Question

The loss L due to a boat accident is exponentially distributed. Boat insurance
policy A covers up to 1 unit for each loss. Boat insurance policy B covers up
to 2 units for each loss.
The probability that a loss is fully covered under policy B is 1.9 times the probability that it is fully covered under policy A:
Calculate the variance of L:

Answer 1

Answer:

The variance for Loss L is 90.1.

Step-by-step explanation:

As per given data

As the distribution is exponential for the Loss L thus the function is

                                             $F(x)=1-e^{-\lambda x}$

Where λ is given as

                                             $\lambda=\sqrt{\frac{1}{var}}$

Now as per the given condition

                                         $F(2)=1.9F(1)\\1-e^{-2\lambda }=1.9 (1-e^{-\lambda })\\\frac{1-e^{-2\lambda }}{1-e^{-\lambda }}=1.9\\\frac{(1-e^{-\lambda })(1+e^{-\lambda })}{1-e^{-\lambda }}=1.9\\1+e^{-\lambda }=1.9\\e^{-\lambda }=1.9-1\\e^{-\lambda }=0.9\\\lambda =-ln(0.9)\\\lambda =0.10536$

Solving for variance

                                       $\lambda=\sqrt{\frac{1}{var}}\\var=\frac{1}{\lambda^2}\\var=\frac{1}{0.10536^2}\\var=90.1$

So the variance is 90.1.