Why Study the Maximum and Minimum?
Many probability questions concern the larger or smaller of two outcomes rather than their sum or difference. For example, a system might be judged by the later of two completion times, while a safety rule might depend on whether the earlier of two measurements falls below a limit. To answer such questions, we need the probability distribution of the maximum or minimum.
The calculations are especially direct when the variables are independent. Independence lets us multiply probabilities for joint events, and cumulative distribution functions turn the calculation into a product. The main care required is to include all possible outcomes and to handle ties without counting them twice.
This independence condition is stated here for discrete random variables, the setting of this tutorial. If a joint probability table is given directly, its entries can be used whether or not the variables are independent. If only the two marginal distributions are given, however, independence is what allows us to calculate the joint probabilities by multiplying.
For a discrete random variable \(X\), its cumulative distribution function is \(F_X(t)=P(X\le t)\). It adds the probabilities of all values no greater than \(t\). The probability mass function gives the probability at each individual value; the cumulative distribution function gives the accumulated probability up to a threshold.
Finding Distributions by Adding Joint Probabilities
The direct method starts with the possible pairs \((x,y)\). For each pair, calculate its joint probability. Then group pairs according to the value of \(\max(x,y)\), or according to the value of \(\min(x,y)\), and add the probabilities in each group. Under independence, each pair has probability \(P(X=x)P(Y=y)\).
A pair with equal values contributes to just one outcome of the maximum and one outcome of the minimum. More generally, when grouping pairs, make sure each pair is assigned once to the appropriate value. The next example shows this method in detail.
Worked Example: Grouping Joint Probabilities
Suppose two independent discrete variables \(X\) and \(Y\) have the following probability mass functions:
| Value | \(P(X=\text{value})\) | \(P(Y=\text{value})\) |
|---|---|---|
| 0 | 0.20 | 0.40 |
| 1 | 0.50 | 0.10 |
| 2 | 0.30 | 0.50 |
Find the probability distributions of \(M=\max(X,Y)\) and \(L=\min(X,Y)\). Independence gives \(P(X=x,Y=y)=P(X=x)P(Y=y)\).
For \(M=0\), both variables must equal 0:
For \(M=1\), the pairs are \((1,0),(1,1),(0,1)\). Their joint probabilities add to
The remaining pairs have maximum 2. Equivalently, \(M=2\) whenever \(X=2\), or when \(Y=2\) and \(X\) is 0 or 1:
For \(L=0\), at least one variable must be 0. Count the pairs with \(X=0\), then add the pairs with \(Y=0\) and \(X>0\), so the pair \((0,0)\) is not counted twice:
For \(L=1\), either \(X=1\) and \(Y\ge1\), or \(Y=1\) and \(X>1\). These cases are disjoint:
Finally, \(L=2\) only when both variables equal 2, giving \(P(L=2)=0.30(0.50)=0.15\). As checks, the maximum probabilities sum to \(0.08+0.27+0.65=1\), and the minimum probabilities sum to \(0.52+0.33+0.15=1\).
A Cumulative-Probability Method
The joint-probability method works well for short lists of possible pairs. A cumulative-probability method is often quicker, especially when the variables have several possible values. The key is to express an event about the maximum or minimum as an event about both variables.
For the maximum to be at most \(t\), both variables must be at most \(t\). Independence therefore gives \(P(X\le t,Y\le t)=P(X\le t)P(Y\le t)\). For the minimum to be at most \(t\), it is enough that at least one variable be at most \(t\). Its complement is that both are greater than \(t\). By independence, the probability of that complement is \(P(X>t)P(Y>t)=[1-F_X(t)][1-F_Y(t)]\). Subtracting from 1 gives the stated formula for the minimum.
These formulas give cumulative probabilities for \(M\) and \(L\), not yet their probabilities at individual values. To obtain a probability mass function, take successive differences. At an attainable value \(w\), the probability \(P(M=w)\) is the cumulative probability through \(w\) minus the cumulative probability below \(w\); the same rule applies to \(L\). This difference step is important when the variables have more than two possible values.
Worked Example: Using Cumulative Probabilities
Suppose independent variables \(X\) and \(Y\) have these distributions:
| Value | \(P(X=\text{value})\) | \(P(Y=\text{value})\) |
|---|---|---|
| 1 | 0.25 | 0.50 |
| 2 | 0.50 | 0 |
| 3 | 0 | 0.50 |
| 4 | 0.25 | 0 |
Find the distributions of the maximum and minimum. The zero entries make the possible values clear: \(X\) can be 1, 2, or 4, while \(Y\) can be 1 or 3. First calculate the cumulative probabilities at the relevant thresholds.
| \(t\) | \(F_X(t)\) | \(F_Y(t)\) | \(P(M\le t)=F_X(t)F_Y(t)\) | \(P(L\le t)=1-[1-F_X(t)][1-F_Y(t)]\) |
|---|---|---|---|---|
| 1 | 0.25 | 0.50 | 0.125 | 0.625 |
| 2 | 0.75 | 0.50 | 0.375 | 0.875 |
| 3 | 0.75 | 1.00 | 0.750 | 1.000 |
| 4 | 1.00 | 1.00 | 1.000 | 1.000 |
For example, at \(t=2\), \(F_X(2)=0.25+0.50=0.75\) and \(F_Y(2)=0.50\). Thus \(P(M\le2)=0.75(0.50)=0.375\). The minimum calculation at the same threshold is
Subtract successive cumulative probabilities to get the probability mass functions. For the maximum, the cumulative probability is 0 before 1, so
For the minimum, the cumulative probability is also 0 before 1:
The minimum cannot be 4 because \(Y\) never exceeds 3. The probabilities for the maximum sum to 1, as do those for the minimum.
