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#1




SOA #229
Hello,
I'm having trouble understanding the SOA solution for this problem. To get the expected value of the second derivative with respect to X. Why Is only the second term multiplied by the density function? Why not 1/theta squared as well? Thank you.
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#2




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Second, you're after the expected value of that second derivative as X varies randomly. But the expected value of 1 / theta^2 equals 1 / theta^2, so there's no need to bother multiplying by the density and integrating. But if you do, you'll get the same answer. By the way, my guess is that such a question is very rare. Prepare for what is typically tested.
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Jim Daniel Jim Daniel's Actuarial Seminars www.actuarialseminars.com jimdaniel@actuarialseminars.com 
#3




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I don't think I've asked my question correctly... I agree that it is the second derivative with respect to theta, not x (I mistyped before). When getting the expected value of this, why is the first term multiplied by theta squared and the second term multiplied by the given density function? I'm just trying to understand conceptually why one term is multiplied by one thing and the other is multiplied by the given density function and then integrated. And yes, I agree that the integral of 1/theta sq multiplied by theta is 1/theta squared. In other words, I think I missing a concept with how to get the expected value. Thank you!
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#4




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#5




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Hi Sorry, I'm still confused about this. Let me ask it this way: If I am trying to get the expected value of the function, if I multiply the term (1/theta squared) by the given density function (theta/(theta + x)^2) I don't get 1/(theta squared) as a result for this term. So I guess I'm back to my original question of why is the first term multiplied by theta squared and the second term multiplied by the given density. Or asked another way, what are you multiplying the first term by in order to get 1/theta squared? Thank you.
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#6




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#7




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Thank you for your patience! I do understand this now
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