Why the Range Has Its Own Sampling Distribution
The sample range is the distance between the largest and smallest observations. It is a simple summary of spread, but its sampling distribution is not determined by the sample size alone: it depends on the population distribution as well. Knowing that distribution lets us calculate probabilities such as the chance that a sample spans at most a specified distance.
Throughout this tutorial, assume \(X_1,\ldots,X_n\) are IID observations from a continuous population with cumulative distribution function \(F\) and density \(f\). “IID” means independent and identically distributed. Assume \(n\geq 2\), and define the sample minimum, maximum, and range by
The range is a statistic: it is calculated from the random sample, so it is itself random before the observations are seen. For a continuous population, ties have probability zero. The range is nonnegative, and its possible upper endpoint depends on the population support. If \(n=1\), the range is always zero, so the results below are for \(n\geq2\).
From the Minimum and Maximum to the Range
The key step is to use the joint density of the sample minimum and maximum. For \(u<v\), one observation must fall near \(u\), a different observation must fall near \(v\), and all other observations must lie between them. There are \(n\) choices for the observation at the minimum and \(n-1\) choices for the one at the maximum. Independence then gives the joint density
This density is zero for \(u\geq v\). The factor \(F(v)-F(u)\) is the probability that one observation falls between \(u\) and \(v\); raising it to \(n-2\) accounts for the remaining observations. This is the continuous-sample joint order-statistic formula applied to the two endpoints.
To find the CDF of the range, add the joint probabilities over all pairs of endpoints whose distance is no more than \(r\). For \(r\geq0\), this means \(u<v\leq u+r\). Integrating first with respect to \(v\) gives
For the second equality, hold \(u\) fixed and use \(\frac{d}{dv}[F(v)-F(u)]^{n-1}=(n-1)f(v)[F(v)-F(u)]^{n-2}\). The inner integral is therefore \([F(u+r)-F(u)]^{n-1}/(n-1)\), which cancels the factor \(n-1\). The formula includes values outside the population’s support by taking \(f=0\) there.
When differentiation under the integral is valid, differentiating the CDF gives the density of the range:
These formulas are useful because they reduce a problem about all \(n\) observations to an integral involving the population CDF and density. They are for an IID continuous sample; they should not be applied unchanged to a discrete population, where ties can have positive probability.
Worked Example: A Range Probability for a Uniform Sample
Four IID measurements are sampled from a hypothetical uniform population on \([0,20]\). Find the probability that the sample range is greater than 3 and no greater than 10. Use the uniform range CDF, and give the probability to four decimal places.
Method. For a uniform population on an interval of width \(L\), \(F(u+r)-F(u)\) varies near the two support boundaries. Applying the range CDF formula and splitting the integral at \(L-r\) gives, for \(0\leq r\leq L\),
Here \(n=4\) and \(L=20\). The event is \(3<R\leq10\), so its probability is \(F_R(10)-F_R(3)\).
At \(r=10\), \(r/L=10/20=0.5\), and \(0.5^3=0.125\), \(0.5^4=0.0625\). Therefore,
At \(r=3\), \(r/L=3/20=0.15\), \(0.15^3=0.003375\), and \(0.15^4=0.00050625\). Therefore,
Subtract the lower-tail probability from the upper CDF value: \(0.3125-0.01198125=0.30051875\). Rounded only at the final step to four decimal places, the answer is 0.3005. As a check, the CDF values are increasing and both lie between zero and one, so the resulting interval probability is positive and no greater than \(F_R(10)\).
Common mistake avoided: using \((r/L)^n\) as the range CDF. That expression does not account for the freedom to place the sample’s minimum anywhere in the interval; the correct CDF here is \(4(r/20)^3-3(r/20)^4\).
Two Useful Special Results
For a uniform population on \([a,b]\), let \(L=b-a\). The range distribution depends on the interval width \(L\), not on its location \(a\). The CDF above applies on \(0\leq r\leq L\); it is zero for \(r<0\) and one for \(r\geq L\). This gives a convenient direct calculation for probabilities and also provides a way to check whether a proposed range is possible.
Another useful case is an exponential population. If \(X_i\) are IID exponential random variables with rate \(\lambda>0\), then \(f(u)=\lambda e^{-\lambda u}\) and \(F(u)=1-e^{-\lambda u}\) for \(u\geq0\). For \(u\geq0\),
Substituting into the range CDF formula yields
The last equality uses \(\int_0^\infty \lambda e^{-n\lambda u}\,du=1/n\). This compact form is specific to the exponential model; it is not a general formula for continuous populations.
