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# Bootstrap Using Chain Ladder Method

This example shows how to apply a chain ladder bootstrap method to generate several `developmentTriangle` objects to estimate the ultimate claims.

Deterministic claim estimation methods produce point estimates of reserve values with no information about the uncertainty of these estimates. The goal of a stochastic claim estimation method is to assess the variability of estimated reserve values. The chain ladder bootstrapping approach is a simulation-based method to randomly modify the `developmentTriangle` data and produce a distribution of estimated reserves that represents the variability of the estimated reserve values. This example is based on the work of Wüthrich and Merz [1].

### Load Data

```load('InsuranceClaimsData.mat'); disp(head(data));```
``` OriginYear DevelopmentYear ReportedClaims PaidClaims __________ _______________ ______________ __________ 2010 12 3995.7 1893.9 2010 24 4635 3371.2 2010 36 4866.8 4079.1 2010 48 4964.1 4487 2010 60 5013.7 4711.4 2010 72 5038.8 4805.6 2010 84 5059 4853.7 2010 96 5074.1 4877.9 ```

### Create `developmentTriangle`

Create a `developmentTriangle` object and use `claimsPlot` to visualize the `developmentTriangle`. For more information on unpaid claims estimation, see Overview of Claims Estimation Methods for Non-Life Insurance.

```dTriangle = developmentTriangle(data); dTriangleTable = view(dTriangle); % visualize the development triangle claimsPlot(dTriangle)```

### Analyze the `developmentTriangle`

The `developmentTriangle` link ratios are estimated using the formula:

`$\stackrel{ˆ}{{\mathit{f}}_{\mathit{j}}}=\frac{\sum _{\mathit{i}=0}^{\mathit{I}-\mathit{j}-1}{\mathit{C}}_{\mathit{i},\mathit{j}+1}}{\sum _{\mathit{i}=0}^{\mathit{I}-\mathit{j}-1}{\mathit{C}}_{\mathit{i},\mathit{j}}}$`

Use `linkRatios` to calculate the age-to-age factors.

`factorsTable = linkRatios(dTriangle);`

Use `linkRatioAverages` to calculate the averages of the age-to-age factors.

```averageFactorsTable = linkRatioAverages(dTriangle); disp(averageFactorsTable);```
``` 12-24 24-36 36-48 48-60 60-72 72-84 84-96 96-108 108-120 ______ ______ ______ ______ ______ ______ _____ ______ _______ Simple Average 1.1767 1.0563 1.0249 1.0107 1.0054 1.0038 1.003 1.002 1.001 Simple Average - Latest 5 1.172 1.056 1.0268 1.0108 1.0054 1.0038 1.003 1.002 1.001 Simple Average - Latest 3 1.17 1.0533 1.027 1.0117 1.0057 1.0037 1.003 1.002 1.001 Medial Average - Latest 5x1 1.1733 1.0567 1.0267 1.0103 1.005 1.004 1.003 1.002 1.001 Volume-weighted Average 1.1766 1.0563 1.025 1.0107 1.0054 1.0038 1.003 1.002 1.001 Volume-weighted Average - Latest 5 1.172 1.056 1.0268 1.0108 1.0054 1.0038 1.003 1.002 1.001 Volume-weighted Average - Latest 3 1.1701 1.0534 1.027 1.0117 1.0057 1.0037 1.003 1.002 1.001 Geometric Average - Latest 4 1.17 1.055 1.0267 1.011 1.0055 1.0037 1.003 1.002 1.001 ```

Display the selected age-to-age factors table and calculate the cumulative development factor (CDF) using `cdfSummary`.

