Re: Is this kind of regression possible?
- From: "Per Sundqvist" <per.sundqvist@xxxxxx>
- Date: Fri, 30 Nov 2007 11:56:04 +0000 (UTC)
"Vivek " <vivek_mutalik@xxxxxxxxx> wrote in message
<fio8nm$6k4$1@xxxxxxxxxxxxxxxxxx>...
Hi,activity.
I m having difficulty in formulating following problem. If
you have any suggestions that'll be great.
Ive set of "aligned DNA sequences" with their activities. I
want to do regression so that i can get weights for each
base (A,C,G,T). This may help me in understanding which
bases are 'important and contribute' towards measured
Example: My activity VS sequence table looks like
(1) 08 ACAG
(2) 10 ATTC
(3) 05 GGTA
(4) 04 CCGT
(5) ... ....
....etc
My solution would be: to minimize the residual sum of square:
(here W is weight of that particular base, which is what im
trying to estimate)
= [8 - (W1A + W2C + W3A + W4G)]^2 + [10 - (W1A + W2T + W3T
+W4C)]^2 + and so on.
to reduce the parameters to be determined, I can substract
weight of 'T' from each of weights and finally add sum of
all 'T' (as if 'T' is in all positions).
That is:
= [8 - (W1A-W1T + W2C-W2T + W3A-W3T + W4G-W4T) +
(W1T+W2T+W3T+W4T)]^2
+
[10 - (W1A-W1T + W4C -W1T) + (W1T+W2T+W3T+W4T) ]^2
+ so on;
Is it making any sense? So by this way, i was thinking of
getting weights for all bases by using some kind of residual
minimizing function. Is it possible ?
F=[8 - (W1A + W2C + W3A + W4G)]^2 + [10 - (W1A + W2T + W3T
+W4C)]^2 + and so on.
dF/dW1A=0, etc...
8+10=(1+1)*W1A+1*W2C+1*W3A+...
b=A*w
Hmm, this minimization looks like it is equal to solving a
linear equation Aw=b. You have 16 unknown right? W1A W2A W3A
W4A,W1C W2C,... =w, the unknown vector. So you need 16
equation at least to get these 16 weights. If you have more
equations you take the backslash-least square solution. You
will get a matrix A, which you have to work out by some
clever way, depending on how your data is arranged. (1+1)
shoul be replaced by the number of sequences that have A at
the first position, and in the element of b you sum the
values of these 8+10+....
Maby it hels you a little,
Per
.
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