Problem #68 from SOA123

[djerry81]

lets assume the problem is the same but the max ben is 100.....How would we solve this problem.....the short cut would be the median property for exp but since the median is higher than the max what would be the correct ans?

[Jo_M.]

In this case, wouldn't the median simply be the maximum benefit (i.e 100)?

[djerry81]

I am not convinced the median could be the maximum/endpoint

[Mosaic]

The median benefit can definitely be the max benefit from a math standpoint. If more than half of the cases require the max benefit, then the median is the max. From a business standpoint, that's not bad since it is successfully minimizing their payout.

[djerry81]

I dont believe that is correct

[url]http://mathworld.wolfram.com/Median.html[/url]

[jthias]

Y is a function of X so it depends on it, thus the median of Y also depends on median of X.

Suppose Y = X for X<2, Y = 2 for X>=2 (Y has cap of 2),

X ~ Uniform(0,1),

then median of Y is value m such that

.5 = Fy(m) = Fx(m) if m < 2
.5 <= Fy(m) = Fy(2) (since maximum value of Y is 2) = P(Y<2) + P(Y=2) = Fx(2) + Sx(2) = 1 if m >= 2

So median for Y is the same as the median for X whenever median of X is less than cap amount of 2, and median of Y is the same as its cap amount whenever median of loss amount is at least the cap amount

In this case m =1/2 , so median benefit is less than cap of 2.

But if we change distribution to X~Uniform(0,4), then m=2 = payout cap (or maximum benefit) which is median benefit in this case.

[jthias]

Cap for Y is fixed at 2. But we can modify distribution of X so that median of Y is at most the same value as the cap amount of Y.