```dTriangle.SelectedLinkRatio = averageFactorsTable{'Volume-weighted Average',:}; currentSelectedFactors = dTriangle.SelectedLinkRatio; dTriangle.TailFactor = 1; selectedFactorsTable = cdfSummary(dTriangle); disp(selectedFactorsTable);```
``` 12-24 24-36 36-48 48-60 60-72 72-84 84-96 96-108 108-120 Ultimate _______ _______ _______ _______ _______ ______ _______ _______ _______ ________ Selected 1.1766 1.0563 1.025 1.0107 1.0054 1.0038 1.003 1.002 1.001 1 CDF to Ultimate 1.3072 1.111 1.0518 1.0261 1.0153 1.0098 1.006 1.003 1.001 1 Percent of Total Claims 0.76501 0.90008 0.95075 0.97453 0.98496 0.9903 0.99402 0.99701 0.999 1 ```

DIsplay the latest diagonal.

`latestDiagonal = dTriangle.LatestDiagonal;`

Compute the projected ultimate claims using `ultimateClaims`.

`projectedUltimateClaims = ultimateClaims(dTriangle);`

Display the full development triangle using `fullTriangle`.

```fullTriangleTable = fullTriangle(dTriangle); disp(fullTriangleTable);```
``` 12 24 36 48 60 72 84 96 108 120 Ultimate ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ________ 2010 3995.7 4635 4866.8 4964.1 5013.7 5038.8 5059 5074.1 5084.3 5089.4 5089.4 2011 3968 4682.3 4963.2 5062.5 5113.1 5138.7 5154.1 5169.6 5179.9 5185.1 5185.1 2012 4217 5060.4 5364 5508.9 5558.4 5586.2 5608.6 5625.4 5636.7 5642.3 5642.3 2013 4374.2 5205.3 5517.7 5661.1 5740.4 5780.6 5803.7 5821.1 5832.7 5838.6 5838.6 2014 4499.7 5309.6 5628.2 5785.8 5849.4 5878.7 5900.8 5918.5 5930.3 5936.3 5936.3 2015 4530.2 5300.4 5565.4 5715.7 5772.8 5804.1 5825.9 5843.4 5855.1 5861 5861 2016 4572.6 5304.2 5569.5 5714.3 5775.4 5806.7 5828.6 5846.1 5857.7 5863.6 5863.6 2017 4680.6 5523.1 5854.4 6000.9 6065.1 6098 6120.9 6139.3 6151.6 6157.7 6157.7 2018 4696.7 5495.1 5804.4 5949.6 6013.3 6045.9 6068.6 6086.8 6099 6105.1 6105.1 2019 4945.9 5819.2 6146.7 6300.5 6367.9 6402.4 6426.5 6445.8 6458.7 6465.2 6465.2 ```

Compute the total reserves using `ultimateClaims`.

```IBNR = ultimateClaims(dTriangle) - dTriangle.LatestDiagonal; IBNR = array2table(IBNR, 'RowNames', dTriangleTable.Properties.RowNames, 'VariableNames', {'IBNR'}); IBNR{'Total',1} = sum(IBNR{:,:}); disp(IBNR);```
``` IBNR ______ 2010 0 2011 5.1857 2012 16.89 2013 34.886 2014 57.583 2015 88.148 2016 149.34 2017 303.29 2018 609.99 2019 1519.3 Total 2784.6 ```

### Bootstrap Chain Ladder

To derive the resampling approaches, the Time Series Model of the distribution-free chain ladder (CL) model is defined as:

`${\mathit{C}}_{\mathit{i},\mathit{j}+1}={\mathit{f}}_{\mathit{j}}{\mathit{C}}_{\mathit{i},\mathit{j}}+{\sigma }_{\mathit{j}}\sqrt{{\mathit{C}}_{\mathit{i},\mathit{j}}}{ϵ}_{\mathit{i},\mathit{j}+1}$`

For the link ratio selected above, Wüthrich [1] and Mack [2] show that the standard deviation is estimated as:

`${\stackrel{ˆ}{{\sigma }_{\mathit{j}}}}^{2}=\frac{1}{\mathit{I}-\mathit{j}-1}\text{\hspace{0.17em}}\sum _{\mathit{i}=0}^{\mathit{I}-\mathit{j}-1}{\mathit{C}}_{\mathit{i},\mathit{j}}{\left(\frac{{\mathit{C}}_{\mathit{i},\mathit{j}+1}}{{\mathit{C}}_{\mathit{i},\mathit{j}}}-\stackrel{ˆ}{{\mathit{f}}_{\mathit{j}}}\right)}^{2}$`

`${\stackrel{ˆ}{{\sigma }_{\mathit{J}-1}}}^{2}=\mathrm{min}\left\{\frac{{\stackrel{ˆ}{{\sigma }_{\mathit{J}-2}}}^{4}}{{\stackrel{ˆ}{{\sigma }_{\mathit{J}-3}}}^{3}}\text{\hspace{0.17em}};{\stackrel{ˆ}{{\sigma }_{\mathit{J}-3}}}^{2}\text{\hspace{0.17em}};{\stackrel{ˆ}{{\sigma }_{\mathit{J}-2}}}^{2}\right\}$`