Worked Example: Range Probabilities for Exponential Observations
A hypothetical reliability study models three independent component lifetimes, in years, as exponential with rate \(\lambda=0.5\) per year. Find the probability that the range of the three lifetimes is greater than 1 year and no greater than 2 years. Give the probability to four decimal places.
Method. For an exponential sample with \(n=3\), use \(F_R(r)=(1-e^{-0.5r})^2\). The requested event is \(1<R\leq2\), so calculate \(F_R(2)-F_R(1)\).
At \(r=1\), \(0.5r=0.5\), \(e^{-0.5}\approx0.60653066\), and \(1-e^{-0.5}\approx0.39346934\). Squaring gives
At \(r=2\), \(0.5r=1\), \(e^{-1}\approx0.36787944\), and \(1-e^{-1}\approx0.63212056\). Thus
The difference is \(0.39957640-0.15481812=0.24475828\). Rounded at the final step to four decimal places, the probability is 0.2448. The result is positive and less than \(F_R(2)\), as required for a probability of a subinterval of \([0,2]\).
Common mistake avoided: treating the three lifetimes as if the range were a single exponential lifetime. The range is formed from both sample extremes, and its CDF is \((1-e^{-0.5r})^2\) for this sample size.
Expected Range in a Uniform Sample
The CDF also gives the expected range when the population is uniform. For a nonnegative random variable bounded by \(L\), the tail-integral identity is \(E[R]=\int_0^L[1-F_R(r)]\,dr\). Substituting the uniform CDF and integrating gives
In the second line, the two power integrals are \(\int_0^L(r/L)^{n-1}dr=L/n\) and \(\int_0^L(r/L)^n dr=L/(n+1)\). The first two terms cancel, leaving the stated result. It has sensible boundary behavior: for \(n=2\) the expected range is \(L/3\), and as \(n\) increases the expected range approaches the full interval width \(L\).
Worked Example: Expected Range From a Uniform Population
Five IID readings are sampled from a hypothetical uniform population on \([8,20]\), measured in the same units as the endpoints. Find the expected sample range.
Method. The interval width is \(L=20-8=12\), and the sample size is \(n=5\). Apply \(E[R]=L(n-1)/(n+1)\):
The expected range is 8 units. This is below the maximum possible range of 12 units, as it must be. The calculation uses the interval width, not its upper endpoint; substituting 20 for \(L\) would incorrectly ignore the lower endpoint.
Common mistake avoided: using \(b\) rather than \(b-a\) as the uniform interval width. Translating the population from \([0,12]\) to \([8,20]\) does not change any range.
Interpretation and Exam Traps
The range is useful when the spread between the two most extreme observations is directly relevant, but it depends only on those two observations. Its distribution must therefore be calculated from the population model and the sample size. A range CDF from one model cannot be transferred to another just because the sample sizes match.
- Reverse the subtraction carefully. For \(a<R\leq b\), use \(F_R(b)-F_R(a)\), not the reverse and not a single CDF value.
- Use the correct scale. For a uniform population on \([a,b]\), the width is \(L=b-a\), and the range cannot exceed \(L\).
- Distinguish range from maximum. The range is maximum minus minimum; it is not the sample maximum and is not the distance from zero to the maximum.
- Check the model assumptions. The formulas here rely on independent, identically distributed continuous observations. Discrete observations may tie, changing the relevant distribution.
- Check probability boundaries. A valid range CDF is nondecreasing, equals zero for negative \(r\), and approaches one as \(r\) grows without bound (or reaches the maximal possible width for a bounded population).
The central calculation is to integrate the joint density of the minimum and maximum over endpoint pairs with separation at most \(r\). That step explains why the range distribution reflects the entire population model, while also making model-specific calculations manageable.
Check Your Understanding
For each problem, identify the range event first, then use the formula appropriate to the stated population. Answers are rounded to four decimal places where necessary.
- Three IID observations come from a uniform population on \([0,15]\). Find \(P(R\leq5)\). (Answer: 0.2593)
- Two IID observations come from a uniform population on \([2,14]\). Find \(P(2<R\leq6)\). (Answer: 0.4444)
- Four IID observations come from an exponential population with rate \(0.25\) per time unit. Find \(P(R\leq4)\). (Answer: 0.2526)
- Five IID observations come from a uniform population on \([0,10]\). Find the expected range. (Answer: \(20/3\), or approximately 6.6667)
- Three IID observations come from an exponential population with rate \(0.2\) per time unit. Find the probability that the range exceeds 5 time units. (Answer: 0.6004)