[SciPy-User] [SciPy-user] Covariance matrix

josef.pktd@gmai... josef.pktd@gmai...
Mon Feb 13 09:41:10 CST 2012


On Mon, Feb 13, 2012 at 10:40 AM,  <josef.pktd@gmail.com> wrote:
> On Mon, Feb 13, 2012 at 10:30 AM, Johannes Eckstein <eckjoh2@web.de> wrote:
>> Haha, Thanks Pauli
>>
>> maybe good, that I don't have that much memory.
>> I just realized that I was confused by the indices...
>> I had it the way round, before... another question:
>>
>> M = X.H * X
>>
>> with X being a de-trended process does give the (unscaled) covariance
>> matrix?
>
> for linear least squares the unscaled covariance matrix is the inverse
> of dot(X.T, X)

unscaled covariance of the parameter estimate, to be explicit

Josef

>
> for nonlinear least squares X is replaced by the Jacobian.
>
> Josef
>
>>
>> cheers, Johannes
>>> 13.02.2012 09:59, Johannes Eckstein kirjoitti:
>>> [clip]
>>>> how can I do a matrix multiplication of a matrix X with shape: (240001, 4)
>>>>
>>>> M = X * X.H
>>>>
>>>> when I do this I get the following:
>>>>      return N.dot(self, asmatrix(other))
>>>> ValueError: array is too big.
>>>>
>>>> What is the best way to avoid this error?
>>> The result would be a 240001 x 240001 matrix that consumes 430 GB of
>>> memory. Do you really have that much available?
>>>
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