From the Cumulative Formula to Individual Probabilities
The cumulative formulas can also be written directly in terms of probabilities at a value. This makes the role of ties explicit. Write \(F_X(m-)=P(X<m)\) and \(F_Y(m-)=P(Y<m)\). The mass of the maximum at \(m\) is
The first product counts pairs with both values at most \(m\); the second counts pairs with both values strictly below \(m\). Their difference is the probability that the maximum is exactly \(m\). This subtraction counts pairs where one value equals \(m\), including the tie \((m,m)\), exactly once.
For the minimum, subtract the cumulative probability below \(m\) from the cumulative probability through \(m\). In terms of the probabilities of exceeding a threshold, this gives
Here the first product is the probability that both variables are at least \(m\); the second is the probability that both are greater than \(m\). Their difference is the probability that the smaller value is exactly \(m\), again including a tie only once.
Worked Example: Applying the Cumulative Formula to a Threshold
Suppose independent variables \(X\) and \(Y\) represent the numbers of checks completed by two hypothetical machines during a fixed interval. Let \(X\) take values 0, 1, and 3 with probabilities 0.30, 0.40, and 0.30. Let \(Y\) take values 1 and 2 with probabilities 0.60 and 0.40. Find the probability that the larger count is at most 2, and the probability that the smaller count is at least 2.
For the maximum to be at most 2, both counts must be at most 2. We have \(F_X(2)=0.30+0.40=0.70\) and \(F_Y(2)=1\). By independence,
Thus, the probability that both machines complete no more than 2 checks is 0.70; equivalently, the maximum count is no more than 2 with probability 0.70.
For the minimum to be at least 2, both counts must be at least 2. The probabilities are \(P(X\ge2)=0.30\) and \(P(Y\ge2)=0.40\). Independence gives
So the probability that the smaller count is at least 2 is 0.12. This also follows from the cumulative probability at 1. Because \(P(X>1)=0.30\) and \(P(Y>1)=0.40\), independence gives \(P(L>1)=0.30(0.40)=0.12\), so \(P(L\le1)=1-0.12=0.88\) and \(P(L\ge2)=1-P(L\le1)=0.12\).
Checks, Assumptions, and Common Pitfalls
Before using products of probabilities, check that independence is part of the model or is supported by the information given. Knowing only the marginal distributions of \(X\) and \(Y\) does not determine the distribution of their maximum or minimum if they may be dependent. For example, multiplying \(P(X\le t)\) by \(P(Y\le t)\) is justified only when the events \(X\le t\) and \(Y\le t\) are independent, as they are when the discrete random variables are independent.
A second check is whether the calculation describes the event asked about. For \(M\le t\), both variables must be at most \(t\). For \(L\le t\), at least one must be at most \(t\); using the product of the two lower-tail probabilities would answer a different question. For a minimum-at-least event, both variables must be at least the threshold.
- Pitfall: multiplying probabilities without an independence assumption. The correction is to use the joint distribution if it is available, or to justify independence before using products.
- Pitfall: confusing “both” with “at least one.” The maximum is at most a threshold only when both variables are; the minimum is at most it when at least one is.
- Pitfall: double-counting ties. When listing pairs directly, use disjoint cases or group each pair once. A pair with \(X=Y=m\) belongs once to the event \(M=m\) and once to the event \(L=m\), but must not be counted twice within either distribution.
- Pitfall: reporting cumulative probabilities as point probabilities. To find \(P(M=m)\) or \(P(L=m)\), subtract the cumulative probability below \(m\) from the cumulative probability through \(m\).
As a final check, the probabilities in each resulting distribution should be nonnegative and sum to 1. If a value is not attainable by either variable, it may still be attainable by their maximum or minimum; identify possible pairs before deciding which values belong in the resulting distribution.
Check Your Understanding
Use the definitions and formulas above to answer the following questions.
- If \(X\) and \(Y\) are independent, express \(P(\max(X,Y)\le t)\) in terms of \(F_X(t)\) and \(F_Y(t)\).
- Why does finding \(P(\min(X,Y)\le t)\) naturally involve a complement?
- Independent variables have \(P(X\le3)=0.70\) and \(P(Y\le3)=0.40\). Find \(P(\max(X,Y)\le3)\).
- Suppose \(P(X>2)=0.30\) and \(P(Y>2)=0.50\), with \(X\) and \(Y\) independent. Find \(P(\min(X,Y)>2)\).
- What calculation converts cumulative probabilities into the probability mass at a particular attainable value?