```estimatedStandardDeviations = currentSelectedFactors; for i=1:width(estimatedStandardDeviations)-1 estimatedStandardDeviations(1,i) = sqrt(sum(((factorsTable{1:end-i,i} - currentSelectedFactors(:,i)).^2).*dTriangleTable{1:end-i,i}) / (height(dTriangleTable)-i-1)); end estimatedStandardDeviations(1,end) = sqrt(min([estimatedStandardDeviations(1,end-1)^4 / estimatedStandardDeviations(1,end-2)^2, estimatedStandardDeviations(1,end-2)^2, estimatedStandardDeviations(1,end-1)^2])); disp(estimatedStandardDeviations);```
``` 0.8667 0.3699 0.2420 0.1310 0.0673 0.0361 0.0001 0.0001 0.0001 ```

To apply the bootstrap method, you need to find the appropriate residuals that allow for the construction of the empirical distribution $\stackrel{ˆ}{{\mathit{F}}_{\mathit{n}}}$ to construct the bootstrap observations.

Consider the following residuals for $\mathit{i}+\mathit{j}\le \mathit{I},\mathit{j}\ge 1$.

$\stackrel{\sim }{{ϵ}_{\mathit{i},\mathit{j}}}=\frac{{\mathit{F}}_{\mathit{i},\mathit{j}}-\stackrel{ˆ}{{\mathit{f}}_{\mathit{j}-1}}}{{\sigma }_{\mathit{j}-1}{\mathit{C}}_{\mathit{i},\mathit{j}-1}^{-1/2}}$ where ${\mathit{F}}_{\mathit{i},\mathit{j}}=\frac{{\mathit{C}}_{\mathit{i},\mathit{j}}}{{\mathit{C}}_{\mathit{i},\mathit{j}-1}}$

Following Wüthrich [1], you can scale the residuals to adjust their variance upwards. Unscaled residuals tend to result in lighter tails in the simulated distribution.

Adjust the residuals such that the bootstrap distribution has an adjusted variance function.

`${\mathit{Z}}_{\mathit{i},\mathit{j}}={\left(1-\frac{{\mathit{C}}_{\mathit{i},\mathit{j}-1}}{\sum _{\mathit{i}=0}^{\mathit{I}-\mathit{j}}{\mathit{C}}_{\mathit{i},\mathit{j}-1}}\right)}^{-\frac{1}{2}}\frac{{\mathit{F}}_{\mathit{i},\mathit{j}}-\stackrel{ˆ}{{\mathit{f}}_{\mathit{j}-1}}}{\stackrel{ˆ}{{\sigma }_{\mathit{j}-1}}{\mathit{C}}_{\mathit{i},\mathit{j}-1}^{-\frac{1}{2}}}$`

You can apply the bootstrap algorithm using three different versions:

• Efron's nonparametric bootstrap for residuals $\stackrel{\sim }{{ϵ}_{\mathit{i},\mathit{j}}}$

• Efron's nonparametric bootstrap for scaled residuals ${\mathit{Z}}_{\mathit{i},\mathit{j}}$

• Parametric bootstrap under the assumption that the residuals have a standard Gaussian distribution, that is ${\mathit{Z}}_{\mathit{i},\mathit{j}}^{*}$ is resampled from $\mathit{N}\left(0,1\right)$

This example uses the second version (Efron's nonparametric bootstrap for scaled residuals) to calculate ${\mathit{Z}}_{\mathit{i},\mathit{j}}$.

