Drawing And Reading Claim Payment Diagrams

Introduction

A claim payment diagram makes the relationship between a loss and the insurer’s payment visible. Rather than beginning with a coverage rule, a reader can inspect the graph to see where payment starts, how quickly it increases, and whether it eventually stops increasing. Those features can reveal important parts of the coverage structure, but only if the axes and the graph’s scale are read correctly.

This tutorial focuses on reading such diagrams. It uses $X$ for the amount of a loss and $Y^L$ for the insurer’s payment per loss. The examples describe hypothetical coverage and do not represent real policies. Unless stated otherwise, amounts on both axes are in dollars, the graph applies to one loss, and the displayed line segments are interpreted as straight between their labelled points.

Definition: A claim payment diagram plots the insurer’s payment $Y^L$ on the vertical axis against the loss amount $X$ on the horizontal axis. The zero-payment region is the range of losses for which the graph lies on the horizontal axis. An attachment point is the loss amount where payment begins to rise above zero. A segment’s slope describes how much the insurer’s payment changes for each additional dollar of loss. A payment cap is indicated when the graph becomes horizontal at a positive payment amount.

How to Read the Axes and Graph Features

First identify what each axis measures and its units. A point $(x,y)$ on the graph means that a loss of $x$ dollars produces an insurer payment of $y$ dollars. The horizontal axis represents loss, not the amount the insured retains, and the vertical axis represents insurer payment, not total loss.

Next, locate the graph’s changes in direction or slope. A horizontal segment at zero means that the insurer payment remains zero over that range of losses. If the graph then rises, the loss value at the start of the rising segment is the attachment point. A rising straight segment means the insurer payment increases as the loss increases. A horizontal segment above zero means payment stays at a fixed amount over that range; when it is the upper plateau of the graph, it indicates a cap within the displayed range.

For a straight segment, its slope is the change in vertical value divided by the change in horizontal value:

\text{slope} = \frac{\text{change in insurer payment}}{\text{change in loss}} = \frac{\Delta Y^L}{\Delta X}.

For example, a slope of $1$ means each additional dollar of loss along that segment corresponds to one additional dollar of insurer payment. A slope of $0.4$ means each additional dollar of loss corresponds to 40 cents of additional payment. A horizontal segment has slope $0$. These interpretations apply to the segment being examined; a slope should not be assumed to remain the same in other parts of the graph.

A Short Justification: Why Segment Slope Matters

Take two points on the same straight segment: $(x_1,y_1)$ and $(x_2,y_2)$. Between those points, the loss increases by $x_2-x_1$, while the payment increases by $y_2-y_1$. The ratio of those changes is the segment’s slope. Because the graph is straight on this interval, the payment changes at a constant rate per dollar of additional loss there.

\text{slope} = \frac{y_2-y_1}{x_2-x_1}, \qquad x_2\ne x_1.

This is why the slope must be calculated from two points on the segment being interpreted. Comparing a point to the origin instead can combine multiple parts of the graph and give a ratio that is not the slope of any one segment.

Worked Examples

Worked Example: Finding an Attachment Point and Reading a Payment

A hypothetical policy’s payment graph has labelled points $(0,0)$, $(1{,}000,0)$, and $(4{,}000,3{,}000)$. The graph is horizontal between the first two points and straight between the second and third. Find the attachment point, the slope of the rising segment, and the payment for a loss of $2{,}500$. State whether the graph shows a payment cap.

Method. Read the zero-payment interval first, then calculate the slope using the endpoints of the rising segment. Use that segment’s rate of increase to read the payment at the specified loss.

Attachment point. Payment is zero from a loss of $0$ through $1{,}000$. The graph begins rising at $X=1{,}000$, so the attachment point is $\$1{,}000$.

Slope. Using $(1{,}000,0)$ and $(4{,}000,3{,}000)$,

\text{slope} = \frac{3{,}000-0}{4{,}000-1{,}000} = \frac{3{,}000}{3{,}000} = 1.

