# Is there a geometric solution to this bivariate normal dist question (Exam P)?

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.

I'm only referring to the part of the question where you need to find $$P(B>A)$$. In the graph red is B's curve and orange is A's Question what area/part of the shaded region are you trying to find or does this question even make sense? I understand how to get the answer by finding $$Var(B-A)$$ and $$E(B-A)$$ for the new normal curve and then finding $$P(X>0)$$ (where $$X=B-A$$), but have no idea what's going on. Can you find the solution using a joint PDF because they are independent?

Your plot of the normal densities does not represent what you want to calculate, so there is no suitable geometric interpretation that you may apply to that plot to obtain the answer. The reason is because the area of any region in your plot will necessarily correspond to an outcome in which both $$A$$ and $$B$$ have the same claim amount, since you have displayed their outcomes along the same axis.
Here, the claim size of $$A$$ is shown in the vertical direction of the image, and the claim size of $$B$$ is shown along the horizontal direction. The probability density is plotted as the height. The set of points for which $$A > B$$ is shown in orange, and the points for which $$B > A$$ is shown in blue. The desired probability, given that both companies made a claim, is the volume underneath the blue shaded region, keeping in mind that the total region extends beyond what is shown in the plot.
• Is the graph you're doing, f(x,y), equal to the 2 normal curves multiplied together, with the integration bounds coming from $-\infty<A<B<\infty$? This is so helpful! I asked an actuary who has worked for 5 years and he didn't know a geometric way of doing this. Thanks so much!! Commented Feb 13, 2023 at 1:08
• @Bobas_Pett Yes: the plot shows the conditional joint density of $(A,B)$, which is bivariate normal. Since $A$ and $B$ are independent, this is simply the product of their univariate normal densities. Specifically, $$\Pr[A < B \mid \text{both had claims}] = \int_{a=-\infty}^\infty \int_{b = a}^\infty f_A(a) f_B(b) \, db \, da$$ where $f_A, f_B$ are the univariate normal densities of $A$ and $B$, respectively. Although strictly speaking we should not integrate over the region for which $A$ or $B$ is negative, the probability in this region is very small. Commented Feb 13, 2023 at 4:24