Why Atoms Change Order-Statistic Calculations
A formula based on a density can be convenient when every observation has a continuous distribution. It is not a complete description when observations can equal a particular value with positive probability. Such a value is called an atom. Atoms can occur at the edge of a variable's support, such as a sensor that reports a fixed lower reading, or in the interior, such as a rounded measurement. For minima and maxima, an atom can create a jump in the order statistic's distribution.
The key is to keep strict and inclusive threshold events separate. For instance, “the maximum is at most \(t\)” and “the maximum is less than \(t\)” differ when an observation can equal \(t\). This distinction is essential for calculations involving boundary atoms and for deciding whether a density formula captures the whole distribution. The methods here are aligned with the SOA Exam P (Probability) syllabus.
Threshold Probabilities for Independent Observations
The event that the maximum is at most \(t\) requires every observation to be at most \(t\). Independence therefore turns the probability of this intersection into a product. The minimum is at most \(t\) if at least one observation is at most \(t\); its complement is the event that every observation is greater than \(t\).
Equivalently, the minimum's inclusive cumulative probability is \(P(X_{(1)}\le t)=1-\prod_i(1-F_i(t))\). The maximum's strict cumulative probability uses \(F_i(t-)\), not \(F_i(t)\), because \(F_i(t-)\) is the probability of being strictly below \(t\). These results require independence. If the observations are dependent, the probabilities of the intersections generally cannot be replaced by products.
For identically distributed independent observations with common distribution function \(F\), the products simplify to powers:
These formulas remain valid whether \(F\) is continuous, discrete, or mixed. What changes when there are atoms is the difference between \(F(t)\) and \(F(t-)\), and therefore the point mass at the threshold.
Worked Example: A Fixed Reading Can Be the Minimum
Four independent sensor readings have the same distribution. Each reading equals \(-1\) with probability \(0.25\); conditional on not equaling \(-1\), it is uniform on \((0,2)\). Find (a) the probability the sample minimum is \(-1\), and (b) the probability the sample maximum is at most \(1\). Use four decimal places for the final probabilities.
Method. Use the atom probability for the minimum and the inclusive one-observation cumulative probability for the maximum, then apply independence. No interest rate or mortality assumption is involved; the distribution is specified by the mixture model in the question.
(a) No reading is less than \(-1\), so the minimum equals \(-1\) exactly when at least one reading equals \(-1\). A single reading is not \(-1\) with probability \(1-0.25=0.75\). Thus
(b) For one reading, the probability of being at most \(1\) includes both the atom at \(-1\) and the continuous part's probability of falling in \((0,1]\). Conditional on the continuous part, that probability is \(1/2\), since the interval has length \(1\) out of \(2\). Therefore
The maximum is at most \(1\) if all four readings are at most \(1\), so
Check. The minimum probability is greater than \(0.5\), as four opportunities to observe the atom make at least one occurrence likely. The maximum probability is below the one-reading probability \(0.625\), as it must be when all four readings must meet the threshold. The common mistake is to use \(0.75^4\) as the probability that the minimum is \(-1\); that is the probability that none equals \(-1\), so it must be subtracted from \(1\).
Deriving the Point Masses
A point mass of an order statistic can be found by subtracting its strict-threshold probability from its inclusive-threshold probability. This works even if the parent observations have different distributions, as long as they are independent.
The event \(X_{(n)}<t\) is contained in \(X_{(n)}\le t\). Their difference is precisely the event \(X_{(n)}=t\).
The inclusive event has probability \(\prod_iF_i(t)\), and the strict event has probability \(\prod_iF_i(t-)\).
Thus \(P(X_{(n)}=t)=\prod_iF_i(t)-\prod_iF_i(t-)\). For the minimum, subtract \(P(X_{(1)}>t)\) from \(P(X_{(1)}\ge t)\), giving \(P(X_{(1)}=t)=\prod_i(1-F_i(t-))-\prod_i(1-F_i(t))\).
