Uniform Severity And The Sample Maximum

Introduction

For a uniform severity model, the density is constant across its possible loss range. If the lower endpoint is known but the upper endpoint is not, every feasible candidate upper endpoint assigns the same density to each observed loss, but a larger endpoint spreads that density over a wider interval. This makes the likelihood favor the smallest upper endpoint that can contain all the observations: the sample maximum.

This is an important boundary case in maximum likelihood estimation. The usual strategy of differentiating the log-likelihood, setting the derivative equal to zero, and solving does not produce the estimate. The parameter also determines the support of the distribution, and the largest observed loss imposes a constraint on its possible values.

Definition: Let \(X_1,\ldots,X_n\) be independent, identically distributed complete loss observations from a uniform distribution on \([a,b]\), where the lower endpoint \(a\) is known and the upper endpoint \(b\) is unknown, with \(b>a\). The sample maximum is \(X_{(n)}=\max(X_1,\ldots,X_n)\). When \(X_{(n)}>a\), the maximum likelihood estimate of \(b\) is \(\hat b=X_{(n)}\), under the usual closed-support likelihood convention.

Likelihood and the Boundary Estimate

Under this model, the density of one observation is \(f_X(x)=1/(b-a)\) for \(a\le x\le b\), and zero otherwise. The observations must be complete, uncensored, and untruncated for this likelihood to apply as written. If even one observation is below \(a\), the model assigns probability zero to the data for every \(b\).

Suppose all observations are at least \(a\), and write \(m=X_{(n)}\). A candidate value of \(b\) can give the data positive likelihood only if \(b\ge m\). For such a candidate, each observation has density \(1/(b-a)\). Independence gives the likelihood:

L(b)=\prod_{i=1}^{n} f_X(x_i) =\frac{1}{(b-a)^n},\qquad b\ge m.

For feasible \(b\), increasing \(b\) increases the length \(b-a\) of the interval and reduces the density assigned to every observation. Equivalently, the likelihood decreases as \(b\) increases. The smallest feasible endpoint is \(m\), so the likelihood is maximized at \(b=m\), provided \(m>a\).

Check the support.
Confirm that every observation is at least the known lower endpoint \(a\). If not, the data are incompatible with the model.

Find the sample maximum.
Set \(m=X_{(n)}\). Any feasible upper endpoint must satisfy \(b\ge m\).

Maximize over feasible values.
Since \(1/(b-a)^n\) decreases as \(b\) increases, choose the smallest feasible value, \(\hat b=m\).

Why There Is No Score Root

For values strictly above the sample maximum, the log-likelihood is \(\ell(b)=-n\log(b-a)\). Its derivative is negative throughout that feasible interior:

\frac{d\ell(b)}{db}=-\frac{n}{b-a}<0,\qquad b>m.

Thus the score equation \(d\ell(b)/db=0\) has no solution. This does not mean there is no maximum. The feasible set begins at \(m\), and the likelihood decreases as one moves away from that boundary. The maximum occurs at the boundary \(b=m\), rather than at a stationary point in the interior.

This reasoning depends on the support constraint. A score equation derived only for interior values cannot express the fact that \(b\) must contain the largest observation. In the usual likelihood convention, the uniform density includes both endpoints of its support, so \(b=m\) is feasible. If a problem instead imposes the strict constraint \(b>m\), the likelihood approaches its supremum as \(b\) decreases toward \(m\), but does not attain it; under that convention there is no MLE, only a supremum. State the parameter-space convention when it affects the conclusion.

Worked Example: Finding the Upper Endpoint from Complete Losses

A hypothetical portfolio has four complete losses, measured in dollars: \(112,119,127,\) and \(131\). Assume they are iid uniform on \([100,b]\), where the lower endpoint is known. Find the MLE of \(b\), write the likelihood, and compare it with the likelihood at \(b=136\).

Method. Use the support constraint and maximize the complete-data likelihood over feasible upper endpoints. This avoids trying to find an interior score root that does not exist.

Step 1: Check feasibility and find the maximum. Every observation is at least \(100\), and the maximum is \(m=131\). Therefore, feasible candidates satisfy \(b\ge131\).

Step 2: Write the likelihood. There are \(n=4\) independent observations and \(b-a=b-100\), so for \(b\ge131\),

L(b)=\frac{1}{(b-100)^4}.

