Monotonicity of Cumulative Distribution Functions

Why Monotonicity Matters

A cumulative distribution function records how much probability lies at or below each threshold. Raising the threshold cannot remove outcomes that were already included; it can only keep the same set of outcomes or include more. This elementary observation has an important consequence: CDF values must never decrease as their input increases.

Monotonicity is useful even when a CDF is not known in full. If its values at two thresholds are known, its value at every threshold between them is trapped between those two values. These bounds can be informative, but they do not generally determine the value exactly. The distinction between a bound and an exact calculation is central to using the property correctly.

Definition: For a real-valued random variable $X$, its cumulative distribution function (CDF) is $F(x)=P(X\leq x)$, for every real number $x$. A function $F$ is nondecreasing if $F(x_1)\leq F(x_2)$ whenever $x_1\leq x_2$.

The Nested-Event Proof

Fix any real numbers $x_1$ and $x_2$ with $x_1\leq x_2$. If an outcome has $X\leq x_1$, then it also has $X\leq x_2$. Thus the threshold events are nested: the event at the lower threshold is a subset of the event at the higher threshold. This argument depends only on the ordering of real numbers and the defining rule for a CDF.

Theorem: If $F$ is the CDF of a real-valued random variable $X$, then $F$ is nondecreasing: for all real $x_1\leq x_2$, $F(x_1)\leq F(x_2)$.

x_1\leq x_2 \quad\Longrightarrow\quad \{X\leq x_1\}\subseteq\{X\leq x_2\} \quad\Longrightarrow\quad P(X\leq x_1)\leq P(X\leq x_2).

The final implication uses monotonicity of probability: whenever events $A$ and $B$ satisfy $A\subseteq B$, $P(A)\leq P(B)$. Applying the definition of $F$ to the two probabilities gives the result.

Order the thresholds.
Take arbitrary real numbers $x_1\leq x_2$.

Compare the events.
Every outcome satisfying $X\leq x_1$ also satisfies $X\leq x_2$, so $\{X\leq x_1\}\subseteq\{X\leq x_2\}$.

Compare their probabilities.
By monotonicity of probability, $P(X\leq x_1)\leq P(X\leq x_2)$. These are $F(x_1)$ and $F(x_2)$, respectively.

The event inclusion is the reason for the result; monotonicity is not an extra assumption about the distribution. The proof applies to discrete, continuous, and mixed distributions alike.

Bounds Between Known CDF Values

The same theorem gives a practical bounding rule. Suppose the CDF is known at two ordered thresholds $a\leq b$, and consider any $x$ between them. Applying monotonicity first to $a\leq x$ and then to $x\leq b$ traps $F(x)$ between the endpoint values.

Consequence: If $a\leq x\leq b$, then $$ F(a)\leq F(x)\leq F(b). $$ These inequalities require only monotonicity and the ordering of the three thresholds.

For completeness, the lower bound follows by applying the theorem to $a\leq x$, which gives $F(a)\leq F(x)$. The upper bound follows by applying it to $x\leq b$, which gives $F(x)\leq F(b)$. Combining the two inequalities proves the stated bounds. In particular, the CDF values at the chosen endpoints must themselves be in the same order: $F(a)\leq F(b)$.

These bounds may be wide. Monotonicity alone says that a value between the endpoint values is possible as a numerical bound; it does not say how probability is distributed between the thresholds. Additional information about the model may narrow the range or give the exact value.

Worked Example: Bounding a CDF Value for a Fair Die

Let $X$ be the outcome of a fair six-sided die, so each of $1,2,3,4,5,6$ has probability $1/6$. Use the values at 2 and 4 to bound $F(3)$, and then compare the bound with the exact value.

By the definition $F(x)=P(X\leq x)$, the event $X\leq2$ consists of outcomes 1 and 2. Therefore $F(2)=2/6=1/3$. The event $X\leq4$ consists of outcomes 1, 2, 3, and 4, so $F(4)=4/6=2/3$.

Since $2\leq3\leq4$, monotonicity gives $$ \frac{1}{3}=F(2)\leq F(3)\leq F(4)=\frac{2}{3}. $$

The model also lets us calculate the exact value: $X\leq3$ consists of outcomes 1, 2, and 3, so $F(3)=3/6=1/2$. This value satisfies the bound because $1/3\leq1/2\leq2/3$. The bound is correct but does not, by itself, identify the exact value.

