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

The variance for Loss L is 90.1.

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{(1)/(var)}`

Now as per the given condition

`F(2)=1.9F(1)\\1-e^(-2\lambda )=1.9 (1-e^(-\lambda ))\\(1-e^(-2\lambda ))/(1-e^(-\lambda ))=1.9\\((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{(1)/(var)}\\var=(1)/(\lambda^2)\\var=(1)/(0.10536^2)\\var=90.1`

So the variance is 90.1.