The minimum argument also follows directly from event inclusion: \(X_{(1)}\ge t\) means every observation is at least \(t\), while \(X_{(1)}>t\) means every observation is strictly greater than \(t\). Their difference is exactly the event that the minimum equals \(t\). This derivation shows why a left limit is needed: \(1-F_i(t-)=P(X_i\ge t)\), whereas \(1-F_i(t)=P(X_i>t)\).
For \(n\) identically distributed independent observations, these become \(F(t)^n-F(t-)^n\) for the maximum and \((1-F(t-))^n-(1-F(t))^n\) for the minimum. These expressions account for ties automatically. In particular, the maximum equals \(t\) if all observations are at most \(t\) and at least one equals \(t\); the minimum equals \(t\) if all observations are at least \(t\) and at least one equals \(t\).
Worked Example: Different Atoms at a Shared Lower Boundary
Two independent readings are nonnegative. Reading \(X_1\) equals \(0\) with probability \(0.20\); otherwise it is uniform on \((0,4)\). Reading \(X_2\) equals \(0\) with probability \(0.50\); otherwise it is uniform on \((0,2)\). Find the probability that (a) the maximum is at most \(1\), (b) the minimum is greater than \(1\), (c) the minimum equals \(0\), and (d) the maximum equals \(0\). Report probabilities to four decimal places.
Method. The observations are independent but not identically distributed, so calculate each marginal probability and multiply only after doing so. The stated mixture models fully specify the distributions; no interest rate or mortality assumption is applicable.
For \(X_1\), the conditional uniform probability of a value at most \(1\) is \(1/4\). Therefore
For \(X_2\), the corresponding conditional uniform probability is \(1/2\), so
(a) Both readings must be at most \(1\):
(b) Both must be greater than \(1\). Their probabilities of being greater than \(1\) are \(1-0.40=0.60\) and \(1-0.75=0.25\):
(c) Because both readings are nonnegative, the minimum equals \(0\) if at least one reading equals \(0\). The probability neither equals \(0\) is \((0.80)(0.50)=0.40\), giving
(d) The maximum equals \(0\) only if both readings equal \(0\):
Check. The minimum's atom at zero is larger than either individual zero probability because either reading can create it. The maximum's atom at zero is smaller than either individual zero probability because both readings must be zero. Confusing these two event structures reverses the products and complements.
Boundary Atoms and the Limits of Density Formulas
A boundary atom deserves the same treatment as any other atom; its location at an endpoint does not remove the jump. Suppose, for example, an observation has a continuous component on \((0,2)\) and a positive probability of equaling \(2\). Then the maximum can equal \(2\) whenever at least one observation takes the atomic value. The minimum equals \(2\) only when every observation takes it. These are distinct events, so the two order statistics can have very different masses at the same boundary.
When the common distribution is continuous and has density \(f\), the familiar formulas for \(n\) independent identically distributed observations are
These are density formulas under a continuous model. They do not report point masses. For a mixed distribution, differentiating the cumulative distribution function where it has a density captures its continuous part, but a jump at an atom is not represented by an ordinary density value. Therefore, using only the displayed density formulas can omit part—or all—of the order statistic's probability distribution.
Worked Example: A Maximum and Minimum Can Both Have a Boundary Atom
Three independent observations have the same distribution: each equals \(2\) with probability \(0.30\); otherwise it is uniform on \((0,2)\). Find (a) the probability the maximum equals \(2\), (b) the probability the minimum equals \(2\), (c) the probability the maximum is at most \(1\), and (d) the probability the minimum is greater than \(1\). Give final probabilities to four decimal places.
Method. Use the jump of the parent distribution at \(2\) for the point masses, and the one-observation cumulative probability at \(1\) for the threshold calculations. The conditional continuous distribution is uniform on \((0,2)\); no financial or mortality assumptions are involved.
