R help - Oct 2016 - simulate AR1 process with uniform distribution and different y0 values

Hello

I need to plot an ar1 graph for process yk=0.75y (k-1) + ek, for y0=1 and
another graph for y0=10.

assume ek is uniformly distributed on interval [-0.5,0.5].

i have the following code but i am not sure how to control y0.

#----------#Start#---------#
rm(list=ls())
library(tseries)
#library(zoo)
set.seed(0)
y<-arima.sim(model=list(ar=.75), n=100, innov = runif(100, 0, 1))
y.1<-y-0.5
ts.plot(y.1)

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On 09/10/16 09:42, Ashwini Patil wrote:
> Hello
>
> I need to plot an ar1 graph for process yk=0.75y (k-1) + ek, for y0=1 and
> another graph for y0=10.
>
> assume ek is uniformly distributed on interval [-0.5,0.5].
>
> i have the following code but i am not sure how to control y0.
>
> #----------#Start#---------#
> rm(list=ls())
> library(tseries)
> #library(zoo)
> set.seed(0)
> y<-arima.sim(model=list(ar=.75), n=100, innov = runif(100, 0, 1))
> y.1<-y-0.5
> ts.plot(y.1)
I don't understand the last three lines; shouldn't "innov" be
equal to
runif(100, -0.5, 0.5), and subtracting 0.5 be skipped?

I think that the following does what you want:

set.seed(42)
y<- arima.sim(model=list(ar=0.75),n.start=1,start.innov=10,n=100,
                  innov=runif(100,-0.5,0.5))

Or you could simply do:

set.seed(42)
e <- runif(100,-0.5,0.5)
yy <- numeric(101)
yy[1] <- 10
for(i in 1:100) yy[i+1] <- 0.75*yy[i] + e[i]
yy <- as.ts(yy[-1])

Same-same.

(Or you could apply filter() to c(10,e) with method="recursive".)

cheers,

Rolf Turner

P.S. I am not convinced that what you want to do makes much sense, but 
the foregoing shows how to do it if you must.

R. T.

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