Independent solution

How to solve this Exponential Stop-Loss Moments question

Setup

Setup

Let the payment be the amount of loss above the deductible. The stated expected-payment reduction implies that a loss exceeds the deductible with probability 0.9.

E[X]=μ,Var⁡(X)=μ2,q=Pr⁡(X>d)=0.9E[X]=\mu,\quad \operatorname{Var}(X)=\mu^2,\quad q=\Pr(X>d)=0.9

Model

Model

By the memoryless property, a positive payment has the same exponential mean and second moment as the original loss. Weighting those moments by the 0.9 chance of a positive payment gives the displayed unconditional moments.

E[Y]=qμ=0.9μ,E[Y2]=2qμ2=1.8μ2E[Y]=q\mu=0.9\mu,\quad E[Y^2]=2q\mu^2=1.8\mu^2

Compute

Compute

Subtracting the squared mean from the second moment leaves 99% of the original variance. The variance reduction is therefore 1%.

Var⁡(Y)=1.8μ2−(0.9μ)2=0.99μ2\operatorname{Var}(Y)=1.8\mu^2-(0.9\mu)^2=0.99\mu^2

Answer

Answer

The variance falls by one percent.

1%(A)\boxed{1\%\quad\text{(A)}}