Date: Thu, 19 Apr 2007 07:52:41 -0700
Reply-To: PBilin <pbilin@GMAIL.COM>
Sender: "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From: PBilin <pbilin@GMAIL.COM>
Organization: http://groups.google.com
Subject: Re: calculating rolling standard deviation
In-Reply-To: <1176976202.663360.79980@q75g2000hsh.googlegroups.com>
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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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