Independent solution

How to solve this Exponential Distribution question

Setup

Setup

A loss is fully covered by a cap u exactly when it does not exceed u.

FL(x)=1−e−λxF_L(x)=1-e^{-\lambda x}
FL(2)=1.9FL(1)F_L(2)=1.9F_L(1)

Model

Model

Substitute the exponential CDF and factor the double-exponent expression.

1−e−2λ=1.9(1−e−λ)1-e^{-2\lambda}=1.9\left(1-e^{-\lambda}\right)
(1−e−λ)(1+e−λ)=1.9(1−e−λ)\left(1-e^{-\lambda}\right)\left(1+e^{-\lambda}\right)=1.9\left(1-e^{-\lambda}\right)

Compute

Compute

Cancel the nonzero factor, recover the rate, and compute the exponential variance.

e−λ=0.9,λ=−ln⁡(0.9)=0.1053605e^{-\lambda}=0.9,\qquad \lambda=-\ln(0.9)=0.1053605
Var⁡(L)=1λ2=90.0832871\operatorname{Var}(L)=\frac1{\lambda^2}=90.0832871

Answer

Answer

The loss variance rounds to 90.1.

90.1(E)\boxed{90.1\quad\text{(E)}}