Combining Tied Outcomes Into CDF Jumps

When Different Outcomes Have the Same Value

A random experiment can have several distinct outcomes that lead to the same numerical value of a random variable. For example, different face values on a die might all be assigned the value 1. The CDF records probabilities for values of the random variable, not separate labels for the outcomes that produced them. Consequently, all outcomes producing one value contribute together to the same increase in the CDF.

This grouping is important when calculating probabilities near a threshold. Just below a value, none of the probability at that value has yet entered $P(X\le x)$; at the value itself, all of it has entered. Just above the value, the CDF is still unchanged until another possible value is reached.

Definition: Suppose a finite set of mutually exclusive outcomes $A_1,\ldots,A_m$ covers all possibilities, with $P(A_i)=p_i$, and the random variable $X$ equals $x_i$ on outcome $A_i$. For each distinct numerical value $v$ taken by $X$, define its combined probability as $$ q(v)=\sum_{i:x_i=v}p_i. $$ The outcomes are tied at $v$ if two or more of them produce that same value. The CDF is $$ F(x)=P(X\le x)=\sum_{v\le x}q(v). $$

The sum in the CDF runs over distinct numerical values $v$, not over distinct outcome labels. Equivalently, one could sum the probabilities of all original outcomes whose assigned values are at most $x$. Grouping first is often clearer, especially when several outcomes are tied.

The Grouping Rule and Its Consequence

The basic rule follows from the event that defines the CDF. The event $\{X\le x\}$ consists of every original outcome whose assigned value is no greater than $x$. Since those outcomes are mutually exclusive, their probabilities add.

Theorem: Under the finite-outcome setup in the definition, for every real $x$, $$ F(x)=\sum_{i:x_i\le x}p_i =\sum_{\substack{v\text{ a distinct value of }X\\v\le x}}\ \sum_{i:x_i=v}p_i. $$ In particular, the probability contributed at a distinct value $v$ is the sum of the probabilities of every outcome tied at $v$.

Proof. The events $A_1,\ldots,A_m$ are mutually exclusive and cover all possibilities. On $A_i$, $X=x_i$, so $A_i$ is part of $\{X\le x\}$ exactly when $x_i\le x$. Thus $\{X\le x\}$ is the disjoint union of the $A_i$ satisfying that inequality. Additivity of probability gives $F(x)=\sum_{i:x_i\le x}p_i$. Partition those indices according to their common value $v=x_i$. Each group contributes $\sum_{i:x_i=v}p_i=q(v)$, and only groups with $v\le x$ are included. This gives the second sum. $\square$

The grouping rule also explains the CDF’s behavior at a tied value. For a finite set of possible values, let $v$ be one of them. The left limit $F(v^-)$ is the probability $P(X<v)$, because values strictly below $v$ have entered the CDF just before $v$, while the value $v$ itself has not. At $v$, all outcomes tied there are included.

Result: If $v$ is a possible value of $X$, then $$ F(v)-F(v^-)=q(v)=\sum_{i:x_i=v}p_i. $$

Derivation. In the finite-outcome setting, $F(v)$ sums the probabilities for outcomes with $x_i\le v$, while $F(v^-)$ sums those with $x_i<v$. Subtracting cancels every term with $x_i<v$, leaving exactly the outcomes with $x_i=v$: $$ F(v)-F(v^-) =\sum_{i:x_i\le v}p_i-\sum_{i:x_i<v}p_i =\sum_{i:x_i=v}p_i =q(v). $$ This is the size of the single CDF jump at $v$, regardless of how many distinct outcomes produce it.

Worked Examples

Worked Example: A Die With Two Tied Values

Roll a fair six-sided die. Define $X=0$ for rolls 1 and 2, $X=1$ for rolls 3, 4, and 5, and $X=3$ for roll 6. Construct the CDF and find $F(0.2)$, $F(1)$, and $P(0.5<X\le2)$.

Each die face has probability $1/6$. The combined probabilities are $$ q(0)=P(\text{roll 1 or 2})=\frac16+\frac16=\frac13, \qquad q(1)=P(\text{roll 3, 4, or 5})=\frac36=\frac12, \qquad q(3)=P(\text{roll 6})=\frac16. $$ They sum to $1/3+1/2+1/6=1$, as required.

The distinct possible values are 0, 1, and 3. Therefore the CDF is $$ F(x)= \begin{cases} 0, & x<0,\\ \frac13, & 0\le x<1,\\ \frac56, & 1\le x<3,\\ 1, & x\ge3. \end{cases} $$ At $x=0.2$, only the value 0 has been included, so $F(0.2)=1/3$. At $x=1$, the values 0 and 1 are included, so $F(1)=1/3+1/2=5/6$.

For the interval probability, the CDF difference gives $$ P(0.5<X\le2)=F(2)-F(0.5)=\frac56-\frac13=\frac12. $$ This is also the combined probability of the die faces 3, 4, and 5, all of which produce $X=1$. No face produces a value strictly between 1 and 3.

Worked Example: Three Outcomes Tied at One Score

A screening rule has five possible recorded outcomes with probabilities $0.12, 0.18, 0.20, 0.30,$ and $0.20$. The rule assigns scores $10,10,20,20,$ and $30$, respectively. Find the combined probabilities and evaluate the CDF at $9.9$, $10$, $10.1$, $20$, and $25$.

