Hi Dakai, can you install the R code somewhere on cedric (or whereever S+ 
is kept). Sanjay, you may be interested in Jaya's comments.
----- Forwarded by Chris Amos/MDACC on 07/28/2004 02:53 PM -----
	
"Satagopan, Jaya M./Epidemiology-Biostatistics" <satagopj@MSKCC.ORG>

07/28/2004 11:15 AM

		 
		
To:
camos@mdanderson.org
cc:
"Satagopan, Jaya M./Epidemiology-Biostatistics" <satagopj@MSKCC.ORG>




Subject:
RE: two stage designs
	


Hi Chris, 
Nice to hear from you. 
 
I am attaching the R code for two-stage design as a text file. This 
pertains to the Genetic Epidemiology
publication of last year. Instructions on how to use this code are given 
in this file. I am assuming 
this is the one you are mentioning in your email. Let me know if this is 
the paper you had in mind. 
Also, please let me know if you have any questions about the code. Thanks.
 
I have not yet applied this method to a data set. Here are my some of my 
thoughts on items 2 and 3. 
Please let me know what you think about this. 
 
Suppose we have accrued n1 cases and n0 controls. We utilize the same 
number of n = n1+n0
subjects in one-stage as well as two-stage designs. The only difference is 
that we genotype fewer loci 
under a two-stage scenario, thus saving on the genotyping cost. Suppose 
the cost of genotyping a single 
locus is the same for a case as well as a control. Say, R1 is the relative 
cost of recruiting a case to the cost 
of genotyping a single polymorphism in a case, and likewise R0 represents 
the relative cost in 
the controls. Then, the cost of the one-stage design would be T1 = n1*R1 + 
n0*R0 + (n1+n0)*m, 
where m is the total number of loci under consideration. The corresponding 
cost of a two-stage 
design would be T2 = n1*R1 + n0*R0 + (n11+n01)*m + (n1-n11 + n0-n01)*m1. 
Here n11 < n1 
and n01<n0 are the case and control subsets utilized in stage 1 for 
genotyping all m loci, and the
promising m1 loci are genotyped in stage 2 on the remaining subjects. 
Thus, when R1 and R0 are 
known, the effective cost equations are: T1* = (n1 + n0)*m and T2* = 
(n11+n01)*m + (n1-n11 + n0-n01)*m1. 
Similar results would then apply. 
 
I believe a related issue would be when a scientist uses arrays for 
genotyping in stage 1, and individually 
genotypes promising loci in stage 2. Say, A1 and A0 are costs of 
ascertaining a single case and a single 
control. Let G be the cost of a single genotype array. Let C1 and C0 be 
the cost of genotyping a single locus 
on a case and a control, respectively. In a one-stage design, we would use 
arrays on all the subjects. Hence, 
the cost is: T1 = n1*A1 + n0*A0 + (n1+n0)*G. In a two-stage design, we use 
arrays for genotyping n11 (< n1) 
cases and n01 (< n0) controls in stage 1. The promising m1 loci are 
genotyped individually in the remaining 
subjects under stage 2. Hence, the cost equation is: 
T2 = n1*A1 + n0*A0 + (n11 + n01)*G + (n1-n11)*m1*C1 + (n0-n01)*m1*C0. When 
C1 = C0 = C, the effective
cost equations are: 
T1* = (n1 + n0) * G
T2* = (n11 + n01) * G + (n1+n0 - n11 - n01) * m1 * C. 
This has a similar form as before. Hence, I believe that similar results 
would apply, although I have not
investigated this in detail. 
 
Best, 
- Jaya
 
 
 

From: camos@mdanderson.org [mailto:camos@mdanderson.org] 
Sent: Monday, July 26, 2004 10:03 PM
To: Satagopan, Jaya M./Epidemiology-Biostatistics
Subject: two stage designs


Hi Jaya, 
  I am finally reading your manuscript on two stage designs for 
association studies. 

This is an extremely helpful paper, but I have a few questions or hoped 
for extensions. 

1.  I would like the code (thanks!).  It will be extremely useful for 
implementing two-stage designs for ongoing collaborations with Genome 
Canada that I have. 

2.  Have you thought about incorporating differing costs for recruiting 
cases versus controls? Sometimes one or the other groups are more 
expensive. 

3.  What if the scientist has far more controls than cases and chooses to 
use n:m matching rather than 1:1 matching to improve power, given limited 
sample sizes (if you assume fixed cost for the n:m type matched unit, then 
I suppose the results are very similar). 

4.  Have you applied the method? 

Sincerely, 


Chris Amos

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