```% Create a copy of the factors table and modify it to create the % residuals table residuals = factorsTable.Variables; colSums = sum(dTriangle.Claims,'omitnan'); for i=1:height(residuals) for j=1:width(residuals) residuals(i,j) = (1 - (dTriangleTable{i,j}/colSums(j)))^-0.5 * (factorsTable{i,j} - currentSelectedFactors(1,j)) / (estimatedStandardDeviations(1,j)*(dTriangleTable{i,j}^-0.5)); end end```

The residuals $\left\{{\mathit{Z}}_{\mathit{i},\mathit{j}},\mathit{i}+\mathit{j}\le \mathit{I}\right\}$ define a bootstrap distribution.

```residualsVector = residuals(:); residualsVector(isnan(residualsVector)) = []; histogram(residualsVector,10) title('Scaled Residuals') xlabel('Residual Value') ylabel('Frequency')```

To simulate a new reserves scenario with the bootstrap method, follow these steps.

#### Step 1: Resample a triangle of residuals from the bootstrap distribution.

Resample the independent and identically distributed (i.i.d.) residuals$\left\{{{\mathit{Z}}^{*}}_{\mathit{i},\mathit{j}},\mathit{i}+\mathit{j}\le \mathit{I}\right\}$ from the bootstrap distribution.

```resampledResiduals = residuals; rng('default'); rng(1); for i = 1:height(residuals)-1 for j = 1:width(residuals)-i+1 resampledResiduals(i,j) = datasample(residuals(~isnan(residuals)), 1); end end disp(resampledResiduals);```
``` -1.5522 -0.5120 -1.2668 0.7776 -1.3649 0.2799 -0.5495 -1.3146 -1.5364 -0.4041 -1.5522 -0.4784 -1.2189 -0.7591 0.2610 -0.4784 -1.5522 NaN -0.4091 -1.3649 -0.5495 -1.6767 -0.8571 -1.3143 -0.4879 NaN NaN -0.7591 1.3226 1.0791 0.2610 0.2861 -0.7591 NaN NaN NaN 0.2799 -1.5522 -0.8571 0.3243 -0.4879 NaN NaN NaN NaN -1.3143 -0.4784 0.5556 -1.2668 NaN NaN NaN NaN NaN 1.9550 0 1.9550 NaN NaN NaN NaN NaN NaN 0.7693 0.5169 NaN NaN NaN NaN NaN NaN NaN 0.2799 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN ```

#### Step 2: Compute bootstrapped claims.

Define ${\mathit{C}}_{\mathit{i},0}^{*}={\mathit{C}}_{\mathit{i},0}$ and, for $\mathit{j}\ge 1$, assume that:

`${\mathit{C}}_{\mathit{i},\mathit{j}}^{*}=\stackrel{ˆ}{{\mathit{f}}_{\mathit{j}-1}}{\mathit{C}}_{\mathit{i},\mathit{j}-1}^{*}+\stackrel{ˆ}{{\sigma }_{\mathit{j}-1}}\sqrt{{\mathit{C}}_{\mathit{i},\mathit{j}-1}^{*}}{\mathit{Z}}_{\mathit{i},\mathit{j}}^{*}$`

This expression represents the new simulated claim values. Using the simulated claim values, you can create a new `developmentTriangle` to estimate new reserve values.

