Actuarial Outpost question on limited expected value
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#1
06-14-2017, 10:06 PM
 degenerateFunction SOA Join Date: May 2017 College: UCLA Posts: 9
question on limited expected value

Say you want to find E [ X + 50 ^ 1000 ]

Is this equivalent to E [ X ^ 950 ] ?
#2
06-14-2017, 10:22 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,227

No. It is a strange thing you are asking about, but I don't think your conclusion is right.

First, I think you must mean E[(X+50)^1000]. Parentheses may not matter, but I assume this is what it meant. Then, suppose (as a counterexample showing the equality is not valid) X is always 2000. X+50 is always 2050, so E[(X+50)^1000] = 1000. But E(X^950) is 950.
#3
06-14-2017, 10:30 PM
 Vorian Atreides Wiki/Note Contributor CAS Join Date: Apr 2005 Location: As far as 3 cups of sugar will take you Studying for ACAS College: Hard Knocks Favorite beer: Most German dark lagers Posts: 64,825

In general, no.

Consider the formula for E[ X ^ d ] where S(y) = 1 - F(y):
$\int_{-\infty}^d x f(x)dx + dS(d)$

Then:
$E[(X+k) \wedge d] = \int_{-\infty}^{d-k} (x+k) f(x)dx + (d-k)S(d-k)$

Do some algebra, then work out the formula for E[ X ^ (d - k)], and you'll find that
$E[(X+k) \wedge d] - E[X \wedge (d-k)] = kF(d-k)$
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Last edited by Vorian Atreides; 06-14-2017 at 10:58 PM.. Reason: adding clarifying ()'s
#4
06-14-2017, 10:56 PM
 degenerateFunction SOA Join Date: May 2017 College: UCLA Posts: 9

Thank you both. And yes, I meant it as you wrote it, Gandalf.

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