The specified loss, $\$2{,}500$, is $\$1{,}500$ above the attachment point. On a segment with slope $1$, payment increases by $\$1$ per additional dollar of loss. Thus the payment is $\$1{,}500$. Equivalently, moving from the point $(1{,}000,0)$ to $X=2{,}500$ gives $Y^L=0+1(2{,}500-1{,}000)=\$1{,}500$.

Cap. No cap is shown on the described graph: the payment is still rising at the final labelled point. The diagram does not establish what happens beyond the plotted range, so it would be incorrect to conclude that the payment must keep rising indefinitely.

Check. The answer lies between the payments at the endpoints of the rising segment: $\$0 < \$1{,}500 < \$3{,}000$. A common mistake is to divide $\$3{,}000$ by the endpoint loss $\$4{,}000$ and call the result the segment slope. That ratio uses the origin, not two points on the rising segment; the correct slope is $\$3{,}000/(\$4{,}000-\$1{,}000)=1$.

Worked Example: Distinguishing a Sloped Segment from a Plateau

A hypothetical payment diagram passes through $(0,0)$, $(2{,}000,1{,}000)$, $(5{,}000,2{,}500)$, and $(7{,}000,2{,}500)$, with straight segments between consecutive points. Find the payment for a loss of $\$4{,}000$, describe the slopes of the segments, and identify whether there is an initial zero-payment region or a displayed cap.

Method. Identify which segment contains the stated loss, calculate the slope on that segment, and then check whether the graph begins with a zero-payment interval or ends in a positive horizontal plateau.

The loss of $\$4{,}000$ lies between $\$2{,}000$ and $\$5{,}000$. On that segment the slope is

\frac{2{,}500-1{,}000}{5{,}000-2{,}000} = \frac{1{,}500}{3{,}000} = 0.5.

The loss increases by $\$4{,}000-\$2{,}000=\$2{,}000$ from the segment’s starting point. The corresponding payment increase is $0.5(\$2{,}000)=\$1{,}000$. Therefore, the payment is $\$1{,}000+\$1{,}000=\$2{,}000$.

The first segment rises from $(0,0)$ to $(2{,}000,1{,}000)$, so its slope is $1{,}000/2{,}000=0.5$. The next rising segment also has slope $0.5$. The final segment is horizontal at $Y^L=\$2{,}500$, so its slope is zero and the graph shows a cap of $\$2{,}500$ over the displayed range. There is no initial zero-payment interval of positive length: payment is already increasing immediately after $X=0$.

Check. At $X=\$4{,}000$, the payment of $\$2{,}000$ is between the payments of $\$1{,}000$ and $\$2{,}500$ at the segment endpoints. A common mistake is to treat the final horizontal segment as a zero-payment region because it has slope zero. Its height matters: this plateau is at $\$2{,}500$, not at zero.

Worked Example: Reading a Graph with Both a Zero Region and a Cap

A hypothetical graph has the following consecutive labelled points, joined by straight segments: $(0,0)$, $(1{,}500,0)$, $(4{,}500,3{,}000)$, $(7{,}500,6{,}000)$, and $(9{,}000,6{,}000)$. Determine the attachment point, the slope on each rising segment, the cap shown, and the payment for a loss of $\$3{,}000$. Also determine the payment for a loss of $\$8{,}000$.

Method. Read the horizontal segment at zero to find where payment begins, calculate each rising segment’s slope separately, and use the final positive horizontal segment to identify the cap.

The graph stays at zero through $X=\$1{,}500$ and then rises, so its attachment point is $\$1{,}500$. The first rising segment has slope

\frac{3{,}000-0}{4{,}500-1{,}500} = \frac{3{,}000}{3{,}000} = 1.

The second rising segment also has slope

\frac{6{,}000-3{,}000}{7{,}500-4{,}500} = \frac{3{,}000}{3{,}000} = 1.

A loss of $\$3{,}000$ is $\$1{,}500$ above the attachment point and lies on the first rising segment. With slope $1$, the payment is $\$1{,}500$. For a loss of $\$8{,}000$, the graph is on its final horizontal segment, which begins at a loss of $\$7{,}500$ and stays at a payment of $\$6{,}000$. Thus the payment is $\$6{,}000$. The displayed cap is $\$6{,}000$.