```bootstrappedClaims = dTriangleTable.Variables; for j = 2:width(bootstrappedClaims) bootstrappedClaims(:,j) = currentSelectedFactors(1,j-1).*bootstrappedClaims(:,j-1) + estimatedStandardDeviations(1,j-1).*sqrt(bootstrappedClaims(:,j-1)).*resampledResiduals(:,j-1); end stackedClaims = reshape(bootstrappedClaims',100,1); stackedClaims = stackedClaims(~isnan(stackedClaims)); newData = data; newData.values = stackedClaims; bootstrappedDevelopmentTriangle = developmentTriangle(newData,'Claims','values');```

#### Step 3: Select a link ratio consistent with the model.

The volume-weighted average is the link ratio that is consistent with the model used in this bootstrap approach.

```bootstrappedAverageFactorsTable = linkRatioAverages(bootstrappedDevelopmentTriangle); bootstrappedDevelopmentTriangle.SelectedLinkRatio = bootstrappedAverageFactorsTable{'Volume-weighted Average',:}; bootstrappedDevelopmentTriangle.TailFactor = 1; bootstrappedSelectedFactorsTable = cdfSummary(bootstrappedDevelopmentTriangle); disp(bootstrappedSelectedFactorsTable);```
``` 12-24 24-36 36-48 48-60 60-72 72-84 84-96 96-108 108-120 Ultimate _______ _______ ______ _______ _______ ______ _______ _______ _______ ________ Selected 1.1751 1.054 1.0253 1.0099 1.0048 1.0036 1.003 1.002 1.001 1 CDF to Ultimate 1.301 1.1072 1.0504 1.0245 1.0145 1.0096 1.006 1.003 1.001 1 Percent of Total Claims 0.76861 0.90321 0.952 0.97609 0.98572 0.9905 0.99403 0.99701 0.999 1 ```

Use `fullTriangle` to display the full development triangle corresponding to the selected link ratio.

```bootstrappedFullTriangle = fullTriangle(bootstrappedDevelopmentTriangle); disp(bootstrappedFullTriangle);```
``` 12 24 36 48 60 72 84 96 108 120 Ultimate ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ________ 2010 3995.7 4616.2 4863.2 4963.4 5023.7 5044.5 5064.1 5079.3 5089.5 5094.6 5094.6 2011 3968 4646.6 4869 4982.8 5024.8 5048.4 5068.1 5083.3 5093.4 5098.5 5098.5 2012 4217 4938.6 5181.1 5301.1 5341.9 5366.6 5383.3 5399.5 5410.2 5415.6 5415.6 2013 4374.2 5103.1 5425.3 5580.2 5642.5 5674.5 5693.8 5710.9 5722.3 5728 5728 2014 4499.7 5310.5 5567.5 5691.3 5755.4 5784.2 5804.8 5822.2 5833.8 5839.6 5839.6 2015 4530.2 5253.5 5536.3 5684.8 5733.2 5761 5781.5 5798.8 5810.4 5816.2 5816.2 2016 4572.6 5494.6 5803.9 5985.1 6044.2 6073.5 6095.1 6113.4 6125.6 6131.7 6131.7 2017 4680.6 5552.6 5879.4 6028.2 6087.7 6117.2 6139 6157.4 6169.7 6175.9 6175.9 2018 4696.7 5542.6 5842 5989.8 6048.9 6078.2 6099.9 6118.2 6130.4 6136.5 6136.5 2019 4945.9 5812 6126 6281 6343 6373.7 6396.4 6415.6 6428.4 6434.8 6434.8 ```

