A policy covers a gas furnace for one year. During that year, only one of three problems can occur:

  1. The igniter switch may need to be replaced at a cost of $60$. There is a $0.10$ probability of this.
  2. The pilot light may need to be replaced at a cost of $200$. There is a $0.05$ probability of this.
  3. The furnace may need to be replaced at a cost of $3000$. There is a $0.01$ probability of this.

Calculate the deductible that would produce an expected claim payment of $30$.

My solution $E[y]=(60-d)*0.1+(200-d)*0.05+(3000-d)*0.01=30$ so $d=100$

Answer d is between $150$ and $200$.

  • 1
    $\begingroup$ If $d>60$, then the first term is zeroed out (insurance doesn't get to take more money from the client!) $\endgroup$ – lulu Aug 29 '16 at 17:39
  • 1
    $\begingroup$ Was my comment enough for you to solve your problem? Again, the point is that your formula for $E[y]$ is not correct. The payouts should be $\max(P-d,0)$ where $P$ is the cost of repair. If, say, $d=3000$ then $E=0$ as there is no scenario in which the policy will have to pay. (side note: not sure what $y$ denotes in your formula). $\endgroup$ – lulu Aug 29 '16 at 18:16
  • $\begingroup$ y denotes claim payment,I get it now..I'm so thankful for all your help I have a test coming up and I can use all the help. $\endgroup$ – Theo Robinson Aug 29 '16 at 20:31

With an ordinary deductible $d$ on an annual policy whose ground-up loss distribution is $X$, with $$\begin{align*} \Pr[X = 0] & = 0.84 \\ \Pr[X = 60] &= 0.10 \\ \Pr[X = 200] &= 0.05 \\ \Pr[X = 3000] &= 0.01, \end{align*}$$ the random claim amount is the random variable $$Y = (X - d)_+ = \max(0, X - d) = \begin{cases} X - d, & X > d \\ 0, & X \le d. \end{cases}$$ Due to the discrete nature of $X$, we need to consider cases for $d$: for example, $$\operatorname{E}[Y \mid d < 60] = (60-d)(0.1) + (200-d)(0.05) + (3000-d)(0.01) = 46 - 0.16d,$$ but in order for this to be equal to $30$, we would need $d = 100$, which contradicts the requirement that $d < 60$, so we know that $d \ge 60$. In this case, the first term drops out and we get $$\operatorname{E}[Y \mid 60 \le d < 200] = (200-d)(0.05) + (3000-d)(0.01) = 40 - 0.06d,$$ and setting this to $30$ gives $d = 166.67$, which does satisfy the interval condition; hence this is the required deductible.

Had the solution led to a contradiction, we would then need to try $\operatorname{E}[Y \mid 200 \le d < 3000] = (3000-d)(0.01) = 30$. Note this one obviously doesn't work since it would require $d = 0$.

  • $\begingroup$ thanks I really appreciate all the help $\endgroup$ – Theo Robinson Aug 29 '16 at 20:34
  • $\begingroup$ @TheoRobinson If you find a response to your question to be satisfactory, please upvote it and/or accept the answer. $\endgroup$ – heropup Aug 29 '16 at 20:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.