--MS_Mac_OE_3007576762_11121614_MIME_Part
Content-type: text/plain; charset="US-ASCII"
Content-transfer-encoding: 7bit
Response to Donald Morrison.
This item gave me fits until a friend and I sat down to work it out. I
apologize in advance if I go into too much detail, but you may find it
instructive to have the derivation of why a Chi-square is used. Here
goes...
Begin with the prob distrib F(x), the distribution of losses.
It is important to note that F(x) is our underlying continuous distribution
that is unknown.
Choose any loss amount x.
For any other loss amount L, Prob (L < x) = F(x)
Define Success as loss L < x
Therefore we have a Bernoulli random variable: Prob (Success) = F(x)
Prob (Failure) = 1 - F(x)
Pick N events (i.e. you observe N losses) this implies N Bernoulli trials
This is a Binomial distribution
Prob(k successes) = Combinatorial(N,k) * (F(x))**k * (1-F(x))**(N-k)
How many of the N observations will be < x?
(Number of losses) * (Prob that L < x) = N * F(x)
Now if F(x) is unknown, what can we use?
We can use the empirical distribution Fn(x) (formed from our observations).
N * Fn(x) is the number of observations < x.
Remember that a loss < x was defined to be a "Success".
If we have k losses that are all < x, then k = N * Fn(x)
k = N * Fn(x) is distributed as a Binomial rand var.
Now continue...
N * Fn(x) is Binomial (N trials, Prob Success = F(x))
Fn(x) is (1/N)Binomial (N trials, Prob = F(x))
E(Fn(x)) = (1/N) * N * F(x) = F(x)
Var(Fn(x)) = (1/N**2) * N * F(x) * (1-F(x)) = (1/N) * F(x) * (1-F(x))
( Fn(x) - E(Fn(x)) ) / ( Var(Fn(x)) )**(1/2) is Normal(0,1) by C.L.T.
( Fn(x) - F(x) ) / ( (1/N)(F(x))(1-F(x)) )^(1/2) is Normal(0,1) by C.L.T.
Square a Normal and you have a Chi-square. Square that thing above and you
have the foundation of equation (4.3) and an explanation of why the weight
function (4.4) was chosen. See page 135.
I hope this is helpful.
- Damon
----------
From: donald_morrison@es.adp.com (Donald Morrison)
To: studygroup4b@lists.casact.org
Date: Apr 21, 1999, 9:03
can someone explain to me this thing it mentioned on page 135 hogg
klugman Fn(X) is a binomial nFn(X)
var(nFn(X)) = n**2 var(Fn(X)) = n * p * (1-p)
var(Fn(X)) = p * (1-p) / n there have several questions that use
thisa concept
--MS_Mac_OE_3007576762_11121614_MIME_Part
Content-type: text/html; charset="US-ASCII"
Content-transfer-encoding: quoted-printable
--MS_Mac_OE_3007576762_11121614_MIME_Part--