Covariance Matrix of Parameter Estimates -- Clark's methods

Function to compute the covariance matrix of the parameter estimates for the ClarkLDF and ClarkCapeCod methods.

Usage

# S3 method for clark
vcov(object, ...)

Arguments

object: object resulting from a run of the ClarkLDF or ClarkCapeCod functions.
...: not used.

Details

The covariance matrix of the estimated parameters is estimated by the inverse of the Information matrix (see Clark, p. 53). This function uses the "FI" and "sigma2" values returned by ClarkLDF and by ClarkCapeCod and calculates the matrix
-sigma2*FI^-1.

References

Clark, David R., "LDF Curve-Fitting and Stochastic Reserving: A Maximum Likelihood Approach", Casualty Actuarial Society Forum, Fall, 2003

Author

Daniel Murphy

Examples


x <- GenIns
colnames(x) <- 12*as.numeric(colnames(x))
Y <- ClarkCapeCod(x, Premium=10000000+400000*0:9, maxage=240)
round(vcov(Y),6) ## Compare to matrix on p. 69 of Clark's paper
#>             ELR     omega     theta
#> ELR    0.002387 -0.002964  0.238498
#> omega -0.002964  0.007804 -0.396186
#> theta  0.238498 -0.396186 32.471686

# The estimates of the loglogistic parameters
Y$THETAG
#>     omega     theta 
#>  1.448797 47.917507 
# The standard errors of the estimated parameters
sqrt(tail(diag(vcov(Y)), 2))
#>      omega      theta 
#> 0.08834184 5.69839331 

# The parameter risks of the estimated reserves are calculated 
# according to the formula on p. 54 of Clark's paper. For example, for
# the 5th accident year, pre- and post-multiply the covariance matrix
# by a matrix consisting of the gradient entries for just that accident year
FVgrad5 <- matrix(Y$FutureValueGradient[, 5], ncol=1)
sqrt(t(FVgrad5) %*% vcov(Y) %*% FVgrad5) ## compares to 314,829 in Clark's paper
#>          [,1]
#> [1,] 312675.8

# The estimated reserves for accident year 5:
Y$FutureValue[5]   ## compares to 2,046,646 in the paper
#> [1] 2041248

# Recalculate the parameter risk CV for all accident years in total (10.6% in paper):
sqrt(sum(t(Y$FutureValueGradient) %*% vcov(Y) %*% Y$FutureValueGradient)) / 
    Y$Total$FutureValue
#> [1] 0.1053735