y = X-1/2$ takes on the values $\pm 1/2$ each with probability $1/2.$

The definition of $F$ causes its values at each jump to be the upper limit, whereas changing $F$ to $1-F(-y)$ rotates its graph around its center at $(0,1/2),$ thereby switching the "up" and "down" directions and causing the graph to attain the lower values at each jump.  We fix that in the definition of $\tilde F$ by sneaking up to the jumps from the left side, as shown in the right hand panel.
We may summarize this English characterization, and re-introduce the center of symmetry $a$ for full generality, by stating

A distribution function $F$ is symmetric around a number $a$ if and only if it coincides with the function obtained from $F$ whose graph (a) uses the lower values at each jump and (b) is rotated around the central point $(a,1/2)$ by 180 degrees.