Check. The payment at $\$3{,}000$ falls between $\$0$ at loss $\$1{,}500$ and $\$3{,}000$ at loss $\$4{,}500$; the $\$8{,}000$ loss is past the point where the graph becomes flat. A common mistake is to extend the rising line past $X=\$7{,}500$ and report a payment above $\$6{,}000$. That contradicts the horizontal plateau shown on the graph.

A Reliable Reading Process

A consistent sequence helps prevent errors, especially when a diagram has several line segments. Use the graph’s labelled values rather than estimating from the apparent visual angle: the axes may use different scales.

Confirm the axes and units.
Check that horizontal values are losses and vertical values are insurer payments. Note whether the axes use dollars, thousands of dollars, or another scale.

Locate zero-payment intervals.
Find any horizontal segment on $Y^L=0$. The loss coordinate where the graph first rises above zero is the attachment point, if such a positive-length interval is shown.

Calculate slopes segment by segment.
Use two points from the same straight segment. Do not use points from different segments or automatically measure from the origin.

Look for a positive plateau.
A horizontal segment above zero shows that payment is constant over that part of the graph. If it is the upper plateau, read its vertical coordinate as the displayed cap.

Place the requested loss on the correct segment.
Use only that segment’s behavior to find the payment. Do not extend a line past a visible change in slope or beyond the range shown.

Common Diagram-Reading Traps

The graph summarizes payment behavior, but it does not excuse careless reading of scale, segment boundaries, or what the diagram actually displays. In particular:

Using the wrong slope interval. A slope must be calculated from two points on the same straight segment. A graph that bends or changes direction has different segment slopes.
Confusing a zero plateau with a cap. Both are horizontal, but a zero plateau lies at payment $0$, while a cap plateau lies at a positive payment amount.
Calling the first rising payment the attachment point. The attachment point is read from the horizontal loss coordinate where payment begins to rise, not from the vertical payment amount.
Ignoring axis scale. A visually steep segment need not have a large numerical slope if the axes use different scales. Use labelled coordinates and units.
Extrapolating beyond the diagram. A graph only supports conclusions over the range shown unless the diagram or accompanying information specifies the behavior outside that range.

Claim payment diagrams are useful because they make changes in insurer payment visible at a glance. Identifying zero-payment regions, attachment points, slopes, and caps is also a practical check on a verbal description or a calculation: a proposed payment should agree with the graph’s direction and the segment containing the loss.

Check Your Understanding

For each hypothetical diagram, use the labelled coordinates and straight-line segments as stated. Amounts are in dollars.

A graph joins $(0,0)$, $(800,0)$, and $(2{,}800,2{,}000)$. What is the attachment point, and what is the slope of the rising segment? (Answer: attachment point $\$800$; slope $1$.)
A graph joins $(0,0)$, $(2{,}000,1{,}000)$, and $(6{,}000,3{,}000)$. What payment does it show for a loss of $\$4{,}000$? (Answer: $\$2{,}000$.)
A graph joins $(0,0)$, $(1{,}000,0)$, $(3{,}000,1{,}000)$, and $(5{,}000,1{,}000)$. Identify the attachment point, the slope of the rising segment, and the displayed cap. (Answer: attachment point $\$1{,}000$; slope $0.5$; cap $\$1{,}000$.)
A graph joins $(0,0)$, $(1{,}500,1{,}500)$, $(4{,}500,4{,}500)$, and $(6{,}500,4{,}500)$. Find the payment for a loss of $\$5{,}500$, and state why the rising-segment slope should not be used there. (Answer: $\$4{,}500$; the loss lies on the horizontal plateau, not a rising segment.)
A graph joins $(0,0)$, $(2{,}000,0)$, $(5{,}000,1{,}500)$, and $(8{,}000,4{,}500)$. Find the slope of each rising segment and the payment at a loss of $\$7{,}000$. (Answer: slopes $0.5$ and $1$; payment $\$3{,}500$.)

What You'll Learn