[SciPy-User] calculate predicted values from regression + confidence intervall
josef.pktd@gmai...
josef.pktd@gmai...
Mon Oct 17 06:54:32 CDT 2011
On Mon, Oct 17, 2011 at 5:59 AM, Johannes Radinger <JRadinger@gmx.at> wrote:
> Hello Josef
> Hello others,
>
> I am not sure if that is what I really want.
> I just want to calculate a new predicted value
> using my regression equation and a variance
> (prediction intervall is statistically correct
> expression) for the new predicted value.
>
> I calculated my regression already in R and want to
> use the results in python manually (without a
> python-R interface).
>
> 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
>
> but how can I get the prediction interval?
> I am not sure if your approach is the one I need
> for that (with my given input) and if yes
> how to use it?
>
> Thank you in advance
>
> /Johannes
>
>
>
>>
>> Message: 2
>> Date: Fri, 14 Oct 2011 18:54:23 -0400
>> From: josef.pktd@gmail.com
>> Subject: Re: [SciPy-User] calculate predicted values from regression +
>> confidence intervall
>> To: SciPy Users List <scipy-user@scipy.org>
>> Message-ID:
>> <CAMMTP+ArodcLXCYtw20Gxye8zqDb6G-0cAj6v-QWAV5aDA3=5w@mail.gmail.com>
>> Content-Type: text/plain; charset=ISO-8859-1
>>
>> On Fri, Oct 14, 2011 at 7:51 AM, Johannes Radinger <JRadinger@gmx.at>
>> wrote:
>> > Hello,
>> >
>> > I did a statistical regression analysis (linear regression) in R which
>> > has following parameters:
>> >
>> > Y = X1 + X2
>> >
>> > from the R-analysis I can get the intercept and slopes for the
>> independent variables so that I get:
>> >
>> > Y = Intercept + slope1*X1 + slope2*X2
>> >
>> > Now I want to use that in Scipy to calculate new predicted Y values.
>> > Generally that is not a problem. I use the new X1 and X2 as input and
>> > the slopes and intercept are predefined. So I can easily calculate Y
>> new.
>> >
>> > Now my question arises:
>> > How can I calculate a kind of uncertainty (e.g. a confidence intervall)
>> for
>> > the new Y? What do I have to extract from R and how do I have to use
>> > the extracted parameters in scipy to calculate such things?
>> > Is that generally possible?
>>
>> Better sandbox then nothing:
>>
>> https://github.com/statsmodels/statsmodels/blob/master/scikits/statsmodels/sandbox/regression/predstd.py#L28
>>
>> This is my version for scikits.statsmodels.OLS (and maybe WLS)
>>
>> usage example
>> https://github.com/statsmodels/statsmodels/blob/master/scikits/statsmodels/examples/tut_ols.py
>>
>> Josef
>>
>> >
>> > Thank you very much
>> > Johannes
>> > --
>> > NEU: FreePhone - 0ct/min Handyspartarif mit Geld-zur?ck-Garantie!
>> > Jetzt informieren: http://www.gmx.net/de/go/freephone
>> > _______________________________________________
>> > SciPy-User mailing list
>> > SciPy-User@scipy.org
>> > http://mail.scipy.org/mailman/listinfo/scipy-user
>> >
>>
>>
>> ------------------------------
>>
>> _______________________________________________
>> SciPy-User mailing list
>> SciPy-User@scipy.org
>> http://mail.scipy.org/mailman/listinfo/scipy-user
>>
>>
>> End of SciPy-User Digest, Vol 98, Issue 22
>> ******************************************
>
> --
> NEU: FreePhone - 0ct/min Handyspartarif mit Geld-zurück-Garantie!
> Jetzt informieren: http://www.gmx.net/de/go/freephone
> _______________________________________________
> SciPy-User mailing list
> SciPy-User@scipy.org
> http://mail.scipy.org/mailman/listinfo/scipy-user
>
More information about the SciPy-User
mailing list