Re: QR vs LU factorisation

[Fredrik Lingvall]

Vic Norton wrote:
> I just used the "inverse" problem as an svd example, Fredrik. I don't 
> generally compute inverses.
>
> Here is the kind of "noisy" data problem I am thinking of. Suppose you 
> are interested in 50 equity "funds" and you have 50 vectors of 39 
> successive 13-week returns. Then you can compute 50 expected returns 
> (assuming some kind of weighting, perhaps uniform, of the individual 
> returns) and 50 vectors of deviations from these expected returns. 
> Think of the deviation vectors as the risks of the individual funds. 
> Your problem: express expected return as a linear function of risk.
>
> For this problem you want to solve
>
>    e' * R = E
>
> for the 39 x 1 expected return vector e, where R is the 39 x 50 risk 
> matrix and E is the 1 x 50 matrix of expected returns. None of your 
> data, being ultimately based on dollars and cents prices per share, 
> has even 6 digit accuracy. This is a problem where the svd approach 
> with svdcut = 1e-6 would seem to be a natural choice. But you 
> definitely would NOT compute the "pseudoinverse" of R to solve the 
> problem.
>
> Regards,
>
> Vic
Hi Vic,

If I understand you correctly then E is your (noisy) data, R is known,  
and e is the parameters you want to find. In that case I would set up 
the problem something like (I use the convention that all vectors are 
column vectors):

E = R'*e + n

where n is a "noise" vector. If you have an idea of the precision 
(variance) of your data, E, then you can assign a  zero mean Gaussian 
distribution for n. You (normally) also have an idea (from previous 
measurements for example) how much the parameters e may vary. Let us for 
simplicity say that you know the variance also here  so we assign a 
Gaussian distribution for e as well (i.e., p(e|I) = N(0,Ce) where Ce is 
the covariance matrix for e). The likelihood of your data will then become,

p(E|e,I) = N(0,Ce) = 1/(2*pi)^N/2 * 1/sqrt(det(Cn)) * exp( -1/2*(E- 
R'*e)'*inv(Cn)*(E-R'*e) )

where Cn is the covariance matrix for n [Cn = sigma_n^2*eye(50,50) if 
the errors are the same for all elements in E].
Bayes rule then gives the pdf for e given the data,

p(e|E,I) = p(e|I) * p(E|e,I) / const.        (1)

You can now find the most likely e by taking the log of (1) and set the 
derivative w.r.t e equal to zero which results in

e_hat = inv( R*inv(Cn)*R' + inv(Ce) )*R*inv(Ce) * E
         = Ce*R*inv(R'*Ce*R+Cn) * E             (2)

If Ce= sigma_e^2*eye(39,39) and Cn = sigma_n^2*eye(50,50) then (2) becomes

e_hat = R*inv(R'*R+sigma_n^2/sigma_e^2*eye(50,50) ) * E

Note that the matrix R'*R+sigma_n^2/sigma_e^2*eye(50,50) always is pos 
def so the inverse is safe.

Hope this helps.

/Fredrik









--
Fredrik Lingvall, PhD                                                              
E-mail: address@hidden, address@hidden
Web:    http://folk.uio.no/fl/