Date: Thu, 19 Apr 2007 12:22:57 -0400
Reply-To: "data _null_;" <datanull@GMAIL.COM>
Sender: "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From: "data _null_;" <datanull@GMAIL.COM>
Subject: Re: calculating rolling standard deviation
In-Reply-To: <1176994361.512818.292920@n76g2000hsh.googlegroups.com>
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A visit to support.sas.com to view examples of the LAG function yields
the following. I'm not sure if it is a rolling STD but it is the STD
of the current and previous 2 observations.
proc plan;
factors
subject=20 ordered
size=10 ordered
y=1 of 100
;
output out=work.rolling(drop=size)
run;
data rollingSTD;
cnt = 0;
do until(last.subject);
set rolling;
by subject;
lagY1=lag1(y);
lagY2=lag2(y);
lagY3=lag3(y);
array _lags[*] lagY:;
cnt + 1;
do i=cnt to dim(_lags);
_lags[i]=.;
end;
if n(of _lags[*]) eq dim(_lags) then std = std(of _lagS[*]);
output;
end;
*drop lagY: cnt i;
run;
proc print;
run;
On 4/19/07, PBilin <pbilin@gmail.com> wrote:
> One more question. I am trying to adjust the proposed code to
> calculate rolling standard deviation for the last 3 observations
> only.
>
> Here is the adjusted code:
>
>
> data Result(drop=SumOfX SumOfX2 xCount);
> retain SumOfX SumOfX2 xCount sumofx_12 sumofx2_12 xcount_12
> xcount_new;
> set my_data;
> by observation_no;
> if first.observation_no then do;
> xCount=1;
> SumOfX=p;
> SumOfX2=p**2;
> end;
> else do;
> xCount=xCount+1;
> SumOfX=SumOfX+p;
> SumOfX2=SumOfX2+p**2;
> end;
> q=sumofx;
> w=sumofx2;
> e=xcount;
> sumofx_12=lag3q);
> sumofx2_12=lag3(w);
> sums=q-sumofx_12;
> sums2=w-sumofx2_12;
> xcount_12=lag3(e);
> xcount_new=xcount-xcount_12;
> PopulationStdDev=((SumOfX2-SumOfX**2/xCount)/xCount)**0.5;
>
> PopulationStdDev_2=((Sums2-Sums**2/xcount_new)/xcount_new)**0.5;
>
> run;
>
> The idea is to take a difference between sumofx for current
> observation and for observation 3 periods before. Unfortunatelly, the
> code is not working properly.
>
> Would appreciate your comments!
>
> Paul
>
> On Apr 19, 10:50 am, PBilin <pbi...@gmail.com> wrote:
> > Thank you all for your help! The solutions work very neatly!
> >
> > Paul
> > On Apr 18, 10:17 pm, Huang...@PRINCIPAL.COM ("Huang, JS") wrote:
> >
> > > Dan:
> >
> > > It is a neat way to calculate sum of square of deviation from the mean
> > > iteratively. For those who are interested in the derivation here is the
> > > proof.
> >
> > > In the following mu_n denote mean of the first n terms and mu_(n-1)
> > > mean of the first n-1 terms.
> >
> > > Sum{(X_i - mu_n)**2, i=1 to n} - Sum{(X_i - mu_(n-1))**2, i=1 to n-1}
> > > = Sum{(X_i - mu_n)**2, i=1 to n-1} + (X_n - mu_n)**2 - Sum{(X_i -
> > > mu_(n-1))**2, i=1 to n-1}
> > > = Sum{((X_i - mu_(n-1)) + (mu_(n-1) - mu_n))**2, i=1 to n-1} + (X_n -
> > > mu_n)**2 - Sum{(X_i - mu_(n-1))**2, i=1 to n-1}
> > > = Sum{(X_i - mu_(n-1))**2 + 2*(X_i - mu_(n-1))*(mu_(n-1) -
> > > mu_n)+(mu_(n-1) - mu_n)**2, i=1 to n-1} + (X_n - mu_n)**2 - Sum{(X_i -
> > > mu_(n-1))**2, i=1 to n-1}
> > > = Sum{(X_i - mu_(n-1))**2, i=1 to n-1} + 2*(mu_(n-1) - mu_n)*Sum{(X_i -
> > > mu_(n-1)), i=1 to n-1} + (n-1)*(mu_(n-1) - mu_n)**2 + (X_n - mu_n)**2 -
> > > Sum{(X_i - mu_(n-1))**2, i=1 to n-1}
> >
> > > The first and last terms cancel each other and Sum{(X_i - mu_(n-1)),
> > > i=1 to n-1} = 0 and hence the above can be simplified to:
> > > (n-1)*(mu_(n-1) - mu_n)**2 + (X_n - mu_n)**2
> > > = (n-1)*(mu_(n-1) - mu_n)*(mu_(n-1) - mu_n) + (X_n - mu_n)**2
> > > = -(X_n - mu_n)*(mu_(n-1) - mu_n) + (X_n - mu_n)**2
> > > (See (***) below for derivation.)
