Q/A bank part 1,question no.-1.25

[Neetu Verma]

Q/A bank part 1,question no.-1.25 part (ii)....
any one tell me .... why we take f(x) instead of f(x) /(1-F(M)) for 50 claim above retention limit in the likelihood function?????

 

[nageshmcl]

Hope it will help

Please have a look at the note given in the solution:

(This was a difficult question. In questions involving XOL arrangements, you need to think carefully whether you are working in terms of the original or the conditional loss distribution, and whether you are working in terms of gross or net claim amounts.)

In the first part we are trying to find the likelihood estimator, only for the reinsured claims(50), then we should use the conditional density function in the likelihood function.

In the second part we are trying to find the liklihood estimator for all the claims(650), then we should use the original density function f(x) only.

The likelihood function is calculated as below:

Out of 650 claims, we have the information like:

1. 600 claims are below the retention, means we don't know the exact claim amounts, so the data is Censored. The probability of each claim amount less than or equal to M is P(X<=M) = F(M).

Likelihood for 600 claims is = F(M) pow 600.

2. We have the information of 50 claims as 'Sum of cuberoot of each claim' = 2,600. We will use this in our expression ' product of f(x) for 50 claims'.

Thanks.

 

[nageshmcl]

Additional Information

The following example given in the material and the question 4.16 will clear more:

Example from the material:
Claims in a portfolio are believed to arise as an Exp(λ ) distribution. There is a
retention limit of 1,000 in force, and claims in excess of 1,000 are paid by the reinsurer. The insurer, wishing to estimate λ , observes a random sample of 100 claims, and finds that the average amount of the 90 claims that do not exceed 1,000 is 82.9. There are 10 claims that do exceed the retention limit. Find the MLE for λ .

In this ex: we have the information about all the claims, thats why we have used the density function f(x).

In the question 4.16, it is asked only for the resinsured claims, so we should use the conditional density function.

Thanks.

 

[gcpgcp]

What is the question ?

My Q bank has some other Q.

 

[Neetu Verma]

thanks

 

[Neetu Verma]

What is the question ?

My Q bank has some other Q.

the question is long ,so i type some part for idea......

An insurer has arranged a reinsurance treaty f or the office's non-life insurance portfolio.Under this treaty the reinsurer pays the excess of indvidual claim that exceeds the retention limit M . so for each claim reported to the insurer ,the reinsurer pays the greater of X-M and zero, where X denotes the gross amount of the claim before reinsurance .
the reinsurer assumes that the random variable X has a weibull distribution with distribution function ( where theata is positive parameter):...........

 

[gcpgcp]

That is Q 1.31 in my Q bank !!!

If you jump to part ii) c) explanation / answer you will get a clearer picture.

Also see http://www.acted.co.uk/forums/showthread.php?t=3095

 

[nageshmcl]

Hi Neetu,

Hope this will resolve the doubts if you have still any on the likelihood formation on 650 claims.

Given

1. 600 claims are less than M:
Probability that a claim less than M = P(X<M) = F(M)
Out of 650 claims 600 claims are less than M, then the probability becomes : L1=(650 c 600) * F(M) pow 600 * (1-F(M)) pow 50.

2. 50 claims are greater than M. Here we need to use the conditional probability.
Probability of a claim greater than M is f(x)/P(x>M) = f(x)/(1-F(M))
For 50 claims L2= Pie(i= 1 to 50) f(xi)/(1-F(M).
Here (1-F(M)) doesn't depend on 'i', so we can write the denominator as
(1-F(M))pow50.

Final likelihood is L = L1 * L2= (650 c 600) * F(M)pow600*Pie(i=1to50)f(xi).
Here (1-F(M)) pow 50 cancelled out.

Thanks,
Nageswara.

 

[Neetu Verma]

Hi Neetu,

Hope this will resolve the doubts if you have still any on the likelihood formation on 650 claims.

Given

1. 600 claims are less than M:
Probability that a claim less than M = P(X<M) = F(M)
Out of 650 claims 600 claims are less than M, then the probability becomes : L1=(650 c 600) * F(M) pow 600 * (1-F(M)) pow 50.

QUOTE]

please tell me ..in this part ... why u use binomial distribution????????

 

[nageshmcl]

By looking at the situation, we can frame the problem description as a Binomial distribution. X~ BIN(n,p) , here
n - total number of trails.
p - probability of success.

Here the 650 claims can be noted as the total number of trails - n,
Probability of a claim less than M is F(M) - p, can be treated as a probability of success.

So X ~ BIN(650,F(M)):

Out of 650 claims, we need to find the the probability that, 600 claims are less than M, i.e,

(650 c 600) * F(M) pow 600 * (1-F(M)) pow 50.

Please suggest if i am wrong.

Thanks
Nageswar.

 

[Neetu Verma]

By looking at the situation, we can frame the problem description as a Binomial distribution. X~ BIN(n,p) , here
n - total number of trails.
p - probability of success.

Here the 650 claims can be noted as the total number of trails - n,
Probability of a claim less than M is F(M) - p, can be treated as a probability of success.

So X ~ BIN(650,F(M)):

Out of 650 claims, we need to find the the probability that, 600 claims are less than M, i.e,

(650 c 600) * F(M) pow 600 * (1-F(M)) pow 50.

Please suggest if i am wrong.

Thanks
Nageswar.

according to the question X~Weibull distribution with parameter theata not binomial distribution...

 

[nageshmcl]

according to the question X~Weibull distribution with parameter theata not binomial distribution...

The X which i considered here in my explanation is a general X which is a RV, that is not the one given in the problem.

 

[Neetu Verma]

The X which i considered here in my explanation is a general X which is a RV, that is not the one given in the problem.

ok... but i ask u why u use here binomial distribution???????...although it is not given in question....

 

[nageshmcl]

Here we don't know which 600 claims are less than M out of 650. So we are trying to find, what is the probability that 600 claims are less than M, given that probability of a claim less than M is F(M).

 

[Neetu Verma]

Here we don't know which 600 claims are less than M out of 650. So we are trying to find, what is the probability that 600 claims are less than M, given that probability of a claim less than M is F(M).

ok....
thanks a lot