[R-SIG-Finance] [R-sig-finance] A VaR question

[RON70]

Thanks Matifou for your reply. Yes I want to simulate VAR with those
parameters. However here my problem is those parameters are estimated
parameters not actual one. Therefore I dont think I can apply TVAR.simin()
directly on those parameters. You see, ch equation for those estimated
parameters doesnt give one unit root, which is the fundamental property for
a co-integrated VAR DGP. 

Therefore I want to do some some tweaks so that I can get a revised
parameter set from those estimated parameters such that ch. equation (based
on revised parameters) gives exactly one unit root. Then I can use
TVAR.simin() to generate data.

Therefore my question is : How can I tweak those estimated parameters?

Best,



matifou wrote:
> 
> Hi RON70
> 
> Not sure about what you want to do... So you want to simulate a VAR with 
> those parameters? There is the function TVAR.simin package tsDyn that 
> allow you to simulate a VAR, see:
> 
> co<-matrix(c(0.985,  0.283, -1.714,  0.125,  1.100, -1.491,0.071, 
> -0.089,  1.388), ncol=3, byrow=TRUE)
> library(tsDyn)
> 
> TVAR.sim(B=cbind(co,co),nthresh=0, inc="none", lag=2)
> 
> Hope this is what you wanted?
> 
> MAT2009
> 
> RON70 a écrit :
>> I think I should be more clear on what I would like to do. From that
>> estimated model I would like to get a "Real" VAR DGP, perhaps by tweaking
>> some of the coeficients. Which I would use to simulate artrificial data,
>> for
>> some study. If somebody shows me some lihgt on how I can achive that, I
>> would be truly grateful.
>>
>> Best,
>>
>>
>>
>> RON70 wrote:
>>   
>>> Hi all,
>>>
>>> My problem seems to be bizzare, however I want to do like that. Here I
>>> have estimated a VECM model from my dataset (seems not stationary) and
>>> once I converted those into a VAR representation I have following
>>> estimates :
>>>
>>>     
>>>> A1; A2; A3; A4
>>>>       
>>>      V1     V2     V3
>>> 1 0.985  0.283 -1.714
>>> 2 0.125  1.100 -1.491
>>> 3 0.071 -0.089  1.388
>>>       V1     V2     V3
>>> 1  0.258 -0.493  1.459
>>> 2  0.252 -0.387  1.165
>>> 3 -0.057  0.076 -0.536
>>>      V1     V2    V3
>>> 1 0.332 -0.459 0.251
>>> 2 0.482 -0.686 0.313
>>> 3 0.112 -0.104 0.218
>>>       V1    V2     V3
>>> 1 -0.532 0.624 -0.006
>>> 2 -0.619 0.714 -0.044
>>> 3 -0.129 0.121 -0.069
>>>
>>> Now I took them as an original DGP process and checked the solution of
>>> it's ch. equation. I got following :
>>>
>>>     
>>>>  library(PolynomF)
>>>>  z = polynom()
>>>>
>>>>
>>>> p11 <- 1 - A1[1,1]*z - A2[1,1]*z^2 - A3[1,1]*z^3 - A4[1,1]*z^4
>>>> p12 <- 0 - A1[1,2]*z - A2[1,2]*z^2 - A3[1,2]*z^3 - A4[1,2]*z^4
>>>> p13 <- 0 - A1[1,3]*z - A2[1,3]*z^2 - A3[1,3]*z^3 - A4[1,3]*z^4
>>>>
>>>>
>>>> p21 <- 0 - A1[2,1]*z - A2[2,1]*z^2 - A3[2,1]*z^3 - A4[2,1]*z^4
>>>> p22 <- 1 - A1[2,2]*z - A2[2,2]*z^2 - A3[2,2]*z^3 - A4[2,2]*z^4
>>>> p23 <- 0 - A1[2,3]*z - A2[2,3]*z^2 - A3[2,3]*z^3 - A4[2,3]*z^4
>>>>
>>>> p31 <- 0 - A1[3,1]*z - A2[3,1]*z^2 - A3[3,1]*z^3 - A4[3,1]*z^4
>>>> p32 <- 0 - A1[3,2]*z - A2[3,2]*z^2 - A3[3,2]*z^3 - A4[3,2]*z^4
>>>> p33 <- 1 - A1[3,3]*z - A2[3,3]*z^2 - A3[3,3]*z^3 - A4[3,3]*z^4
>>>>
>>>> p <- p11*(p22*p33 - p23*p32) - p12*(p21*p33 - p23*p31) + p13*(p21*p32 -
>>>> p22*p31)
>>>>
>>>>
>>>> abs(solve(p))
>>>>       
>>>  [1] 1.521516 2.102119 2.102119 4.912478 4.912478 1.000233 1.000233
>>> 1.502034 1.502034 1.228100 2.536582 5.342635
>>>     
>>> Now if I assume (upto a few significant digits) "1.000233 1.000233 "
>>> both
>>> equal to "1" then, I am actually getting two unit roots here. Therefore
>>> I
>>> am wondering how to tackle it as VAR is defined on max one unit root
>>> process. 
>>>
>>> Am I missing anything? Can anyone please help me?
>>>
>>> Best
>>>
>>>     
>>
>>
> 
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