Step 3: Maximize. The likelihood decreases as \(b\) increases, so the smallest feasible value gives the maximum: \(\hat b=131\). The maximized likelihood is \(L(131)=1/31^4=1/923521\).

At \(b=136\), \(L(136)=1/36^4=1/1679616\), which is smaller. In fact, the likelihood ratio is \(L(136)/L(131)=(31/36)^4<1\), confirming the comparison. The estimate contains the largest observation, and increasing the endpoint reduces the likelihood, as required.

The Sample Maximum as a Random Variable

The estimate \(\hat b=X_{(n)}\) varies from sample to sample. Its distribution explains why it tends to fall below the true endpoint. Assume the same iid uniform model with true upper endpoint \(b\). For \(a\le t\le b\), the event \(X_{(n)}\le t\) occurs exactly when every observation is at most \(t\). Independence therefore gives:

P(X_{(n)}\le t) =\prod_{i=1}^{n}P(X_i\le t) =\left(\frac{t-a}{b-a}\right)^n.

Differentiating this cumulative distribution function with respect to \(t\) gives the density of the sample maximum on \([a,b]\):

f_{X_{(n)}}(t)= \frac{n(t-a)^{n-1}}{(b-a)^n},\qquad a\le t\le b.

The maximum is more likely to be near \(b\) than a single observation is, because all \(n\) observations must lie below any proposed value of the maximum. Nevertheless, it cannot exceed \(b\), so it is systematically low on average when \(n\) is finite.

Result: For an iid sample of size \(n\) from the uniform distribution on \([a,b]\), \(E[X_{(n)}]=a+\frac{n}{n+1}(b-a)\). Therefore the MLE \(X_{(n)}\) has bias \(E[X_{(n)}]-b=-(b-a)/(n+1)\).

To see why the expectation formula holds, let \(Y=X_{(n)}-a\), so \(0\le Y\le b-a\). For a nonnegative random variable bounded above by \(b-a\), its expectation is the integral of its survival probability over that range. The event \(Y>y\) means at least one observation exceeds \(a+y\); the probability that all observations are at most \(a+y\) is \((y/(b-a))^n\). Hence:

\begin{aligned} E[X_{(n)}] &=a+E[Y]\\ &=a+\int_0^{b-a}P(Y>y)\,dy\\ &=a+\int_0^{b-a}\left[1-\left(\frac{y}{b-a}\right)^n\right]dy\\ &=a+(b-a)-\frac{b-a}{n+1}\\ &=a+\frac{n}{n+1}(b-a). \end{aligned}

This derivation describes the sampling behavior of the MLE; it does not change the likelihood maximization. An estimator can maximize the likelihood and still be biased. The bias also suggests a simple unbiased estimator when the model assumptions hold: \(a+\frac{n+1}{n}(X_{(n)}-a)\). That estimator is not the MLE, and it may exceed the true endpoint for a particular sample.

Worked Example: Expected Maximum and Bias

For a hypothetical study, suppose four independent losses follow a uniform distribution on \([200,500]\). A particular sample has maximum \(452\). Find the MLE of the upper endpoint, the expected sample maximum under the stated model, its bias as an estimator of \(b\), and the corresponding unbiased estimator evaluated at this sample maximum.

Method. Use the sample maximum for the MLE, then apply the sampling expectation formula. Keeping the two calculations separate avoids mistaking an unbiased adjustment for a maximum likelihood estimate.

Step 1: MLE for this sample. The observed maximum is \(452\), so \(\hat b=452\).

Step 2: Expected sample maximum. Here \(a=200\), \(b-a=500-200=300\), and \(n=4\). Therefore,

E[X_{(4)}] =200+\frac{4}{5}(300) =200+240 =440.

Step 3: Bias. The true upper endpoint for this sampling calculation is \(500\), so the bias is \(440-500=-60\). This also agrees with \(-(b-a)/(n+1)=-300/5=-60\).

Step 4: Evaluate the unbiased adjustment. The adjusted estimator is \(200+\frac{5}{4}(X_{(4)}-200)\). For the observed maximum \(452\), \(200+\frac{5}{4}(452-200)=200+\frac{5}{4}(252)=200+315=515\). Thus this sample gives an adjusted estimate of \(515\), while the MLE remains \(452\). The adjusted estimate exceeding the true endpoint in this particular sample does not contradict unbiasedness, which is a statement about the estimator’s average over repeated samples. The calculations check: the expected maximum is below \(500\), with bias \(-60\), and the unbiased adjustment has expectation \(500\).

