View Full Version : Why the median is not affected by a benefit limit?

Starry Night

04-26-2005, 10:04 PM

Sample question #68:

An insurance policy reimburses dental expense, X, up to a maximum benefit of $250. The probability density function for X is exponential with mean $250.

Calculate the median benefit for this policy.

The solution ignores the benefit limit totally.

Why?

If I had a BL = $5, and the mean for the loss = $100,000, would the median benefit still be equal $100,000*ln2 ??

felipe_R

04-26-2005, 10:17 PM

I think it has something to do with the fact that both these sets have the same median:

{1,2,3} and {1,2,500}

as long as the benefit limit is greater than the median of the expense, the benefit and expense distributions should have the same median.

Been There Done That

04-26-2005, 10:18 PM

Well, if the limit is greater than the 50th percentile, then the median (=50th percentile) is not changed. If it is lower, then it is changed.

Starry Night

04-26-2005, 10:25 PM

Thank you.

invictus

11-20-2007, 12:43 AM

In ASM manual, as long as P(X > 250) < 50%, the median of X is the same as the median benefit of the policy. What if P(X > 250) > 50%, how are we going to deal with this? Do you have any idea how to solve for the median in this case?

tommie frazier

11-20-2007, 01:43 AM

you should try to get these things straight, esp since median and mean are often thrown around loosely as "average" in some places (places where people don;t know enough to be more clear).

median is unaffected by extreme values-it's a head count.

mean is always affected by all values, which is why lognormal (for example) is skewed and mean>median.

daaaave

11-20-2007, 09:17 AM

In ASM manual, as long as P(X > 250) < 50%, the median of X is the same as the median benefit of the policy. What if P(X > 250) > 50%, how are we going to deal with this? Do you have any idea how to solve for the median in this case?

Whenever X>250, the actual payment will be the benefit limit of 250. That means that if P(X>250)>.5, then we have P(payment=250)>.5 which will in turn imply that 250 is the median.

More generally, since deductibles and benefit limits are what are called monotonic transformations, it will turn out that the median payment after applying deductibles and benefit limits is equal to the payment associated with the median loss. If the median loss is 250 and the benefit limit is >250, then the payment for a loss of 250 will still be 250, and hence the median payment will be 250. If the median loss is 250 and the benefit limit is 200, then the payment for a loss of 250 will be the limit of 200 and hence the median payment will be 200. The fact that this works is a peculiar property of medians (and is very much not true for means!) and the types of transformations that are common in insurance.

Gandalf

11-20-2007, 10:05 AM

In ASM manual, as long as P(X > 250) < 50%, the median of X is the same as the median benefit of the policy. What if P(X > 250) > 50%, how are we going to deal with this? Do you have any idea how to solve for the median in this case?

Then the median is 250, since Pr(benefit<=250)>=50% and Pr(benefit>=250)>=50%.

Here, it's not relevant, but for a different distribution of a random variable Y, you might find Pr(Y<=k)>=50% and Pr(Y>=k)>=50% for more than one value of k. In that situation, different books might disagree on what the median is. E.g. Roll one die; outcome is the number showing. Some would say median = 3.5; some would say the is no unique median, and any k with 3<=k<4 is a median.

instableoxymoron

11-02-2011, 05:57 PM

I was having trouble with this one until I read Guo's explanation which resulted in the following formula

-ln(50%) / 0.004 = 173.28

This was more helpful to me than 250ln2 even though both produce the same answer.

I solved the same question today, thought the median should change but after not being able to figure out a way to state how it should be different, I simply went and solved the problem the usual way. I'm terrible at percentiles but the way I see it, median is P(X>x)<0.5 (for continuous variables) right? X is the amount the insurance company pays. So, median is the amount where the percentage of the amount paid by the insurer is %50 (not sure if I'm making myself clear). So it doesn't matter if there is an upper limit, %50 of the payments made are below the median. At least this is what I got from the solution. Correct me if I'm wrong.

instableoxymoron

11-02-2011, 06:40 PM

A badly drawn diagram might be helpful.

Benefit payable

|250 - - - - _____y____

|............../

|.......... /

|....... /

|...../

|../

|0________250____x_____

Dental Expense

Y is a non-decreasing function of X. Copying Guo "250 is the 63.2 percentile of Y. Because Y is always $250 once X >= 250, any percentile higher than 63.2 is always 250.

riskdiscjockey

06-24-2014, 11:38 AM

I got stuck on this problem. What helped is I pictured it from a discrete standpoint. There are the same number of people standing in the benefit line as there are those standing in the claims line. The lines have the same length. Scientifically speaking, they have the same number of measurements. Irrespective of the measurement's value, for every unique (unique because of continuous number line so differential measurements per spot in line) claim measure there is a specific (not necessarily unique in this case due to 250 limit) benefit measure. For any spot in line, picture the benefit guy giving the claim guy a specific benefit payment measure. If there are N measurements, then the middle measurement (or median) is at the N/2 spot in line. The claims measurement, at its N/2 spot is 173.29. Because the upper limit on benefit payment measures is 250 and the 173.29 claim is less than 250, then the N/2 spot (or median measure) in the benefit line is also 173.29. If the N/2 spot in the claims line is at or above 250, the N/2 spot in the benefits line will remain at 250 based on the upper limit as well.

pquach

07-01-2014, 08:37 PM

In ASM manual, as long as P(X > 250) < 50%, the median of X is the same as the median benefit of the policy. What if P(X > 250) > 50%, how are we going to deal with this? Do you have any idea how to solve for the median in this case?

I was wondering, can we plot the distribution of the benefit, and use that? Based on the earlier comments by Gandalf and daaave (from 7 years ago ^^), I was considering a mixture of an exponential distribution and a discrete point (at y=250, the benefit limit), which terminates the latter. The exponential distribution for the benefit mirrors the loss, up to the benefit limit only. The discrete point is necessary since the benefit no longer varies after 250.

Now if P(Y >= 250)=P(Y=250)< 50% or F(250)>=50%, then the benefit median is the same as the loss median ("because there's enough room").

If P(Y >= 250)=P(Y=250)> 50%, that means that the discrete point has >50% so that it contains the median.

(For example, if P(Y<250)=40%, then P(Y=250)=60%. Since the median cannot lie within 0<y<250, the median is captured when Y=250).

rgreenlee

07-01-2014, 11:33 PM

you should try to get these things straight, esp since median and mean are often thrown around loosely as "average" in some places (places where people don;t know enough to be more clear).

median is unaffected by extreme values-it's a head count.

mean is always affected by all values, which is why lognormal (for example) is skewed and mean>median.

I asked this same question in my own study...very succinct answer, thanks!

vBulletin® v3.7.6, Copyright ©2000-2014, Jelsoft Enterprises Ltd.