# Why is cumulative distribution function monotone non-decreasing?

If you have a quantity $${X}$$ that takes some value at random, the cumulative distribution function $${F(x)}$$ gives the probability that $${X}$$ is less than or equal to $${x}$$, that is: $$\begin{equation*} F(x)= P(X \leq x) \end{equation*}$$ $${F(x)}$$ is bounded below by $${0}$$, and bounded above by $${1}$$ (because it doesn't make sense to have a probability outside $${[0,1]}$$) and that it has to be non-decreasing in $${x}$$.

My question is explain why the cumulative distribution function has to be monotone non-decreasing in $${x}$$?

• $P(A_1\cup A_2)\geq P(A_1)$ Commented Aug 25, 2020 at 3:17
• ...and, as you increase $x$, you are adding zero or more possible events in addition to the ones you already have. Commented Aug 26, 2020 at 4:49
• It's the sum of all probabilities below $x$, and you can't have negative probabilities. Commented Apr 8, 2021 at 2:01

Because if $$x \leq y$$, then if $$X \leq x$$, it follows that $$X \leq y$$. Therefore, $$P(X \leq x) \leq P(X \leq y)$$.

More generally, probabilities are monotone in the sense that if $$A$$ and $$B$$ are events and $$A \subseteq B$$, then $$P(A) \leq P(B)$$. This follows from writing $$B$$ as the disjoint union of $$A$$ and $$B \setminus A$$, whence by the probability axioms $$P(B) = P(A) + P(B \setminus A) \geq P(A)$$ (since $$P(B \setminus A) \geq 0$$).

In the case of cumulative distribution functions with $$x \leq y$$, take $$A = \{X \leq x\}$$ and $$B = \{X \leq y\}$$.

For a function $$f$$ to be monotonically non-decreasing, we must have: $$f(x+\epsilon)\ge f(x)$$ for any non-negative $$\epsilon$$.

Let's check this for the CDF. We have: $$F(x) = \Pr(X \le x)$$ $$F(x+\epsilon) = \Pr(X \le x+\epsilon)$$ We can rewrite the right hand side of that last equation as: $$\Pr(X\le x+\epsilon) = \Pr(X \le x) + \Pr(x < X \le x+\epsilon)$$ That is, the probability that $$X$$ is smaller than or equal to $$x+\epsilon$$ is equal to the probability that it is smaller than or equal to $$x$$, plus the probability that it is between $$x$$ and $$x+\epsilon$$.

Using the definition of $$F$$, we can rewrite the equation as:

$$F(x+\epsilon) = F(x) + \Pr(x < X \le x+\epsilon)$$ Since $$\Pr(x < X \le x+\epsilon)$$ is a probability and must therefore be non-negative, this implies: $$F(x+\epsilon) \ge F(x)$$

which is what we set out to prove.

Increasing $$x$$ may change the claim $$X \leq x$$ from false to true, but there's no way for it to go from true to false. Thus, it's a non-increasing function. Suppose $$x$$ is how long you've been waiting on hold, and $$F(X)$$ is the probability that after $$X$$ seconds, you've been helped. The longer you wait, the more likely you'll be helped. There's no way for the probability to go down from waiting longer.

Suppose $$x_2 > x_1$$. Consider the following three possibilities:

(A) $$x \leq x_1$$
(B) $$x_1 < x \leq x_2$$
(C) $$x_2 < x$$

These are mutually exclusive possibilities, so when we combine their probabilities, we can just add. That is, $$P(A \lor B) = P(A)+P(B)$$. But $$P(A \lor B)$$ is the same as $$P(x \leq x_2)$$, which is the same as $$F(x_2)$$. And $$P(A) = F(x_1)$$. So we have $$F(x_1)+P(B) = F(x_2)$$. Since $$P(B) \geq 0$$, it follows that $$F(x_2) \geq F(x_1)$$.