Thank you All for the response,
acf do not accept 2 variables so naturally
These may not work for me.

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Sudheer Joseph
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Ministry of Earth Sciences, Govt. of India
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From: "josef.pktd@gmail.com" <josef.pktd@gmail.com>
To: Discussion of Numerical Python <numpy-discussion@scipy.org>
Sent: Tuesday, 19 March 2013 1:51 AM
Subject: Re: [Numpy-discussion] Numpy correlate

On Mon, Mar 18, 2013 at 1:10 PM, Skipper Seabold <jsseabold@gmail.com> wrote:
> On Mon, Mar 18, 2013 at 1:00 PM, Pierre Haessig <pierre.haessig@crans.org>
> wrote:
>>
>> Hi Sudheer,
>>
>> Le 14/03/2013 10:18, Sudheer Joseph a écrit :
>>
>> Dear Numpy/Scipy experts,
>>                                              Attached is a script which I
>> made to test the numpy.correlate ( which is called py plt.xcorr) to see how
>> the cross correlation is calculated. From this it appears the if i call
>> plt.xcorr(x,y)
>> Y is slided back in time compared to x. ie if y is a process that causes a
>> delayed response in x after 5 timesteps then there should be a high
>> correlation at Lag 5. However in attached plot the response is seen in only
>> -ve side of the lags.
>> Can any one advice me on how to see which way exactly the 2 series are
>> slided back or forth.? and understand the cause result relation better?( I
>> understand merely by correlation one cannot assume cause and result
>> relation, but it is important to know which series is older in time at a
>> given lag.
>>
>> You indeed pointed out a lack of documentation of in matplotlib.xcorr
>> function because the definition of covariance can be ambiguous.
>>
>> The way I would try to get an interpretation of xcorr function (& its
>> friends) is to go back to the theoretical definition of cross-correlation,
>> which is a normalized version of the covariance.
>>
>> In your example you've created a time series X(k) and a lagged one : Y(k)
>> = X(k-5)
>>
>> Now, the covariance function of X and Y is commonly defined as :
>>  Cov_{X,Y}(h) = E(X(k+h) * Y(k))  where E is the expectation
>>  (assuming that X and Y are centered for the sake of clarity).
>>
>> If I plug in the definition of Y, I get Cov(h) = E(X(k+h) * X(k-5)). This
>> yields naturally the fact that the covariance is indeed maximal at h=-5 and
>> not h=+5.
>>
>> Note that this reasoning does yield the opposite result with a different
>> definition of the covariance, ie. Cov_{X,Y}(h) = E(X(k) * Y(k+h))  (and
>> that's what I first did !).
>>
>>
>> Therefore, I think there should be a definition in of cross correlation in
>> matplotlib xcorr docstring. In R's acf doc, there is this mention : "The lag
>> k value returned by ccf(x, y) estimates the correlation between x[t+k] and
>> y[t]. "
>> (see http://stat.ethz.ch/R-manual/R-devel/library/stats/html/acf.html)
>>
>> Now I believe, this upper discussion really belongs to matplotlib ML. I'll
>> put an issue on github (I just spotted a mistake the definition of
>> normalization anyway)
>
>
>
> You might be interested in the statsmodels implementation which should be
> similar to the R functionality.
>
> http://nbviewer.ipython.org/urls/raw.github.com/jseabold/tutorial/master/tsa_arma.ipynb
> http://statsmodels.sourceforge.net/devel/generated/statsmodels.tsa.stattools.acf.html
> http://statsmodels.sourceforge.net/devel/generated/statsmodels.graphics.tsaplots.plot_acf.html

we don't have any cross-correlation xcorr, AFAIR
but I guess it should work the same way.

Josef

>
> Skipper
>
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