The first two outcomes are both assigned 10, so their probability combines to $q(10)=0.12+0.18=0.30$. The outcomes assigned 20 combine to $q(20)=0.20+0.30=0.50$. The remaining outcome gives $q(30)=0.20$. The check $0.30+0.50+0.20=1$ confirms that all probability has been accounted for.

At $9.9$, no possible score is at most the threshold, so $F(9.9)=0$. At 10, both outcomes tied at 10 enter, giving $F(10)=0.30$. At 10.1, no new score has been reached, so $F(10.1)=0.30$. At 20, the combined probability at 20 is added: $F(20)=0.30+0.50=0.80$. At 25, the score 30 has still not entered, so $F(25)=0.80$.

Thus the CDF is $$ F(x)= \begin{cases} 0, & x<10,\\ 0.30, & 10\le x<20,\\ 0.80, & 20\le x<30,\\ 1, & x\ge30. \end{cases} $$ The jump at 10 is $0.30$, not either individual probability $0.12$ or $0.18$. The jump at 20 is $0.50$, the sum of the two probabilities assigned that score.

Worked Example: Tied Sums From Two Independent Trials

Two independent tests each have probability $0.4$ of a positive result and probability $0.6$ of a negative result. Let $X$ be the number of positive results. Find its CDF and calculate $P(0<X<2)$.

The four ordered outcome pairs have probabilities $$ P(0,0)=0.6(0.6)=0.36,\quad P(0,1)=0.6(0.4)=0.24,\quad P(1,0)=0.4(0.6)=0.24,\quad P(1,1)=0.4(0.4)=0.16. $$ The middle two pairs are distinct outcomes, but both give $X=1$. Hence the combined probabilities are $q(0)=0.36$, $q(1)=0.24+0.24=0.48$, and $q(2)=0.16$. Their sum is $0.36+0.48+0.16=1$.

The CDF is $$ F(x)= \begin{cases} 0, & x<0,\\ 0.36, & 0\le x<1,\\ 0.84, & 1\le x<2,\\ 1, & x\ge2. \end{cases} $$ The event $0<X<2$ is exactly the event $X=1$, so its probability is $0.48$. The CDF calculation agrees: $$ P(0<X<2)=F(2^-)-F(0)=0.84-0.36=0.48. $$ The contribution at 1 is the sum from both ordered pairs, not the probability of just one of them.

Using the Grouped CDF at Nearby Thresholds

Once the distinct values and their combined probabilities are listed, evaluating the CDF requires only checking which values are at most the threshold. If $a<b$ are consecutive possible values, then the CDF is constant from just after $a$ up to just before $b$. At $b$, the entire combined probability $q(b)$ is added. This follows directly from the defining condition $X\le x$.

For instance, if the only values around a threshold are 10 and 20, then $F(9.9)$ includes neither value, $F(10)$ includes the full mass at 10, and $F(10.1)$ includes exactly the same values as $F(10)$. The CDF does not assign only part of a tie at the threshold: all outcomes producing 10 satisfy $X\le10$, and none of them satisfy $X\le9.9$.

For interval probabilities, the grouped CDF can be used through the usual difference formula, with endpoints handled according to the event being asked about. For example, $P(a<X\le b)=F(b)-F(a)$. If a value is tied at $b$, its full combined probability is included; if a value is tied at $a$, it is excluded from this particular interval. The strict and weak inequalities are part of the event, so check them before subtracting.

Common Mistake and Why Grouping Matters

Wrong: Give separate jumps at 10 of sizes $0.12$ and $0.18$ because two outcomes produce 10. Correction: A CDF is a function of the numerical value $X$, so it has one jump at 10. Its size is $0.12+0.18=0.30$.
Wrong: At a threshold equal to a tied value, include only the probability of one outcome producing that value. Correction: The event $X\le v$ includes every outcome for which $X=v$, so it includes their full combined probability.
Wrong: Treat a threshold just above a tied value as if it had reached the next possible value. Correction: If there are no possible values between $v$ and the next value $w$, then $F(x)=F(v)$ for $v\le x<w$.
Wrong: Add probabilities for outcome labels without checking whether those outcomes satisfy the threshold. Correction: First identify which numerical values are at most $x$, then include every outcome tied at each of those values.

The same method applies whenever a rule maps several raw outcomes to one number: a score, a count, a category code treated as a numerical value, or a transformed measurement. Grouping makes the CDF reflect the distribution of the quantity actually being studied, rather than the bookkeeping labels of the experiment.

Key takeaway: To construct a CDF from outcome-level information, list the distinct values of $X$, add together the probabilities of all outcomes tied at each value, and accumulate those combined probabilities at thresholds that include the value.

Check Your Understanding

For each question, identify the tied values before calculating.

Three mutually exclusive outcomes have probabilities $0.2, 0.3,$ and $0.5$, and produce values $4,4,$ and $7$. What are the combined probabilities at 4 and 7?
For the outcomes in question 1, find $F(3.9)$, $F(4)$, and $F(6)$.
Four equally likely outcomes produce values $0,1,1,$ and $2$. What is the jump at 1, and what is $F(1)$?
Explain why two distinct outcomes with the same value do not create two separate CDF jumps at that value.
If the distinct possible values are 2 and 5, what values of $x$ satisfy $F(x)=F(2)$ when $2\le x<5$?