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Some questions about CMP CS2 CH8 to CH12

Y

ykai

Ton up Member
1.CMP CS2-CH8-question 8.1-(iii)
What is "the force of mortality for the male is 56% higher than the force of mortality for the female" comes from?

"the force of mortality for the female is 36% less than the force of mortality for the male"
I can understand "mortality for female"/"mortality for male"=0.64= mortality for famale is only 64% of that for male.

but I can't understand why "the force of mortality for the male is 56% higher than the force of mortality for the female".

2.CS2 Assignments X3-3.5-(ii)
Why Z2 make M as 1? Is it because of 1.86*1.45>2.32?

3.CS2 Assignments X3-3.11-(i)-(c)
Where is 3 in explanation of solution 3.11 come from?
 
4.What is meaning of t positive groups of grouping of sign test?
Since n_1 is the number of positive outcomes and n_2 is the number of negative outcomes, what is the meaning of t?
What is logic of this test and formula?

5.Why r_j ~N(0,1/m) in serial correlations test? Where is it comes from?
j= lag
Is there any proof or derivation from other theories?
For example,CLT and so on.
 
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1. Ch8 8.1(iii): if we look at males over females so with mu0(x) cancelling, we are left with e^0.416/e^-0.030 = 1.56, so males 56% higher than females
 
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2. Assignment X3 Q3.5(ii): If the baseline hazard is for female non-smokers, so we have additions for smokers and males so have z1 = 1 for smokers and z2=1 for males
 
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3. Assignments X3 Q3.11(i)(c):
The 3 is the number of runs of positive standard deviations, so if you look at the z_x's in (i)(a), we have 3 positive runs one from 34-36, one at 40 and a final one 44
 
Thank you for your reply.
I have totolly understood for 4 questions,but I still want to know if
I have found out answer for last one.
 
Hello

See some comments below:

4. As Andrea says, a group of positive signs is a run of 1 or more consecutive positive deviations. Does her example for 3.11 help? I know you said you were happy with the first 4 questions but wanted to make sure.

5. This relates to a similar result that we have in the time series chapters. In Section 2.2 of Chapter 14, we have that the autocorrelations are approximately normally distributed with mean 0 and variance 1/n (for a large enough sample size). We are effectively working the same quantities here in graduation, as the r_j are autocorrelations of the z's, which under H0 are independent sample values from the N(0,1) distribution.

The derivation of this distributional result is not covered in CS2 and is not trivial though there is literature out there if you want some further reading.

If you're interested, the true variance is actually given by (n-j) / (n*(n+2)), which is approximated in our results here by 1/n, which is reasonable for large n and small j in comparison to n.

This is also where the structure of the Ljung-Box test comes from. Using the more accurate result:

r_j ~ approx N(0, (n-j) / (n*(n+2)))
means:
r_j * sqrt(n*(n+2)) / sqrt(n-j) ~ approx N(0,1)

and:

[r_j * sqrt(n*(n+2)) / sqrt(n-j)]^2 ~ approx chi-squared(1)

ie:

r_j^2 * (n*(n+2)) / (n-j) ~ approx chi-squared(1)

and so n*(n+2) * sum(j = 1 to s) (r_j^2 / (n-j)) ~ approx chi-squared(s)

This result is given in Section 2.2 of Chapter 14 for the Ljung-Box test.

Hope this helps!

Andy
 
Thank you for your reply!

For question 4, I find out that it is a range not a single value. I did find my blind spot from the reply of Andrea Groude.

For question 5,may I ask what is the literature name?I really want to understand it more deeply.I don't care how complex it is and I am very willing to understand it totally.
Although there are many recommended book titles on the IFOA website, I really don't know where is it. When entering CS2, it is not as easy to find the correct information from the Internet as CS1.

I saw other definitions of CLT on the Internet: the distribution of anything tends to N(0,1) under large samples, so I though r~N(0,1) here,too.
I saw the numerator being divided by n-j and the denominator being divided by n, and I mistakenly thought that's why the variance was transformed from the variance formula of joint distribution of CS1 to 1/n.
Thank you for your response that made me realize I still need to find the exact source of information to make sure it's correct.
 
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