At the boundary \(2\), the left limit is \(F(2-)=0.70\), because only the continuous component lies strictly below \(2\). The inclusive value is \(F(2)=1\), since the entire distribution is at or below \(2\).
(a) The maximum equals \(2\) if at least one observation equals \(2\):
(b) The minimum equals \(2\) only if all three observations equal \(2\):
For the threshold \(1\), there is no atom at \(1\). Conditional on the continuous component, the probability of being at most \(1\) is \(1/2\), so
The atom at \(2\) contributes zero to this threshold event. (c) All three observations must be at most \(1\):
(d) Each observation is greater than \(1\) with probability \(1-0.35=0.65\), so
Check. The maximum's mass at \(2\) is much larger than the minimum's: one occurrence is enough for the maximum, while three are needed for the minimum. A density-only calculation would fail to include these respective masses of \(0.6570\) and \(0.0270\).
Common Mistakes to Avoid
- Replacing “at most” with “less than.” For an atom at \(t\), \(F(t)\) includes the mass at \(t\), while \(F(t-)\) excludes it. Use the version matching the event exactly.
- Using a one-observation probability for an order statistic. For a maximum threshold, every observation must meet the threshold; for a minimum threshold, use the complement that every observation exceeds it.
- Using identical-distribution powers for unequal observations. If distributions differ, use \(\prod_i F_i(t)\) or the appropriate product of tail probabilities rather than \(F(t)^n\).
- Treating a density as a complete mixed distribution. A density describes continuous probability, not jumps. Calculate point masses from inclusive and strict probabilities.
- Assuming a minimum and maximum have the same atom rule. An atom at a lower boundary can make the minimum occur there if any observation hits it; an upper-boundary maximum can occur there if any observation hits it. For the opposite order statistic, all observations may need to hit the boundary.
A reliable exam approach is to translate the requested event into a statement about each observation, choose strict or inclusive marginal probabilities carefully, and use independence only for the resulting intersection. When a point mass is requested, subtract the strict threshold probability from the inclusive one. These steps keep boundary atoms visible rather than losing them in a density calculation.
Check Your Understanding
For each question, identify whether the event requires at least one observation or every observation to meet the stated condition. Round probabilities to four decimal places where needed.
- Three independent devices each equal \(0\) with probability \(0.20\); otherwise each is uniform on \((0,5)\). What is the probability their minimum equals \(0\)? (Answer: 0.4880)
- Independent \(X_1\) and \(X_2\) are nonnegative. \(X_1\) equals \(0\) with probability \(0.40\), otherwise it is uniform on \((0,2)\). \(X_2\) equals \(0\) with probability \(0.20\), otherwise it is uniform on \((0,4)\). Find \(P(\max(X_1,X_2)\le1)\). (Answer: 0.2800)
- Five independent observations each equal \(1\) with probability \(0.20\); otherwise each is uniform on \((0,1)\). Find the probability their maximum equals \(1\). (Answer: 0.6723)
- Independent \(A\) and \(B\) each have an atom at \(1\). \(A\) equals \(1\) with probability \(0.30\), otherwise it is uniform on \((1,3)\); \(B\) equals \(1\) with probability \(0.50\), otherwise it is uniform on \((1,5)\). Find (a) \(P(\min(A,B)=1)\) and (b) \(P(\max(A,B)\le2)\). (Answer: (a) 0.6500; (b) 0.4063)
- Independent \(X_1\) and \(X_2\) each have an atom at \(1\). \(X_1\) equals \(1\) with probability \(0.20\), otherwise it is uniform on \((0,2)\). \(X_2\) equals \(1\) with probability \(0.50\), otherwise it is uniform on \((0,4)\). Find (a) \(P(\max(X_1,X_2)<1)\) and (b) \(P(\max(X_1,X_2)=1)\). (Answer: (a) 0.0500; (b) 0.3250)