Actuarial Outpost Can someone please explain this to me?
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CATASTROPHE MODELING JOBS

#1
09-28-2009, 11:31 PM
 mdavis567 Member CAS SOA Join Date: Sep 2009 Studying for FM Favorite beer: The free kind Posts: 39
Can someone please explain this to me?

This problem is giving me a solution that I simply don't understand...

The pf is P[X=x] = 2/3^x, and I'm supposed to find the probability that X is even.

I get that its the sum of the probabilities of X = 2,4,6...but what I don't get is how the book is getting this:

P[X=2] + P[X=4] +... = 2/3 * (1/3 + 1/3^3 + 1/3^5+....) = 2/3^2 * 1/(1-(1/3^2)) = 1/4

I'm not following the simplification, how do they get the 2nd part from the first, and the 3rd from the 2nd?
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-Michelle
#2
09-29-2009, 12:13 AM
 jprep Member Join Date: Aug 2009 Location: Buffalo Studying for VEEs & FAP College: University at Buffalo Alumni Favorite beer: Corona w/ Lime Posts: 1,559

This is a geometric series.

P[X=2] = 2/3^2
P[X=4] = 2/3^4
P[X=6] = 2/3^6 and so on.

Now you add all these up. That's

2/3^2 + 2/3^4 + 2/3^6 + 2/3^8 + ...

Now factor the first term out of every term (factor out 2/3^2, that is), so that you get this:

2/3^2 * (1 + 1/3^2 + 1/3^4 + 1/3^6 + ...) which is equivalent to:

2/9 * (1 + [1/9]^1 + [1/9]^2 + [1/9]^3 + ...)

The bolded part is special, here's the rule (yes, you'll just have to memorize this, there is a proof though):

(1 + [1/r]^1 + [1/r]^2 + [1/r]^3 + ...) = 1 / (1 - r)

In your question, r is 1/9. So the above bolded part becomes:

2/9 * (1 / (1 - 1/9)) = 2/9 * (1 / (8/9)) = 2/9 * 9/8 = 2/8 = 1/4

It's not supposed to actually look that hard, I just don't know how to use the Tex thingy on this forum. These kind of geometric series show up a lot in the practice exams, but they don't ask many questions like this on the real exam - they could, though. That problem was solved weirdly - they skipped some algebra, and the second step looks weird.

Good luck!
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Quote:
 Originally Posted by campbell REPENT THE END IS NIGH
Spoiler:
Quote:
 Originally Posted by Aluan You sound way too excited about studying for 7. A bit of studying will fix that, though.
Quote:
 Originally Posted by MyKenk Nope. 7 is the must pass of all must passes next spring. Can leave absolutely nothing to chance. Study schedule starts on November 15th. Is it nuts? Probably. Will I pass? Definitely. Will it be worth it? Of course.
Quote:
 Originally Posted by MyKenk's blog A bit of an early start for a second sitting, but this exam is a must pass, and there's no such thing as too much studying for a must pass
#3
09-29-2009, 12:22 AM
 mdavis567 Member CAS SOA Join Date: Sep 2009 Studying for FM Favorite beer: The free kind Posts: 39

You cleared that right up for me.

As for the clarity, im kind of a dunce as well, and it was fine. I copied it down so it looked nicer, thanks!
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-Michelle
#4
09-29-2009, 12:50 AM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 5,327

A few key commands:

\frac{1}{2} will produce $\frac{1}{2}$

\int_{-\infty}^{\infty} will produce $\int_{-\infty}^{\infty}$

\sum_{i=0}^{n} will produce $\sum_{i=0}^{n}$

You can usually hit quote if you see anyone writing in TeX and see what "code" they used.
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 Originally Posted by Buru Buru i'm not. i do not troll.
#5
09-29-2009, 01:03 AM
 jprep Member Join Date: Aug 2009 Location: Buffalo Studying for VEEs & FAP College: University at Buffalo Alumni Favorite beer: Corona w/ Lime Posts: 1,559

Quote:
 Originally Posted by Actuarialsuck jprep: http://www.actuarialoutpost.com/actu...ad.php?t=74727 A few key commands: \frac{1}{2} will produce $\frac{1}{2}$ \int_{-\infty}^{\infty} will produce $\int_{-\infty}^{\infty}$ \sum_{i=0}^{n} will produce $\sum_{i=0}^{n}$ You can usually hit quote if you see anyone writing in TeX and see what "code" they used.
Thanks! Now that's $\int_{-\infty}^{\infty} inte-great!$

Error: Calculation too lame.
__________________
P FM MFE MLC C VEE Stats CorpFinance Econ FAP

Quote:
 Originally Posted by campbell REPENT THE END IS NIGH
Spoiler:
Quote:
 Originally Posted by Aluan You sound way too excited about studying for 7. A bit of studying will fix that, though.
Quote:
 Originally Posted by MyKenk Nope. 7 is the must pass of all must passes next spring. Can leave absolutely nothing to chance. Study schedule starts on November 15th. Is it nuts? Probably. Will I pass? Definitely. Will it be worth it? Of course.
Quote:
 Originally Posted by MyKenk's blog A bit of an early start for a second sitting, but this exam is a must pass, and there's no such thing as too much studying for a must pass
#6
09-29-2009, 03:34 AM
 atomic Member CAS Join Date: Jul 2006 Posts: 4,088

Quote:
 Originally Posted by mdavis567 This problem is giving me a solution that I simply don't understand... The pf is P[X=x] = 2/3^x, and I'm supposed to find the probability that X is even. I get that its the sum of the probabilities of X = 2,4,6...but what I don't get is how the book is getting this: P[X=2] + P[X=4] +... = 2/3 * (1/3 + 1/3^3 + 1/3^5+....) = 2/3^2 * 1/(1-(1/3^2)) = 1/4 I'm not following the simplification, how do they get the 2nd part from the first, and the 3rd from the 2nd?
A simpler approach would be to observe that since

$X \in \{1, 2, 3, ...\}$

in order for the given mass function to correspond to a valid probability function, we see that

$\Pr[X = 2k] = \frac{1}{3} \Pr[X = 2k-1], \quad k = 1, 2, 3, ...$.

Therefore the probability that X is even is simply 1/3 the probability that X is odd, and because the probability that X is either even or odd is 1, we obtain

$\Pr[X = {\rm even}] = 1/4$.
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Think of it like a movie. The Torah is the first one, and the New Testament is the sequel. Then the Qu'ran comes out, and it retcons the last one like it never happened. There's still Jesus, but he's not the main character anymore, and the messiah hasn't shown up yet.

Jews like the first movie but ignored the sequels, Christians think you need to watch the first two, but the third movie doesn't count, Moslems think the third one was the best, and Mormons liked the second one so much they started writing fanfiction that doesn't fit with ANY of the series canon."

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#7
09-29-2009, 12:54 PM
 Colymbosathon ecplecticos Member Join Date: Dec 2003 Posts: 4,987

Cute!

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