[SciPy-User] calculate predicted values from regression + confidence intervall
josef.pktd@gmai...
josef.pktd@gmai...
Mon Oct 17 14:39:33 CDT 2011
On Mon, Oct 17, 2011 at 3:29 PM, Johannes Radinger <jradinger@gmx.at> wrote:
> Hello again,
>
> it is not a problem to get the covariance matrix of the parameter
> estimates (with the vcov() function of R) and of course I can use
> that also for the calculations...
>
> The covariance matrix is:
> (Intercept) L uT stream.order
> (Intercept) 1.33059472 -0.246834193 -0.0307302435 0.0267554775
> L -0.24683419 0.054481286 0.0014007745 -0.0105982957
> uT -0.03073024 0.001400774 0.0053472652 -0.0007137384
> stream.order 0.02675548 -0.010598296 -0.0007137384 0.0091419997
>
> I probably just have to save that in the form of a python matrix (Just have
> to look up how to do that).
>
> But how can I know proceed?
>
> I found following description of prediction intervals (Thats the thing
> I want :) here: http://statmaster.sdu.dk/courses/st111/module05/module.pdf (page 11).
> But can that be realised in python using the covariance matrix resp. the
> other estimates and std. errors of my regression parameters?
page 12: replace sigma^2 * (X'X)^{-1} by cov_beta, the first
expression is cov_beta for OLS
Josef
>
> /Johannes
>
>
> Am 17.10.2011 um 19:00 schrieb scipy-user-request@scipy.org:
>
>>> The R Coefficients are as follows:
>>> ? ? ? ? ? ? Estimate Std. Error t value Pr(>|t|)
>>> (Intercept) ?-9.00068 ? ?1.15351 ?-7.803 8.26e-12 ***
>>> Variable X1 ?1.87119 ? ?0.23341 ? 8.017 ?2.95e-12 ***
>>> Variable X2 ?0.39193 ? ?0.07312 ? 5.360 ?5.92e-07 ***
>>> Variable X3 ?0.27870 ? ?0.09561 ? 2.915 ?0.00445 **
>>>
>>> Can I use these results to manually calculate
>>> a predicted value of Y with a give set of new Xs? like
>>> X1 = 200
>>> X2 = 150
>>> X3 = 5
>>>
>>> I can easily calculate the predicted Y as
>>> Y = -9 + 200*1.87 + 150*0.39 + 5*0.28
>>
>> I don't think this is enough information to get the prediction
>> confidence interval. You need the entire covariance matrix of the
>> parameter estimates.
>>
>> Roughly (I would need to check the details):
>> the parameter estimate is from a multivariate normal distribution,
>> your y is a linear transformation, so the prediction should be normal
>> distributed with mean y = Y = X*beta, and var(y) = X' * cov_beta * X +
>> var_u_estimate (dot products for appropriate shapes)
>>
>> Without knowing the covariance matrix of the parameter estimates, you
>> would have to assume that cov_beta is diagonal which is almost surely
>> not the case.
>>
>> Josef
>
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