SOA Sample #28

[JamesBond]

Help!!! I'm having trouble getting this problem particularly the limited expected value for this grouped data. please help...

28. You are given:
Claim Size(X) Number of Claims
(0,25] 25.00
(25,50] 28.00
(50,100] 15.00
(100,200] 6.00


Assume a uniform distribution of claim sizes within each interval.

Estimate E(X^2) - E[(X (angle) 150))^2]
(A) Less than 200
(B) At least 200, but less than 300
(C) At least 300, but less than 400
(D) At least 400, but less than 500
(E) At least 500

[stchrist]

If you divide it up into integrals you get:

E[x^2] = integral of x^2 f(x) from 0 to 200. (The probability of claims > 200 = 0).
E[(x min 150)^2] = intergral of x^2 f(x) from 0 to 150 PLUS intergral of 150^2 f(x) from 150 to 200.

If you subtract the two of them, the two integrals with x^2 f(x) become just the intergral of x^2 f(x) from 150 to 200. And you are left with:

Integral of x^2 f(x) from 150 to 200 minus the integral of 150^2 f(x) from 150 to 200. You can combine them into 1 integral to get:

the integral of ( x^2 - 150^2 ) f(x) from 150 to 200

The second part is to figure out what f(x) is. In the interval from 100 to 200 (from the chart), f(x) = 6/7400 (or 6/74 * 1/100, since the interval from 100 to 200 has a width of 100). Plug this in for f(x) to get:

the integral of (x^2 - 150^2) (6/7400) from 150 to 200.

[SamCat]

The second part is to figure out what f(x) is. In the interval from 100 to 200 (from the chart), f(x) = 6/7400 (or 6/74 * 1/100, since the interval from 100 to 200 has a width of 100).


Often, when between ranges, we would use linear interpolation. Yet f(x) = 6/74 * 1/100, and not 3/74 * 1/100 Is this because we then integrate from 150 to 200?

Makes sense to me, but can anyone confirm?

[The Half]

yes.