[R-SIG-Finance] [R-sig-finance] determine non-linear correlation

[manulemalin]

things are not that easy
in your example, your data are non stationnary

( it should be clearer if you have a look at the unit roots of your model(s)
)

as already said, having a look at cointegration can be a good start

moreover, ( as already said!), there is not ONE function to determine
nonlinearity; it's up to you to determine, and test, the kind of your
nonlinearities
if you have got no idea / your nonlinearities, you can use nonparametric
modelling, which can work, for instance with neural networks, as a black box
( which can also be misleading...)

hope it helps

regards

manu





Mark Breman-2 wrote:
> 
> Here is the thesis I was talking about:
> www.jos.org.cn/ch/reader/download_pdf.aspx?file_no=20001203&year_id=2000&quarter_id=12&falg=1
> 
> I did some experiments based on the replies to my post which pointed me in
> the right direction (thank you for that):
> 
> let's say I have a timeseries a:
> 
> 
>> a
>            priceA
> 2009-06-01      1
> 2009-06-02      2
> 2009-06-03      3
> 2009-06-04      4
> 
> And a timesieries b:
> 
>> b
>            priceB
> 2009-06-01     11
> 2009-06-02     12
> 2009-06-03     13
> 2009-06-04     14
> 
> priceB is derived from priceA by the simple linear function priceB = 10 +
> priceA.
> As expected all correlation methods agree that timeseries B is highly
> correlated to A:
> 
>> cor(a,b, method="pearson")
>        priceB
> priceA      1
>> cor(a,b, method="kendall")
>        priceB
> priceA      1
>> cor(a,b, method="spearman")
>        priceB
> priceA      1
> 
> Now suppose I have another timeseries F:
> 
>> f
>            priceF
> 2009-06-01      9
> 2009-06-02     24
> 2009-06-03     45
> 2009-06-04     72
> 
> priceF is derived from priceA by the nonlinear function: 3*(priceA^2) +
> 6*priceA.
> If I'm correct than this is known a "second order kwadratic function"
> which is not a linear function. (If you would draw it in a graph it would
> be shown as a parabolic line)
> 
> Now let's look at the different correlation methods for timeseries A and
> F:
> 
>> cor(a,f, method="pearson")
>           priceF
> priceA 0.9919354
>> cor(a,f, method="kendall")
>        priceF
> priceA      1
>> cor(a,f, method="spearman")
>        priceF
> priceA      1
> 
> Note that Pearson obviously does not know about the nonlinear function
> between priceA and priceF as it does not reports a correlation of 1.
> Kendall and Spearman on the other hand seem to know about the nonlinear
> function as they report a correlation of 1 between A and F.
> So my conclusion (correct me if I am wrong): kendall and spearman can
> handle nonlinear functions as the basis for the corellation, pearson can
> not. (one of you already pointed this out in a reply I think)
> 
> I'm still left with one questions thought:
> Is it possible (with R) to obtain the function that forms the basis for
> the correlation as reported by kendall and spearman? I mean: if I have two
> real timeseries which are highly correlated according to kendall and
> spearman (i.e. cor(x,y) = 1) I would like to know the (nonlinear) function
> that form the basis for the correlation.
> 
> Regards,
> 
> -Mark- 
> 
> 
> 
> 
>  
> 
> ________________________________
> 
> 
> Sent: Friday, June 5, 2009 1:18:55 PM
> Subject: Re: [R-SIG-Finance] determine non-linear correlation
> 
> 
> correlation by definition is a linear measure of assocation. So what is
> the presice definition of non-linear correlation?
> can you give the thesis......I am curious to know....
>  Kaushik Bhattacharjee
> 
> 
> 
> 
> ________________________________
> 
> To: r-sig-finance at stat.math.ethz.ch
> Sent: Thursday, June 4, 2009 12:45:17 AM
> Subject: [R-SIG-Finance] determine non-linear correlation
> 
> I would like to know if two financial time-series are nonlinear
> correlated, and if so, what that correlation function is.
> Is there an easy way to do this with R?
> 
> I have read a thesis about the "high order correlation coefficient to
> solve the nonlinear correlation problem" but I'm not able to translate
> this into a solution for my problem. All these statistics are very
> interesting but also challenging for me...  
> 
> Kind regards,
> 
> -Mark-
> 
> 
>       
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