Worked Example: Comparing Feasible Endpoints

A hypothetical dataset consists of complete losses \(46,52,\) and \(59\), modeled as iid uniform on \([40,b]\). Determine the MLE, evaluate the log-likelihood derivative at \(b=65\), and explain why setting that derivative to zero would be the wrong method.

Method. First identify the feasible boundary, then use monotonicity of the likelihood. The derivative is useful for confirming the direction of change, not for locating an interior root.

Step 1: Determine the feasible region. All observations exceed \(40\), and \(X_{(3)}=59\). Hence \(b\ge59\).

Step 2: Write and maximize the likelihood. For feasible \(b\), \(L(b)=1/(b-40)^3\), so the likelihood decreases with \(b\). It is maximized at \(\hat b=59\), with \(L(59)=1/19^3=1/6859\).

Step 3: Check the derivative at the proposed larger value. The log-likelihood is \(\ell(b)=-3\log(b-40)\), giving \(d\ell/db=-3/(b-40)\). At \(b=65\), the derivative is \(-3/25=-0.12\), which is negative. Moving to larger \(b\) reduces the log-likelihood; moving down toward the feasible boundary increases it.

The score equation has no root because \(-3/(b-40)\) is negative for every feasible interior value. The boundary estimate \(59\) is therefore correct. At \(b=65\), the likelihood is \(1/25^3=1/15625\), less than \(1/6859\). This confirms both the boundary argument and the numerical comparison.

Common Mistakes and Model Limits

Looking only for a score root. The score is negative throughout the feasible interior. Check the parameter constraints and evaluate the boundary.
Choosing an endpoint below the sample maximum. Such a candidate assigns zero likelihood to at least one observation. The feasible region begins at \(X_{(n)}\).
Forgetting the known lower endpoint. The likelihood is \(1/(b-a)^n\), not \(1/b^n\), unless \(a=0\).
Calling the MLE unbiased. The sample maximum is biased downward by \((b-a)/(n+1)\). Maximum likelihood does not imply unbiasedness.
Using the complete-data likelihood for incomplete records. Truncation or censoring changes what the observations contribute to the likelihood. The formula here requires complete observations.

There is also a degenerate data case worth recognizing. If all observed losses equal the known lower endpoint, then \(m=a\). As \(b\) decreases toward \(a\), the likelihood \(1/(b-a)^n\) grows without bound, so no finite \(b>a\) maximizes it. Such a sample has probability zero under a continuous uniform model, but an exam question may use it to test whether the usual formula’s condition \(X_{(n)}>a\) has been checked.

Key takeaway: For complete iid uniform losses on \([a,b]\) with known \(a\), the data require \(b\ge X_{(n)}\), and the likelihood decreases over that feasible range. When \(X_{(n)}>a\), the upper-endpoint MLE is the boundary value \(\hat b=X_{(n)}\), not a solution to the score equation.

The boundary-likelihood argument is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. Its essential steps are to state the support, form the likelihood for feasible parameter values, and compare the boundary with the interior rather than applying an interior score condition mechanically.

Check Your Understanding

For each problem, identify the feasible endpoint values before calculating the estimate.

Complete losses \(9,13,12,\) and \(20\) are iid uniform on \([7,b]\). Find the MLE of \(b\) and state the feasible region. (Answer: \(\hat b=20\), with \(b\ge20\).)
Three complete losses are \(31,38,\) and \(44\), from a uniform model on \([25,b]\). Find the MLE and the likelihood ratio \(L(50)/L(\hat b)\). (Answer: \(\hat b=44\); \(L(50)/L(44)=(19/25)^3=0.438976\).)
Three complete observations all equal the known lower endpoint \(3\). The model is uniform on \([3,b]\), with \(b>3\). Does a finite MLE exist? (Answer: No; the likelihood \(1/(b-3)^3\) is unbounded as \(b\) decreases toward \(3\).)
A sample of size five comes from a uniform distribution on \([10,70]\). Find the expected sample maximum and its bias as an estimator of \(b\). (Answer: \(E[X_{(5)}]=60\); bias \(=60-70=-10\).)
Four complete losses are \(52,61,73,\) and \(69\), modeled as uniform on \([50,b]\). Is \(b=72\) feasible? Find the MLE and its maximized likelihood. (Answer: \(b=72\) is infeasible; \(\hat b=73\) and \(L(\hat b)=1/23^4=1/279841\).)

What You'll Learn