[BioC] Re: lm.series, design matrix

[James Wettenhall]

Dear Chunrong,

[I've included a few maths formulae in this email, which will 
 only line up correctly in a fixed-width font like Courier.]

The linear model says that for each gene, the Expected Value of 
the "observed" log-ratios vector is equal to the design matrix 
multipled by the vector of log-ratio(s) (parameters) to be 
estimated.

Sometimes estimating the log-ratio(s) of interest is 
simply a matter of averaging the "observed" log-ratios from the 
raw data.  (But it's also nice to estimate confidence statistics 
based on correlation between replicates.)  

For the Swirl Zebrafish dye-swap example (in the Limma manual), 
for each gene, there is only one log-ratio to be estimated 
(Swirl Mutant versus Wild Type) so there is only one column 
in the design matrix.  Estimating the log-ratio for each gene 
can be done by taking a weighted average using -1's for dye 
swaps.

For the ApoAI KO data set, the design matrix would be easier 
to construct if we chose to estimate parameters (log-ratios) for 
(Wild Type versus Reference) and (ApoAI Knockout versus Reference), 
because this is the way the arrays were hybridized.  Then the 
design matrix would look like this: (16 rows in total)

1 0
: :
1 0
0 1
: :
0 1

but because we are really interested in estimating log-ratios 
for (Wild Type vs Reference) and (ApoAI Knockout vs Wild Type), 
we have to post-multiply our design matrix above by the 
matrix 
1 0
1 1

[ PLEASE READ THIS IN COURIER FONT ]

because 

 M (WT vs Ref)        1    0    M (WT vs Ref)
[             ]   = [        ] [             ]
 M (KO vs Ref)        1    1    M (KO vs WT)

where M is a log ratio being estimated.

With the simpler parameterization, we had :

E{M_observed1(WT vs Ref)}     1 0
:                             : :   M (WT vs Ref)
E{M_observed8(WT vs Ref)}  = [1 0] [
E{M_observed1(KO vs Ref)}     0 1   M (KO vs Ref)
:                             : :
E{M_observed8(KO vs Ref)}     0 1

where E{*} is the Expected Value of *

So now we have:

E{M_observed1(WT vs Ref)}     1 0
:                             : :   1 0    M (WT vs Ref)
E{M_observed8(WT vs Ref)}  = [1 0] [   ]  [             ]
E{M_observed1(KO vs Ref)}     0 1   1 1    M (KO vs WT)
:                             : :
E{M_observed8(KO vs Ref)}     0 1


so :

E{M_observed1(WT vs Ref)}     1 0
:                             : :    M (WT vs Ref)
E{M_observed8(WT vs Ref)}  = [1 0]  [             ]
E{M_observed1(KO vs Ref)}     1 1    M (KO vs WT)
:                             : :
E{M_observed8(KO vs Ref)}     1 1


Regards,
James