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 Short-Term Actuarial Math Old Exam C Forum

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#1
06-08-2018, 04:34 PM
 SweepingRocks Member SOA Join Date: Jun 2017 College: Bentley University Posts: 78
Dumb Mixture Question

[Disclaimer: I'm studying for STAM, not C]

The question is 4.7 from the latest C manual: Losses for an insurance coverage follow a distribution which is a mixture of an exponential distribution with mean 5 and an exponential distribution with mean X. The mean loss size is 7.5. The variance of loss size is 75.

Determine the coefficient of skewness of the loss distribution.

Okay so I know variation for an exponential is mean squared and E[X^2]-E[X]^2 = Var[X]. So for an exponential, E[X^2] = 2*(Mean^2). So with w as the weight of the exponential with mean 5, 5w+X(1-w)=7.5. Simplifying gets us w=(7.5-X)/(5-X).
We can also find that 50w+2(X^2)(1-w)=131.25, because 131.25 is the E[X^2] of the mixture (75 + (7.5^2)) and the weighted E[X^2] should be equal to this. Simplifying gives us w=(131.25-2(X^2))/(50-2(X^2)).

Setting the two w expressions equal to find X, I eventually simplify to the quadratic: 50(X^2) - 81.25X + 281.25 = 0. This quadratic has no solution. The solution in the book used a completely different approach so I can't look there to see where I messed up. Any help would be appreciated
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#2
06-08-2018, 04:53 PM
 ARodOmaha Member SOA Join Date: May 2016 Location: Omaha, NE College: University of Nebraska (alma mater) Favorite beer: Captain Morgan Posts: 197

Quote:
 Originally Posted by SweepingRocks [Disclaimer: I'm studying for STAM, not C] The question is 4.7 from the latest C manual: Losses for an insurance coverage follow a distribution which is a mixture of an exponential distribution with mean 5 and an exponential distribution with mean X. The mean loss size is 7.5. The variance of loss size is 75. Determine the coefficient of skewness of the loss distribution. Okay so I know variation for an exponential is mean squared and E[X^2]-E[X]^2 = Var[X]. So for an exponential, E[X^2] = 2*(Mean^2). So with w as the weight of the exponential with mean 5, 5w+X(1-w)=7.5. Simplifying gets us w=(7.5-X)/(5-X). We can also find that 50w+2(X^2)(1-w)=131.25, because 131.25 is the E[X^2] of the mixture (75 + (7.5^2)) and the weighted E[X^2] should be equal to this. Simplifying gives us w=(131.25-2(X^2))/(50-2(X^2)). Setting the two w expressions equal to find X, I eventually simplify to the quadratic: 50(X^2) - 81.25X + 281.25 = 0. This quadratic has no solution. The solution in the book used a completely different approach so I can't look there to see where I messed up. Any help would be appreciated
You are correct that you should equate the two moments and not equate the variance.

5w + (1-w)*x = 7.5
50w + (1-w)*(2x^2) = 131.25

I solved for x = (7.5 - 5w)/(1-w). Plug this into the second equation to get w=0.6 and x=11.25.
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#3
06-08-2018, 05:06 PM
 SweepingRocks Member SOA Join Date: Jun 2017 College: Bentley University Posts: 78

Quote:
 Originally Posted by ARodOmaha You are correct that you should equate the two moments and not equate the variance. 5w + (1-w)*x = 7.5 50w + (1-w)*(2x^2) = 131.25 I solved for x = (7.5 - 5w)/(1-w). Plug this into the second equation to get w=0.6 and x=11.25.
I tried that and got the solution you got. Still curious as to why mine didn't work
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#4
06-08-2018, 05:07 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,475

You could use the variance equation:

25w + (1-w)x^2 + (x-5)^2 w(1-w) =75

If you plug in x in terms of w from the mean equation you get a linear equation in w after some cancelation.
#5
06-09-2018, 02:19 PM
 Jim Daniel Member SOA Join Date: Jan 2002 Location: Davis, CA College: Wabash College B.A. 1962, Stanford Ph.D. 1965 Posts: 2,710

Quote:
 Originally Posted by SweepingRocks [Disclaimer: I'm studying for STAM, not C] The question is 4.7 from the latest C manual: Losses for an insurance coverage follow a distribution which is a mixture of an exponential distribution with mean 5 and an exponential distribution with mean X. The mean loss size is 7.5. The variance of loss size is 75. Determine the coefficient of skewness of the loss distribution. Okay so I know variation for an exponential is mean squared and E[X^2]-E[X]^2 = Var[X]. So for an exponential, E[X^2] = 2*(Mean^2). So with w as the weight of the exponential with mean 5, 5w+X(1-w)=7.5. Simplifying gets us w=(7.5-X)/(5-X). We can also find that 50w+2(X^2)(1-w)=131.25, because 131.25 is the E[X^2] of the mixture (75 + (7.5^2)) and the weighted E[X^2] should be equal to this. Simplifying gives us w=(131.25-2(X^2))/(50-2(X^2)). Setting the two w expressions equal to find X, I eventually simplify to the quadratic: 50(X^2) - 81.25X + 281.25 = 0. This quadratic has no solution. The solution in the book used a completely different approach so I can't look there to see where I messed up. Any help would be appreciated
You have the coefficient of X^2 wrong in your simplified quadratic---it should be 5 rather than 50. Then the quadratic has two solutions, X=5 (which is impossible by your first equation involving w and X), and X=11.25. Then you get w = 0.6.
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