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#1
07-09-2008, 08:19 PM
 coldplay Join Date: Dec 2007 Posts: 25
What is Darth Vader Rule in ASM?

I did google and wiki. they give me bunch of star wars stuff and rule of two. unfortunately, still dont understand it. Could anybody please explain it in Math. Thx
#2
07-09-2008, 08:30 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,455

Why search there? If you do an AO search for "darth", you'll find some explanations of the rule.

Basically, if you have a survival function for a non-negative random variable, its expected value is equal to the integral from 0 to infinity of s(x) dx [or, equivalently, of [1-F(x)] dx.

If there is a maximum value of the random variable, you can use that as the upper limit of the integral, since s(x) = 0 above that. 0 is always the lower limit. (There are variations that don't use 0 as the lower limit, but they get the same result as using 0 as the lower limit.)
#3
07-09-2008, 08:42 PM
 coldplay Join Date: Dec 2007 Posts: 25

Thank you, Gandalf. But why Darth Vader? why not Frodo Baggins Rule ?
#4
07-09-2008, 08:45 PM
 Tight Tin Foil Hat Member CAS SOA Join Date: Jul 2007 Location: Cheecagoland Studying for 4/C Favorite beer: Jooky Posts: 733

Here's my contribution. All those math-inclined members of AO feel free to poke holes in my explanation.

For any fully continuous random variable X where all possible values of X are >= 0
$EX=\int_{all x} s(x)dx$

For any discrete random variable X where all possible values of X are >= 0 AND integers

$EX =\sum_{n=0}^{+\infty} Pr(X > n)$

If you have a mixture you mix accordingly.

What does this have to do with Darth Vader? I guess it's some esoteric actuarial humor...
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#5
07-09-2008, 11:19 PM
 carrytheCrøss Member CAS Join Date: May 2005 Posts: 2,761

Quote:
 Originally Posted by Tight Tin Foil Hat What does this have to do with Darth Vader? I guess it's some esoteric actuarial humor...
Correct.
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a wonderful post
#6
07-09-2008, 11:21 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 26,455

Quote:
 Originally Posted by coldplay Thank you, Gandalf. But why Darth Vader? why not Frodo Baggins Rule ?
See this thread for more about the rule, and in particular post 6. Krzysio gave it the name (I think), so I guess he's the ultimate authority on the reason.
#7
07-18-2008, 03:13 AM
 booyaolian Member Join Date: Feb 2007 Posts: 125

Whether a R.V. is continuous or discrete, and it's ">=0," its expected value is the integral from 0 to infinity of Sx(X), which is the survival function.

For a discrete R.V. and it is ">=1," its expected value is the summation, from n=1 to n=infinity, of Sx(X).
#8
07-24-2008, 09:53 AM
 cj101 Member Join Date: Jul 2008 Posts: 40

Can someone give an example in which you would use the Darth Vader Rule and also give the calculation for solving such a problem.
#9
07-24-2008, 10:04 AM
 daaaave David Revelle Join Date: Feb 2006 Posts: 2,487

Quote:
 Originally Posted by cj101 Can someone give an example in which you would use the Darth Vader Rule and also give the calculation for solving such a problem.
Suppose losses have an exponential distribution with mean 3, and there is a benefit limit of 10. Then the expected payment can either be calculated as:
$\int_0^{10} x f(x) dx + 10 P[X=10] = \int_0^{10} \frac{x}{3} e^{-x/3} dx + 10 e^{-10/3}$
which I don't feel like computing, or you could use the survival function and get
$\int_0^\infty P[X>x] dx = \int_0^{10} e^{-x/3} dx = 3-3 e^{-10/3}$
(note that the second integral only goes to 10 because the probability of a payment being greater than 10 is 0).

The point is that using the survival function is often quicker when dealing with mixed distributions, which includes problems with continuous loss amounts and either a deductible or a benefit limit. It also saves you a step of calculus when dealing with exponential distributions and Pareto distributions.
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#10
07-24-2008, 10:31 AM
 cj101 Member Join Date: Jul 2008 Posts: 40

Wow that looks extremely handy. And all along I was wasting my time with the integration formula for an exponential times x. That will be nice for the exam because I won't have to calculate xe^ax/a^2-e^ax/a. Thanks a bunch daaave

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