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#1
06-02-2018, 06:45 PM
 royevans Member SOA Join Date: Jul 2017 Posts: 462
im baffled... help with series?

I've been working on something and I dont understand where I'm going wrong.... please correct me on the following.

Let F be the cumulative distribution function of an exponential distribution with theta = 3.

define w = [F(2) - F(1)] + 2[F(3) - F(2)] + 3[F(4) - F(3)] + ...

w must be positive because each bracket is positive (the difference of 2 distribution functions with a higher x value minus lower x value would be a positive value)

rearranging the above terms we have:

w = -F(1) - F(2) - F(3) + .....

Now w is a negative value?!?!? Each F is a positive value and we're adding up negative values...

similarly we know S(x) = 1 - F(x)

substituting F(X) with S(X) in the original equation, we get

w = [{1 - S(2)} - {1 - S(1)}] + 2[{1 - S(3)} - {1 - S(2)}] + 3[{1 - S(4)} - {1 - S(3)}] + ...
w = [S(1) - S(2)] + 2*[S(2) - S(3)] + 3*[S(3) - S(4)] + ....
w = S(1) + S(2) + S(3) +...

WHICH IS AGAIN A POSITIVE VALUE?

where am I messing up?

i remember something like this in a calc class I took where if u rearrange terms in an infinite series differently, you'll get different results... how do I get the correct answer then? How do you know how to CORRECTLY rearrange terms to get the correct answer?
__________________
Quote:
 Originally Posted by ao fan I just don't want to see some dumb punk from some school like Penn state suddenly be handed asa when they couldn't pass the prelims to save their life.

Last edited by royevans; 06-02-2018 at 06:48 PM..
#2
06-02-2018, 07:01 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,913

I believe your problem is that the series, in any of the forms you present (or in any other form), does not converge, and therefore there is no answer to what the sum equals.
#3
06-02-2018, 07:22 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,057

If the absolute values of all the terms does not converge the series is conditionally convergent. Conditionally convergent series can be rearranged to equal any value.
Your series does not converge absolutely as the F's are increasing to one.

https://en.wikipedia.org/wiki/Riemann_series_theorem
#4
06-02-2018, 07:27 PM
 royevans Member SOA Join Date: Jul 2017 Posts: 462

Quote:
 Originally Posted by Academic Actuary If the absolute values of all the terms does not converge the series is conditionally convergent. Conditionally convergent series can be rearranged to equal any value. Your series does not converge absolutely as the F's are increasing to one. https://en.wikipedia.org/wiki/Riemann_series_theorem
does the last format converge (written as S's).... here is an answer key to a problem I was looking at.. it seems the asnwer implies that it converges if written in terms of S?

__________________
Quote:
 Originally Posted by ao fan I just don't want to see some dumb punk from some school like Penn state suddenly be handed asa when they couldn't pass the prelims to save their life.
#5
06-02-2018, 07:41 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,057

Quote:
 Originally Posted by royevans does the last format converge (written as S's).... here is an answer key to a problem I was looking at.. it seems the asnwer implies that it converges if written in terms of S?
It's not intuitive at all, but the series written out in S's is absolutely convergent so the terms can be rearranged, but the series written out in F's is not so the terms cannot be rearranged.
#6
06-03-2018, 01:17 AM
 royevans Member SOA Join Date: Jul 2017 Posts: 462

Quote:
 Originally Posted by Academic Actuary It's not intuitive at all, but the series written out in S's is absolutely convergent so the terms can be rearranged, but the series written out in F's is not so the terms cannot be rearranged.
can i ask you how to solve the following problem then?

The EV of the whole mixture is the expected value of the primary and secondary distribution.... the secondary distribution's (exponential) makes sense when written out in terms of the first equation I had which seems to be infinite?

How would you go about finding the expected value of the billable hours? The solution just starts out with the series involving S(x) which isn't intuitive to me.

__________________
Quote:
 Originally Posted by ao fan I just don't want to see some dumb punk from some school like Penn state suddenly be handed asa when they couldn't pass the prelims to save their life.
#7
06-03-2018, 10:33 AM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,057

Your original series in F's is convergent (conditionally). The first term F(2) - F(1) = exp(-1/3) - exp(-2/3).

Factor it out then sum the remaining terms which look the the series for an increasing perpetuity due from compound interest. Cancel to simplify. you are left with:

exp(-1/3)/[1-exp(-1/3)] which is the same as the sum of the S(x)'s. You run into problems if you reorder the sequence.
#8
06-04-2018, 08:24 AM
 Michael Mastroianni SOA Join Date: Jan 2018 Posts: 20

Quote:
 Originally Posted by royevans can i ask you how to solve the following problem then?
So you have the number of hours in interval $i$ is $X_{i}\sim \text{Exponential}(\lambda=\frac{1}{2})$

The number of hours billed for this interval is dropping the decimal portion which is
$\left \lfloor{X_{i}}\right \rfloor \sim \tex{Geometric}(p=1-e^{-\lambda}=0.39347)$ so we know the expectation $E[\left \lfloor{X_{i}}\right \rfloor]=\frac{1-p}{p}$

We also have the number of intervals as a zero-truncated geometric with $\beta=4$. Call that $N$.
That has expectation $E[N]=\frac{\beta}{1-1/(1+\beta)}=\frac{4}{1-1/(1+4)}=5$

We want the expectation of $S=500\cdot \left \lfloor{X_{1}}\right \rfloor+\dots+500\cdot \left \lfloor{X_{N}}\right \rfloor$

Then the answer assuming independence is
$E[S]=E[E[S\mid N]]=E[E[500\cdot\left \lfloor{X_{1}}\right \rfloor] \cdot N]=E[500\cdot \frac{1-0.39347}{0.39347}\cdot N]\approx770.7449\cdot E[N]=770.7449\cdot 5\approx \3853.72$

No series manipulation necessary
That's how I see it anyway.
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Michael Mastroianni, ASA
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