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 ${x}$?

  • 8
    $\begingroup$ $P(A_1\cup A_2)\geq P(A_1)$ $\endgroup$
    – Glen_b
    Commented Aug 25, 2020 at 3:17
  • $\begingroup$ ...and, as you increase $x$, you are adding zero or more possible events in addition to the ones you already have. $\endgroup$ Commented Aug 26, 2020 at 4:49
  • $\begingroup$ It's the sum of all probabilities below $x$, and you can't have negative probabilities. $\endgroup$
    – naught101
    Commented Apr 8, 2021 at 2:01

3 Answers 3


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 $F(x_2) \geq F(x_1)$.


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