x******m
发帖数: 736 | |
l*******l
发帖数: 204 | |
x******m
发帖数: 736 | 3
我的phenotype是多少天大鼠得糖尿病，范围在１－２００天。如果有的大鼠２００天
内没有症状，就认为他们没有糖尿病。(kinda of right-cencored survival).
问题就在这个，现有package都要求phenotype是normal或者binary。即使有的能分析这
种数据，like R/qtl,或者是non-parametric,function也有限。不能作locus
interaction.
不知道有没有这方面的建议,many thanks
【在 l*******l 的大作中提到】
: Where is the question
|
l*******l
发帖数: 204 | 4 simple way is to do one marker at a time using survival analysis.Most software
will do.
More precise and accurate way is to do mixture model. I do not know if there
is software for this purpose. You may have to write your own program. |
x******m
发帖数: 736 | 5 Could you tell me what's the mixture model for my data? Do you mean
seperating the phenotype into two groups (or more), like normal part (random
), binary part (0 or 1)? |
s*r
发帖数: 2757 | 6 probably longitudinal obs of blood sugear
random
【在 x******m 的大作中提到】
: Could you tell me what's the mixture model for my data? Do you mean
: seperating the phenotype into two groups (or more), like normal part (random
: ), binary part (0 or 1)?
|
x******m
发帖数: 736 | 7 I don't think it is longitudinal. A longitudinal study is a correlational
research study that involves repeated observations of the same items over
long periods of time. Here is not repeated obs. of the same items over long
time.
【在 s*r 的大作中提到】
: probably longitudinal obs of blood sugear
:
: random
|
l*******l
发帖数: 204 | 8 f(t|qtl): pdf of event time given QTL, possible exponential or
Weibull.
q(qtl|m1,m2): distribution of QTL given two franking markers, marker 1 and marker
2.
likelihood of individual with event= Sum_{qtl} f(t|qtl)q(qtl|m1,m2). Sum over all possible QTL genotype give genotype of m1 and m2.
likelihood of individual with right censor= Sum_{qtl} S(t|qtl)q(qtl|m1,m2), where S is the survival function of f(t|qtl)
log likelihood = sum (log( likelihood of individual))
However it may not easy to minimize |
