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#1
02-19-2009, 07:12 PM
 Sophie H. Member Join Date: Jan 2009 Studying for FAP FA Posts: 572
Counting degree of freedom

Case I: If given some distribution and provide parameters, df=n-1

Case II: If given some distribution and provide parameters, fitted from the data, df= n-r-1
From here,
given binomial n and q, df=n-2-1
given Pareto alpha and theta, df= n-2-1
given negative binomial, df= n-2-1

what if uniform distribution, so Ej will be total and equally divided, then df= n-??-1....?
----------------------------------------------------------------------------------------------
Let's assume 40.14 and 40.10 are really the case II. Then, what will their df's be?

40.14 is as follow:

The observed # of claims for a group of 50 risks has been recorded as:
# of claims // # of risks
0 // 7
1 // 10
2 // 12
3 // 17
4 // 4
Null hypothesis: the # of claims per risk follows a uniform distribution on 0,1,2,3 and 4.
So 50 risk equally divided into 5 groups is 10 each group. Can I assume 10 is a given parameter? So, df=5-1-1=3?

--------------------------------------------------------------------------------------------
40.10
100 observed losses have been recorded in thousands of dollars and are grouped as follow:
Interval // # of Losses
(0,1) // 15
[1,5) // 40
[5,10) // 20
[10,15) // 15
[15, infinity)// 10
Random variable X underlying observed losses, in thousands, is believed to have density function
f(x) = 0.2 e^(-0.2x)

Can I assume it's given as an exponential with theta=1/.2=5,
so df= 5-1-1=3?

Last edited by Sophie H.; 02-19-2009 at 07:52 PM..
#2
02-19-2009, 07:38 PM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 5,334

How many parameters are fitted for uniform?

Maybe you should post 40.14
__________________
Quote:
 Originally Posted by Buru Buru i'm not. i do not troll.
#3
02-19-2009, 07:42 PM
 Sophie H. Member Join Date: Jan 2009 Studying for FAP FA Posts: 572

Another confusion -> 40.19.

"null hypothesis, H0, is the number of claims per risk follows a Poisson Distribution.
.........
min chi-square estimate of the mean of the Poisson distribution is .3055."

Case I: This is what I got. 4 groups down to 3 groups. df=3-1=2 because it doesn't said it's fitted from the data so it doesn't minus the given parameter (Poisson mean).

Case II: which is the solution. Is it because of this sentence? "min chi-square estimate of the mean of the Poisson distribution is .3055"

So, can I draw a conclustion that whenever a distribution parameter is given along with key words such as "chi-square estimate", "fitted by data", ...,etc, then we use df=n-r-1.
#4
02-19-2009, 08:21 PM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 5,334

__________________
Quote:
 Originally Posted by Buru Buru i'm not. i do not troll.
#5
02-19-2009, 08:24 PM
 Actuarialsuck Member Join Date: Sep 2007 Posts: 5,334

Quote:
 Originally Posted by Sophie H. Another confusion -> 40.19. "null hypothesis, H0, is the number of claims per risk follows a Poisson Distribution. ......... min chi-square estimate of the mean of the Poisson distribution is .3055." Case I: This is what I got. 4 groups down to 3 groups. df=3-1=2 because it doesn't said it's fitted from the data so it doesn't minus the given parameter (Poisson mean). Case II: which is the solution. Is it because of this sentence? "min chi-square estimate of the mean of the Poisson distribution is .3055" So, can I draw a conclustion that whenever a distribution parameter is given along with key words such as "chi-square estimate", "fitted by data", ...,etc, then we use df=n-r-1.
You are told that a parameter is fitted! "chi square estimate of the mean..." and how is Poisson dist. determined? By lambda. And what does the mean = ?
__________________
Quote:
 Originally Posted by Buru Buru i'm not. i do not troll.
#6
02-20-2009, 12:54 PM
 Sophie H. Member Join Date: Jan 2009 Studying for FAP FA Posts: 572

Quote:
 Originally Posted by Actuarialsuck Have you had a chance to read this http://www.actuarialoutpost.com/actu...hlight=freedom

I'm still not getting everything for chi-square!

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