Actuarial Outpost ASM 3.18
 Register Blogs Wiki FAQ Calendar Search Today's Posts Mark Forums Read
 FlashChat Actuarial Discussion Preliminary Exams CAS/SOA Exams Cyberchat Around the World Suggestions

 D.W. Simpson & Company International Actuary Jobs   Canada  Asia  Australia  Life  Pension  CasualtyBermuda, United Kingdom, Europe, Asia, Worldwide

#1
02-23-2006, 05:31 PM
 rebeccap Member Join Date: Sep 2004 Location: Napa, CA Posts: 494
ASM 3.18

I am temporarily forgetting why, in the last line of the solution, the second 50 needs to go inside of the square root sign rather than outside? Any short reminder will be appreciated.
#2
02-23-2006, 05:35 PM
 BallaActuary Member CAS Join Date: Jan 2005 Posts: 15,061

Quote:
 Originally Posted by rebeccap I am temporarily forgetting why, in the last line of the solution, the second 50 needs to go inside of the square root sign rather than outside? Any short reminder will be appreciated.
What is the problem for people without the ASM?
__________________
Quote:
 Originally Posted by Dr. John Zoidberg Hey guys my first professional wager was "Will This Thread Become About Buru?" I took Yes at really crappy odds but I knew it was a lock. ALL ABOARD THE TRAIN TO PROFIT TOWN! CHOO CHOO!
#3
02-23-2006, 07:05 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 6,194

This is the discussion in Section 2.1, particularly at the bottom of page 21 to the top of page 22. The variance of the SUM of 50 independent random variables is 50 times the variance of one, whereas the variance of 50 times a random variable is 2500 times the variance of one. The problem states that the employees are mutually independent.
#4
02-23-2006, 08:16 PM
 rebeccap Member Join Date: Sep 2004 Location: Napa, CA Posts: 494

Quote:
 Originally Posted by Abraham Weishaus This is the discussion in Section 2.1, particularly at the bottom of page 21 to the top of page 22. The variance of the SUM of 50 independent random variables is 50 times the variance of one, whereas the variance of 50 times a random variable is 2500 times the variance of one. The problem states that the employees are mutually independent.
Ah, I knew I learned it somewhere...Thank you

Bella: This is a typical trap to the sort of the following--

If all the Xi's are independent and have identical distributions, then
Var [sum of all Xi's, from i=0 to i=n] = n*Var(X)
#5
02-24-2006, 11:43 AM
 BallaActuary Member CAS Join Date: Jan 2005 Posts: 15,061

Quote:
 Originally Posted by rebeccap Bella: This is a typical trap to the sort of the following--
Is Bella supposed to be me?
__________________
Quote:
 Originally Posted by Dr. John Zoidberg Hey guys my first professional wager was "Will This Thread Become About Buru?" I took Yes at really crappy odds but I knew it was a lock. ALL ABOARD THE TRAIN TO PROFIT TOWN! CHOO CHOO!
#6
02-24-2006, 02:09 PM
 rebeccap Member Join Date: Sep 2004 Location: Napa, CA Posts: 494

Quote:
 Originally Posted by BallaActuary Is Bella supposed to be me?
Oops, typo. Should be BAlla. Sorry.
#7
02-24-2006, 03:19 PM
 BallaActuary Member CAS Join Date: Jan 2005 Posts: 15,061

Quote:
 Originally Posted by rebeccap Oops, typo. Should be BAlla. Sorry.
I was confused because I wanted to know what the problem was. I already passed M, so I don't have the M ASM manual.

Key formula (know it and love it):

Var(aggregate losses) = E(N)Var(Y) + [E(Y)^2]Var(N)
where N is from your frequency distribution and Y is from your severity distribution (and they are independent). This works when N is not fixed.

If N is fixed, the formula breaks down to: N[Var(Y)]

Neato!
__________________
Quote:
 Originally Posted by Dr. John Zoidberg Hey guys my first professional wager was "Will This Thread Become About Buru?" I took Yes at really crappy odds but I knew it was a lock. ALL ABOARD THE TRAIN TO PROFIT TOWN! CHOO CHOO!
#8
02-24-2006, 03:48 PM
 rebeccap Member Join Date: Sep 2004 Location: Napa, CA Posts: 494

Quote:
 Originally Posted by BallaActuary I was confused because I wanted to know what the problem was. I already passed M, so I don't have the M ASM manual. Key formula (know it and love it): Var(aggregate losses) = E(N)Var(Y) + [E(Y)^2]Var(N) where N is from your frequency distribution and Y is from your severity distribution (and they are independent). This works when N is not fixed. If N is fixed, the formula breaks down to: N[Var(Y)] Neato!
That's right. I thought you have passed already...some of these questions are pretty long, so I will do my best in typing them out.

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off

All times are GMT -4. The time now is 06:35 PM.

 -- Default Style - Fluid Width ---- Default Style - Fixed Width ---- Old Default Style ---- Easy on the eyes ---- Smooth Darkness ---- Chestnut ---- Apple-ish Style ---- If Apples were blue ---- If Apples were green ---- If Apples were purple ---- Halloween 2007 ---- B&W ---- Halloween ---- AO Christmas Theme ---- Turkey Day Theme ---- AO 2007 beta ---- 4th Of July Contact Us - Actuarial Outpost - Archive - Privacy Statement - Top