Let the density function of random variable X be symmetric with respect to x = u, and prove that its distribution function satisfies the following requirements: F (U + x) + F (u-x) = 1 (x is between positive and negative infinity)

Let the density function of random variable X be symmetric with respect to x = u, and prove that its distribution function satisfies the following requirements: F (U + x) + F (u-x) = 1 (x is between positive and negative infinity)

In the simplest way
Let f (x) be the density function
F (x) = ∫ f (T) DT (integral interval from - ∞ to x)
f(u+x) = f(u-x)
There is f '(x) = f (x)
that
Let g (x) = f (U + x) + F (u-x)
Let g '(x) = f (U + x) - f (u-x) = 0
It is shown that G (x) = f (U + x) + F (u-x) = C is a constant function
Let x → + ∞ be positive on both sides
Left = f (+ ∞) + F (- ∞) = 1-0 = 1 = C
therefore
F(u+x)+F(u-x)=1