For Company A there is a 60% chance that no claim is made during the coming year. If one or more claims are made, the total claim amount is normally distributed with mean 10,000 and standard deviation 2,000.

For Company B there is a 70% chance that no claim is made during the coming year. If one or more claims are made, the total claim amount is normally distributed with mean 9,000 and standard deviation 2,000.

The total claim amounts of the two companies are independent.

Calculate the probability that, in the coming year, Company B’s total claim amount will exceed Company A’s total claim amount.

• 0.180
• 0.185
• 0.217
• 0.223
• 0.240

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: D

Let

[math]I_A[/math] = Event that Company A makes a claim

[math]I_B[/math] = Event that Company B makes a claim

[math]X_A[/math] = Expense paid to Company A if claims are made

[math]X_B [/math] = Expense paid to Company B if claims are made

Then we want to find

[[math]] \begin{align*} \operatorname{P}[ I^c_A ∩ I_B ] ∪ [( I_A ∩ I_B ) ∩ ( X_A \lt X_B ) ] &= \operatorname{P}[ I^c _A ∩ I_B ] + \operatorname{P}[( I_A ∩ I_B ) ∩ ( X_A \lt X_B )] \\ &= \operatorname{P}[ I^c_A ]\operatorname{P}[ I_B ] + \operatorname{P}[ I_A ] \operatorname{P}[ I_B ] \operatorname{P}[ X_A \lt X_B ] \\ &= ( 0.60 )( 0.30 ) + ( 0.40 )( 0.30 ) \operatorname{P}[ X_B − X_A ≥ 0] \\ &= 0.18 + 0.12 \operatorname{P}[ X_B − X_A ≥ 0] \end{align*} [[/math]]

Now [math]X_B − X_A[/math] is a linear combination of independent normal random variables. Therefore, [math]X_B − X_A[/math] is also a normal random variable with mean

[[math]] M = \operatorname{E} [ X_B − X_A ] = E [ X_B ] − E [ X_A ] = 9, 000 − 10, 000 = −1, 000 [[/math]]

and standard deviation

[[math]] \sigma = \sqrt{Var ( X_B ) + Var ( X_A )} = \sqrt{(2000)^2 + (2000)^2} = 2000 \sqrt{2}. [[/math]]

It follows that

[[math]] \begin{align*} \operatorname{P}[ X B − X A ≥ 0] &= \operatorname{P}[ Z ≥ \frac{1000}{2000 \sqrt{2}} ] \\ &= \operatorname{P}[Z \geq \frac{1}{2\sqrt{2}}] \\ &= 1 - \operatorname{P}[Z \lt \frac{1}{2\sqrt{2}}] \\ &= 1 − \operatorname{P}[ Z \lt 0.354] \\ &= 1 − 0.638 = 0.362 \end{align*} [[/math]]

Finally,

[[math]] \operatorname{P}[[ I^c_A ∩ I_B ]] ∪ [( I_A ∩ I_B ) ∩ ( X_A \lt X_B ) ]] = 0.18 + ( 0.12 )( 0.362 ) = 0.223. [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.