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Thread: Median Benefit - ACTEX Manual

1. Median Benefit - ACTEX Manual

Hi,

Please refer to the following question in the ACTEX Manual (Practice Exam 1 Qn.12) and also in the SOA sample exam P-09-05 Qn.68:

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

f(x) = ce**-.004x for x>=0
= 0 otherwise

where c is a constant.

Calculate the median benefit for this policy.

The correct answer is, of course, 173. It is arrived at by integrating f(x) from 0 to m and equating it to 0.5. However, if, say, the maximum benefit was 100 (instead of 250), would not the median benefit be different? What I am trying to say is that the calculation of median 'benefit' must involve using the maximum benefit amount also. Does this make sense or is it just my overworked gray matter exploding into nonsense?

Thanks!
Good Luck!
Rathi

2. Rathi,

The conventional meaning of "median" is that it
is the 50th percentile of a distribution. For a continuous
random variable, as you have noted, there would be a
single point m at which F(m) = .5 . If a distribution is
discrete, or is partly discrete, it is possible that 50 percent
of the probability is reached as a result of a discrete point
of probability. That would be the case in your modified
example with a limit of 100. If X is the dental expense,
and Y is the amount remibursed with a limit of 100, then
Y = 100 is a discrete point with a probability mass, and
in this example that would be a probability of more than .5 .
We could then say that 100 is the median, but it is also
true that 100 is the 60-th, 70-th, 99-th etc.
percentile, since 60% of the probability is at Y<=100
and so is 70 percent, etc. Percentile and median becomes
a less meaningful concept in this situation.

A practical way of dealing with discrete points of probability
is do apply some sort of interpolation between points to
find percentiles.

Sincerely,

Sam Broverman

3. Dear Dr. Broverman,

Thank you for your prompt and detailed response. The percentiles make more sense now.

Regards
Rathi

4. Rathi

An easier way of working this is to identify that f(x) is an exponential function and thus the mean is equal to 250.

From this you can work out that the median = ln2 x mean = ln2 x 250 = 173

Just makes for quicker calculation.

Hope this helps,

Miles

5. Hi Miles,

'''...that works out nicely. In my university classes, it was critical to identify the distribution to work out a solution. I guess it would be good to do that for these problems too. Thanks!

Rathi

6. Hi,

I noticed the same problem. The question asks to calculate the median benefit, but then the answer is the median dental expense.

These expenses include those that are above 250 dollars in the set of expenses. Since the benefits stop at 250, but are distributed the same as the expense, since every expense above 250 can be mapped to 250 and still be greater than the values below 250. (1000 -> 250 which is greater than 100 still). Since medians don't "actually" take into account the amount of difference between values, only whether or not one is greater than the other. ( 1000 and 250 are both greater than 100, and if there are 250 250s then they will all be greater than 100) So the median should still be the same in the case of this specific answer where the median expense is mapped as an equivalent value to the median benefit.

For when the calculated expense is greater than the maximum benefit, than the median benefit would be the maximum benefit. This is because the median expense is greater than the max benefit and therefore the median benefit can only be the max benefit, since the insurance company ( or apparently the people who do their work), will map the max benefit onto the distribution of expenses anywhere above the max benefit .

I know this logic is a bit technical, but I believe it is correct, not just what one is suppose to do. Please provide corrections if there need be some.

Dirk

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