Worked Example: Bounding Within a Discrete Model

Suppose $X$ takes values $-1,1,3$ with probabilities $0.2,0.4,0.4$, respectively. Bound $F(0.5)$ using the CDF values at 0 and 2.

At threshold 0, only the outcome $-1$ is included. Thus $F(0)=P(X=-1)=0.2$. At threshold 2, the outcomes $-1$ and 1 are included, giving $F(2)=0.2+0.4=0.6$. Since $0\leq0.5\leq2$, the bound is $$ 0.2\leq F(0.5)\leq0.6. $$

In this particular model, the only possible value of $X$ at or below $0.5$ is $-1$. Direct calculation therefore gives $F(0.5)=0.2$. The lower bound happens to be attained, but monotonicity alone would not reveal that; the extra calculation uses the specified support and probabilities.

Using Bounds When the Full Distribution Is Not Known

In applications, one may know a few CDF values from a probability model, a validated table, or a calculation, while other values remain unavailable. Monotonicity then gives guaranteed bounds without requiring a formula for the whole CDF. The guarantee is useful only if the known values are themselves correct and the thresholds are ordered as stated.

Worked Example: Bounding a Threshold Probability From a Table

A random measurement $X$ has a distribution for which a trusted calculation gives $F(10)=0.18$ and $F(14)=0.63$. No other details of the distribution are available. Find bounds for $F(12)$, and state what can and cannot be concluded.

The thresholds satisfy $10\leq12\leq14$. Therefore $$ F(10)\leq F(12)\leq F(14), \qquad\text{so}\qquad 0.18\leq F(12)\leq0.63. $$

Thus the probability $P(X\leq12)$ is at least 0.18 and at most 0.63. No exact value within that range follows from these two table entries alone. For example, linear interpolation would assign a particular value by imposing an additional shape assumption; monotonicity does not justify that assumption.

What Monotonicity Does—and Does Not—Say

Nondecreasing means that a larger threshold cannot produce a smaller CDF value. It does not mean that the CDF must increase at every step. Two different thresholds can have the same CDF value, and an inequality $F(a)\leq F(x)\leq F(b)$ can be too broad to be useful for a precise numerical answer.

It is also important not to confuse ordering with a formula for change. If $x$ lies between $a$ and $b$, monotonicity does not imply that $F(x)$ is halfway between $F(a)$ and $F(b)$, or that it follows a straight line between them. Such conclusions require more information than the ordering property supplies.

Common mistake: “Because $a<x<b$, I can estimate $F(x)$ by taking the same fraction of the change from $F(a)$ to $F(b)$.” This is linear interpolation, not a consequence of monotonicity. The guaranteed conclusion is only $F(a)\leq F(x)\leq F(b)$. To find an exact value or justify a more precise estimate, use additional facts about the distribution.

A useful check when reading or calculating CDF values is therefore to compare any two thresholds in order. If $x_1\leq x_2$, a reported pair of values with $F(x_1)>F(x_2)$ cannot be correct for a CDF. This check can catch mistakes in threshold comparisons or cumulative probability calculations.

Key takeaway: The CDF is nondecreasing because its threshold events are nested. Between two known thresholds, its value is bounded below by the lower-threshold CDF value and above by the higher-threshold value; monotonicity alone does not give an exact value between them.

Check Your Understanding

Use the nested-event argument and the bounding rule to answer the following.

In one sentence, explain why $\{X\leq2\}\subseteq\{X\leq5\}$ and state the resulting inequality for $F(2)$ and $F(5)$.
If $F(3)=0.25$ and $F(8)=0.70$, what bounds follow for $F(6)$?
A proposed CDF has $F(1)=0.4$ and $F(4)=0.3$. What property does this violate?
If $F(a)=0.2$, $F(b)=0.8$, and $a<x<b$, can monotonicity alone determine whether $F(x)=0.5$? Explain.
For a fair six-sided die, calculate $F(1)$ and $F(3)$, and verify that their order agrees with monotonicity.