![]() |
|
|
|||||||
| FlashChat | Actuarial Discussion | Preliminary Exams | CAS/SOA Exams | Cyberchat | Around the World | Suggestions |
![]() |
|
|
Thread Tools | Display Modes |
|
#1
|
|||
|
|||
|
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
|
||||
|
||||
|
Why search there?
Here is one thread about it; there are others. 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
|
|||
|
|||
|
Thank you, Gandalf. But why Darth Vader? why not Frodo Baggins Rule ?
|
|
#4
|
||||
|
||||
|
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 For any discrete random variable X where all possible values of X are >= 0 AND integers 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... |
|
#5
|
|||
|
|||
|
Quote:
__________________
a wonderful post |
|
#6
|
||||
|
||||
|
Quote:
|
|
#7
|
|||
|
|||
|
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
|
|||
|
|||
|
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
|
|||
|
|||
|
Quote:
which I don't feel like computing, or you could use the survival function and get (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. |
|
#10
|
|||
|
|||
|
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
|
![]() |
| Thread Tools | |
| Display Modes | |
|
|