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 Short-Term Actuarial Math Old Exam C Forum

#1
03-19-2009, 12:23 PM
 JamesBond Member CAS AAA Join Date: Jan 2007 Posts: 56
SOA Sample #28

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
#2
03-24-2009, 06:57 PM
 stchrist Wiki Contributor CAS Join Date: Apr 2006 Location: Salem, Oregon Studying for CAS Exam 5 College: U of I Alum Posts: 11

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.
#3
04-29-2009, 02:51 PM
 SamCat Member CAS Join Date: Feb 2007 Posts: 120

Quote:
 Originally Posted by stchrist 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?
#4
04-29-2009, 02:57 PM
 The Half Member Join Date: Sep 2008 College: Arizona State University Posts: 90

yes.
#5
06-07-2013, 02:16 AM
 Jet0L19 Member SOA Join Date: Jun 2010 College: East Carolina Favorite beer: 312 Posts: 74

Quote:
 Originally Posted by SamCat 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?

That's not really the reason. You could use 3/74, but then that's only referring to the range between 150 and 200, not 100 and 200. So f(x) = (3/74)(1/50), which equals the (6/74)(1/100) from above. Basically it's an extra step for no reason.
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#6
09-12-2018, 07:33 PM
 gauchodelpaso Member SOA Join Date: Feb 2012 College: Eastern Michigan U Posts: 94

I've seen a couple of interpretations that clarified a lot for me
1) If we have to integrate (X^150)^2 f(x), we integrate X^2 f(x) as long as X< 150, and then integrate 150^2 f(x) when X>150, as the expression says to use the smallest of the 2.
2) this from the video on WI school of Business, that (X^150)^2 is the same thing than (x^2 ) ^ (150^2), for the same reason
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#7
09-13-2018, 12:37 AM
 Jim Daniel Member SOA Join Date: Jan 2002 Location: Davis, CA College: Wabash College B.A. 1962, Stanford Ph.D. 1965 Posts: 2,685

Quote:
 Originally Posted by 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
This is a mixture problem. The random variable is
Unif((0,25) with probability 25/74
Unif(25,50) with probability 28/74
Unif(50,100) with probability 15/74
Unif(100,200) with probability 6/74.

For the X^150 calculation, split the last of these Uniforms as
Unif(100,150) with probability 3/74, and
Unif(150,200) with probability 3/74.

By the mixing method, you calculate the desired expression for each Unif, and then mix the answers together with the stated probabilities. If you are unfamiliar with mixture distributions, take a look at the free short study note on mixtures at my website.
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