Berquist-Sherman adjustments
Derivation of Exhibit F. The last diagonal in Exhibit F is the same as the last diagonal in Exhibit A, since we are deflating the case reserves from the last diagonal backwards. Work through several columns. Go to Exhibit B and detrend the last diagonal straight upwards at 15% per annum. Multiply by the number of open claims in Exhibit D to get the cumulative case reserves outstanding. Be careful: aggregate losses are in thousands of dollars and individual losses are in dollars. Then add the cumulative paid losses from Exhibit E to get Exhibit F.
Consider an example. Look at the 48 month column in Exhibit F. The last figure in the column, $73,733,000 (accident year 1973) is taken directly from Exhibit A. To get the 48 month figure for accident year 1972, we first go to Exhibit B. The average case reserve for accident year 1973 (not 1972) is $32,266. We detrend this at 15% per annum to get $33,266 / 1.15 = $28,927. We multiply by the number of claims for 1972, or 1894 (see Exhibit D) to get $28,927 * 1894 = $54,788,000. We add the cumulative paid losses from Exhibit E of $9,771,000 for accident year 1972 at 48 months to get $64,559. That’s the figure that appears in Exhibit F for accident year 1972 at 48 months.
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Berquist-Sherman: The Interpolation
Many candidates get stumped by the interpolation. Past exam problems have used simpler interpolation methods, such as linear interpolation. For your study, make sure that you understand the interpolation in the paper.
The first paragraph on page 135 says: "Since the coefficient of determination (.99821) of the exponential curve fit is exceptionally high, interpolation by means of only a two point curve fit seems appropriate." Let’s work through the procedure in the text of the paper.
We first determine the "ultimate claims disposed ratios" that we will use for the restated settlement rates. We will use the last diagonal, just as we did for the first Berquist-Sherman adjustment. For 36 month column, we use 88.55%, shown in Exhibit K, last entry in the 36 month column. This figure is the ratio of cumulative closed claims in Exhibit I (6,916) to ultimate reported claims in Exhibit J (7,810): 6,916 / 7,810 = 0.8855. Note that the denominator is the ultimate claims from the last column in Exhibit J, not the reported claim count as of 36 months. (The authors don’t show how they derive the ultimate claims in Exhibit J.)
We now determine the 6,926 claims in Exhibit M (the first entry in the 36 month column, or accident year 1969). This is 88.55% of the ultimate claim count from Exhibit J, or 7,822: 88.55% * 7,822 = 6,926.
Now we come to the interpolation. We use the actual paid claim figures and cumulative paid loss figures from Exhibits H and I. First make sure you can find the four figures in the interpolation example on page 135 in Exhibits H and I. Exhibit H shows the cumulative paid losses (look at the accident year 1969 row) and Exhibit I shows the cumulative paid claims. Now look at Exhibit L, where these figures are carried to column X and column Y.
Question: Must we do this exponential curve fitting separately for each accident year?
Answer: Yes. But that’s actually good. Since the exponential curve fitting takes time, you probably won’t be asked to replicate the procedure on the exam. It is more likely that you will be given the parameters or simply asked to do a two point linear interpolation. Nevertheless, for your exam preparation, make sure you can follow what the authors do.
Exhibit L also shows the parameters of the exponential curve fit. We use the 7 point fit, not the 8 point fit. The authors allude to this in the second full paragraph on page 134: "This coefficient [of determination] increases to 0.99821 when the first point is dropped." On page 135, they note again that the coefficient of determination is 0.99821, so it is clear that they are using the 7 point fit. You can see the coefficients of determination in the third to the last row of Exhibit L as well.
The parameters of the 7 point exponential fit are "a" = $150.625 and "b" = 0.000542. You might say: "Can we plug the 6,926 claims corresponding to an 88.55% settlement point into the exponential equation?" Let’s try: $150.625e(6,926 * 0.000542) = $6,429,714. That’s close, but it’s not quite the figure that the authors get on page 135.
The problem is that we must use the two point exponential fit, not the 7 point exponential fit. The two point exponential fit is simple. We have two simultaneous equations in two unknowns:
$5,398,000 = a * eb * 6,616
$7,496,000 = a * eb * 7,192
To solve this, we take logarithms of both sides:
ln(5,398,000) = ln (a) + b * 6,616
ln(7,496,000) = ln (a) + b * 7,192
Now we have linear equations, and we can solve for ln(a) and for "b." From ln(a) we derive the value of "a." We then use the new values of "a" and "b" to derive the cumulate paid losses that correspond to cumulative paid claims of 6,296.