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Old 06-10-2018, 09:35 PM
RockOn RockOn is offline
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Hello,
Can someone please advise me if I can use the following formula (instead of the simple integration) to answer SOA #28?

E[X^a] = (u^3 - d^3) / 3 (u - d)

Where u is the high end of each interval and d is the low end of each interval?

Thank you!
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Old 06-10-2018, 10:15 PM
Academic Actuary Academic Actuary is offline
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You would get a quicker response if you would post the question.
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Old 06-10-2018, 11:35 PM
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You would get a quicker response if you would post the question.
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Old 06-11-2018, 10:57 AM
bravesandfalcons bravesandfalcons is offline
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Originally Posted by RockOn View Post
Hello,
Can someone please advise me if I can use the following formula (instead of the simple integration) to answer SOA #28?

E[X^a] = (u^3 - d^3) / 3 (u - d)

Where u is the high end of each interval and d is the low end of each interval?

Thank you!
Going by the formula you just gave, E[X^2] = E[X^3] = E[X^4].

Did you mean to perhaps substitute the 3's on the right hand side with a's?
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Old 06-11-2018, 11:01 AM
bravesandfalcons bravesandfalcons is offline
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You would get a quicker response if you would post the question.
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Couldn't get the picture to post. Hopefully, the attachment is there.

Looks like that worked!
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Old 06-11-2018, 11:26 AM
ARodOmaha ARodOmaha is offline
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Couldn't get the picture to post. Hopefully, the attachment is there.

Looks like that worked!
Yes, the correct approach is to use E[X^2] = (b^3 - a^3) / (3(b-a)) for each interval. For the limited expected value piece, first rewrite the intervals with a maximum value of 150. Just be sure to weight the intervals by their empirical probabilities.
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Old 06-11-2018, 12:41 PM
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Your formula gives a conditional expectation: E[X^2| d < X < u]
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Old 06-12-2018, 01:49 PM
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Originally Posted by ARodOmaha View Post
Yes, the correct approach is to use E[X^2] = (b^3 - a^3) / (3(b-a)) for each interval. For the limited expected value piece, first rewrite the intervals with a maximum value of 150. Just be sure to weight the intervals by their empirical probabilities.
Hi ARodOmaha,
Thank you for the response. I did try that but could not seem to get the right answer. For the last part of the interval above 150 would it be: [3 * (150^2)]/78?

Thank you!
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Old 06-12-2018, 01:50 PM
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Quote:
Originally Posted by bravesandfalcons View Post
Couldn't get the picture to post. Hopefully, the attachment is there.

Looks like that worked!
Thank you for posting the question, bravesandFalcons!
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Old 06-12-2018, 02:44 PM
ARodOmaha ARodOmaha is offline
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Originally Posted by RockOn View Post
Hi ARodOmaha,
Thank you for the response. I did try that but could not seem to get the right answer. For the last part of the interval above 150 would it be: [3 * (150^2)]/78?

Thank you!
Yes, the last part is a little different because you are essentially going from 150 to 150. So it is E[X^2] = 150^2. Or think of E[X^2] as Var(X) + E[X]^2. The variance of 150 here is 0. The expected value is 150^2. Weight this with 3/74.
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