#### Step 4: Compute the total reserves.

Compute the total reserves from the simulated `developmentTriangle`.

```bootstrappedDevelopmentTriangleTable = view(bootstrappedDevelopmentTriangle); bootstrappedIBNR = ultimateClaims(bootstrappedDevelopmentTriangle) - bootstrappedDevelopmentTriangle.LatestDiagonal; bootstrappedIBNR = array2table(bootstrappedIBNR, 'RowNames', bootstrappedDevelopmentTriangleTable.Properties.RowNames, 'VariableNames', {'IBNR'}); bootstrappedIBNR{'Total',1} = sum(bootstrappedIBNR{:,:}); disp(bootstrappedIBNR);```
``` IBNR ______ 2010 0 2011 5.0881 2012 16.188 2013 34.197 2014 55.485 2015 83.048 2016 146.61 2017 296.45 2018 593.94 2019 1489 Total 2720 ```

You can repeat the previous steps many times to genreate a full, simulated, distribution of reserves. The simulation produces reserves for each year and for the total reserves.

### Simulate Multiple Bootstrapped Scenarios

Create `1000` bootstrapped development triangles and calculate the incurred-but-not-reported (IBNR) for each `developmentTriangle`.

```n = 1000; simulatedIBNR = zeros(10,n); for i = 1:n simulatedResiduals = residuals; for j = 1:height(residuals)-1 for k = 1:width(residuals)-j+1 simulatedResiduals(j,k) = datasample(residuals(~isnan(residuals)),1); end end simulatedClaims = dTriangleTable.Variables; for j = 2:width(simulatedClaims) simulatedClaims(:,j) = currentSelectedFactors(1,j-1).*simulatedClaims(:,j-1) + estimatedStandardDeviations(1,j-1).*sqrt(simulatedClaims(:,j-1)).*simulatedResiduals(:,j-1); end simulatedClaims = reshape(simulatedClaims',100,1); simulatedClaims = simulatedClaims(~isnan(simulatedClaims)); simulatedData = data; simulatedData.ReportedClaims = simulatedClaims; simulatedDevelopmentTriangle = developmentTriangle(simulatedData); simulatedAverageFactorsTable = linkRatioAverages(simulatedDevelopmentTriangle); simulatedDevelopmentTriangle.SelectedLinkRatio = simulatedAverageFactorsTable{'Volume-weighted Average',:}; simulatedDevelopmentTriangle.TailFactor = 1; simulatedLatestDiagonal = simulatedDevelopmentTriangle.LatestDiagonal; simulatedProjectedUltimateClaims = ultimateClaims(simulatedDevelopmentTriangle); simulatedIBNR(:,i) = simulatedProjectedUltimateClaims - simulatedLatestDiagonal; end simulatedIBNR(end+1,:) = sum(simulatedIBNR);```

Select a year to plot the distribution of the IBNR, calculate the mean, and compare that mean to a calculated deterministic value.

```originYear =5; histogram(simulatedIBNR(originYear+1,:)); hold on; plot(mean(simulatedIBNR(originYear+1,:)),0,'O','LineWidth',2) plot(IBNR{originYear+1,1},0,'X','LineWidth',2); legend('Simulated IBNR',['Simulated mean : ' num2str(round(mean(simulatedIBNR(originYear+1,:)),2))],['Deterministic IBNR : ' num2str(round(IBNR{originYear+1,1},2))]); hold off;```

Plot a histogram of the totals for IBNRs, simulated means, and deterministic values.

```histogram(simulatedIBNR(11,:)); hold on; plot(mean(simulatedIBNR(11,:)),0,'O','LineWidth',2) plot(IBNR{11,1},0,'X','LineWidth',2); legend('Simulated Total IBNR',['Simulated mean : ' num2str(round(mean(simulatedIBNR(11,:)),2))],['Deterministic Total IBNR : ' num2str(round(IBNR{11,1},2))]); hold off;```

### References

1. Wüthrich, Mario, and Michael Merz. Stochastic Claims Reserving Methods in Insurance. Hoboken, NJ: Wiley, 2008

2. Mack, Thomas. "Distribution-Free Calculation of the Standard Error of Chain Ladder Reserve Estimates." Astin Bulletin. Vol. 23, No. 2, 1993.