Apologies for thickness - I'm sure that this operates as documented and with good reason. However... My understanding of arima.sim() is obviously imperfect. In the example below I assume that x1 and x2 are similar white noise processes with a mean of 5 and a standard deviation of 1. I thought x3 should be an AR1 process but still have a mean of 5 and a sd of 1. Why does x3 have a mean of ~7? Obviously I'm missing something fundamental about the burn in or the innovations. x1 <- rnorm(1e3,mean=5,sd=1) summary(x1) x2 <- arima.sim(list(order=c(0,0,0)),n=1e3,mean=5,sd=1) summary(x2) x3 <- arima.sim(list(order=c(1,0,0),ar=0.3),n=1e3,mean=5,sd=1) summary(x3) # why does x3 have a mean of ~7?
arima.sim: innov querry
3 messages · Rolf Turner, Andy Bunn
On 22/11/11 13:04, Andy Bunn wrote:
Apologies for thickness - I'm sure that this operates as documented and with good reason. However... My understanding of arima.sim() is obviously imperfect. In the example below I assume that x1 and x2 are similar white noise processes with a mean of 5 and a standard deviation of 1. I thought x3 should be an AR1 process but still have a mean of 5 and a sd of 1. Why does x3 have a mean of ~7? Obviously I'm missing something fundamental about the burn in or the innovations. x1<- rnorm(1e3,mean=5,sd=1) summary(x1) x2<- arima.sim(list(order=c(0,0,0)),n=1e3,mean=5,sd=1) summary(x2) x3<- arima.sim(list(order=c(1,0,0),ar=0.3),n=1e3,mean=5,sd=1) summary(x3) # why does x3 have a mean of ~7?
X_t = 0.3 * X_{t-1} + E_t
where E_t ~ N(5,1).
So E(X_t) = 0.3*E(X_{t-1}) + E(E_t), i.e
mu = 0.3*mu + 5, whence
mu = 5/0.7 = 7.1429 approx. = 7
So all is in harmony. OMMMMMMMMMMMMMMMMMM! :-)
cheers,
Rolf Turner
P. S. If you want the population mean of x3 to be 5, add 5 *after*
generating
x3 from innovations with mean 0.
R. T.
On 22/11/11 13:04, Andy Bunn wrote:
Apologies for thickness - I'm sure that this operates as documented
and with good reason. However...
My understanding of arima.sim() is obviously imperfect. In the
example below I assume that x1 and x2 are similar white noise processes with a mean of 5 and a standard deviation of 1. I thought x3 should be an AR1 process but still have a mean of 5 and a sd of 1. Why does x3 have a mean of ~7? Obviously I'm missing something fundamental about the burn in or the innovations.
x1<- rnorm(1e3,mean=5,sd=1) summary(x1) x2<- arima.sim(list(order=c(0,0,0)),n=1e3,mean=5,sd=1) summary(x2) x3<- arima.sim(list(order=c(1,0,0),ar=0.3),n=1e3,mean=5,sd=1) summary(x3) # why does x3 have a mean of ~7?
X_t = 0.3 * X_{t-1} + E_t
where E_t ~ N(5,1).
So E(X_t) = 0.3*E(X_{t-1}) + E(E_t), i.e
mu = 0.3*mu + 5, whence
mu = 5/0.7 = 7.1429 approx. = 7
Of course, stupid of me. I should not send r-help requests out at the end of the day. But now I do have a more nuanced question. I'm trying to simulate an ARMA(1,1) process where the underlying distribution is log normal. At the end of the process, can I use arima.sim in conjunction with rlnorm with the parameters below? Thanks in advance for advice. -A mu <- -0.935338 sigma <- 0.4762476 # the dist'n I want but with white noise x1 <- rlnorm(1e5,meanlog=mu,sdlog=sigma) # how can I add these arima coefs to a log-normal distn yet keep parameters mu and sigma? ar1 <- 0.6621 ma1 <- -0.1473 # This is not it: x2 <- arima.sim(list(order=c(1,0,1),ar=ar1,ma=ma1), n = 1e3, rand.gen=rlnorm, meanlog=mu, sdlog=sigma)
So all is in harmony. OMMMMMMMMMMMMMMMMMM! :-)
cheers,
Rolf Turner
P. S. If you want the population mean of x3 to be 5, add 5 *after*
generating
x3 from innovations with mean 0.
R. T.