> > > = (X_n - mu_n)*(-(mu_(n-1) - mu_n) + (X_n - mu_n))
> > > = (X_n - mu_n)*(X_n - mu_(n-1))
> > > = (X_n - mu_(n-1))*(X_n - mu_n)
> >
> > > We obtained
> > > Sum{(X_i - mu_n)**2, i=1 to n} - Sum{(X_i - mu_(n-1))**2, i=1 to
> > > n-1} = (X_n - mu_(n-1))*(X_n - mu_n)
> > > Or
> > > Sum{(X_i - mu_n)**2, i=1 to n} = Sum{(X_i - mu_(n-1))**2, i=1 to
> > > n-1} + (X_n - mu_(n-1))*(X_n - mu_n)
> >
> > > This is what is used in Dan's statement
> >
> > > v + (value_x - prior_mean) * (value_x - mean);
> >
> > > QED.
> >
> > > (***)
> > > (n-1)*(mu_n-1) - mu_n)
> > > = (n-1)*mu_(n-1) - n*mu_n + mu_n
> > > = Sum{X_i, i=1, n-1} - Sum{X_i, i=1, n} + mu_n
> > > = -X_n + mu_n
> > > = -(X_n - mu_n)
> >
> > > Replacing the first
> >
> > > -----Original Message-----
> > > From: SAS(r) Discussion [mailto:S...@LISTSERV.UGA.EDU] On Behalf Of
> >
> > > Nordlund, Dan (DSHS/RDA)
> > > Sent: Wednesday, April 18, 2007 3:12 PM
> > > To: S...@LISTSERV.UGA.EDU
> > > Subject: Re: calculating rolling standard deviation
> >
> > > > -----Original Message-----
> > > > From: SAS(r) Discussion [mailto:S...@LISTSERV.UGA.EDU] On Behalf Of
> > > > PBilin
> > > > Sent: Wednesday, April 18, 2007 10:45 AM
> > > > To: S...@LISTSERV.UGA.EDU
> > > > Subject: calculating rolling standard deviation
> >
> > > > hello,
> >
> > > > I am trying to calculate a rolling standard deviation for a variable
> > > > value_x:
> >
> > > > date obs_number value_x stdandard_deviation
> > > > 01012000 1 1 std(1)
> > > > 01022000 1 1.2 std(1,1.2)
> > > > 01032000 1 12 std(1,1.2,12)
> > > > 01042000 1 5 std(1,1.2,12,5)
> > > > 01052000 1 6 ....
> > > > 01012000 2 42
> > > > 01022000 2 1.422
> > > > 01032000 2 15232
> > > > 01042000 2 65
> > > > 01052000 2 63
> >
> > > > As you see, each next value of standard_deviation uses one extra
> > > > observation of value_x.
> >
> > > > Would appreciate your help!
> >
> > > Paul,
> >
> > > I see you have at least one solution, let me add another (which I
> > > learned some time ago from the BMDP software documentation). Depending
> > > on the number of observations and the nature of the data this method
> > > could be more accurate than using the common computational formula,
> > > sum(x**2)-sum(x)**2/N, for the sums of squares. You can drop whatever
> > > variables you don't want to keep. It would need to be modified slightly
> > > if you have any missing data, or if you wanted to use weighted data.
> >
> > > data want;
> > > set your_data;
> > > retain n mean v 0;
> > > n + 1;
> > > prior_mean = mean;
> > > mean + (value_x - mean)/n;
> > > v + (value_x - prior_mean) * (value_x - mean);
> > > standard_deviation = (v/(n-1))**.5 ;
> > > run;
> >
> > > Hope this is helpful,
> >
> > > Dan
> >
> > > Daniel J. Nordlund
> > > Research and Data Analysis
> > > Washington State Department of Social and Health Services Olympia, WA
> > > 98504-5204
> >
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