
#1




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(1w)=7.5. Simplifying gets us w=(7.5X)/(5X). We can also find that 50w+2(X^2)(1w)=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=(131.252(X^2))/(502(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




Quote:
5w + (1w)*x = 7.5 50w + (1w)*(2x^2) = 131.25 I solved for x = (7.5  5w)/(1w). Plug this into the second equation to get w=0.6 and x=11.25.
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#4




You could use the variance equation:
25w + (1w)x^2 + (x5)^2 w(1w) =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




Quote:
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