probability expected payment

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 $200$.

• If $d>60$, then the first term is zeroed out (insurance doesn't get to take more money from the client!) – lulu Aug 29 '16 at 17:39
• 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). – lulu Aug 29 '16 at 18:16
• 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. – 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 $d = 0$.