It does! thank you very much.
Sigalit.


On 10/17/07, Peter McMahan <peter.mcmahan@gmail.com> wrote:
>
> There are two things that will help you out.
>
> First, instead of getting the individual z1 and x1 one at a time and
> then multiplying them you can call rbinom(200,1,p) to get a vector of
> 200 independent draws. So you could replace your for loop with:
>
> z1 <- rbinom(200,1,.6)
> x1 <- rbinom(200,1,.95)
> y1 <- z1*x1
>
> R will multiply vectors element-wise unless you use matrix multiply
> so this will get you the same result much faster and easier.
>
> Second, if you don't care about saving the z1 and x1 vectors, the
> replicate() function ("help(replicate)") can run the "experiment" any
> number of times and return it to you as yet another vector. These two
> lines
>
> yMeans1 <- replicate(200,mean(rbinom(200,1,.6)*rbinom(200,1,.95)))
> yMeans2 <- replicate(200,mean(rbinom(200,1,.6)*rbinom(200,1,.8)))
>
> will give you the end results you're looking for, with yMeans1 being
> a vector of means of 200 independent runs of the full 200 draws. The
> first argument is the number of times to replicate, the second is the
> expression you want to run multiple times.
>
> Hope this helps.
>
> On Oct 17, 2007, at 3:03 AM, sigalit mangut-leiba wrote:
>
> >  hello,
> > I want to simulate 200 times the mean of a joint probability (y1)
> > and 200
> > times the mean of another joint distribution (y2),
> > that is I'm expecting to get 200 means of y1 and 200 means of y2.
> > y1 and y2 are probabilities that I calculate from the marginal
> > prob. (z1 and
> > z2 respectively) multiple by the conditional prob. (x1 and x2
> > respectively),
> > which I generaterd from the binomial distribution.
> > here are the results for one mean of y1 and one mean of y2:
> >
> >> y1 <- 0
> >> for (i in 1:200){
> > + z1[i] <- rbinom(1, 1, .6)
> > + x1[i] <- rbinom(1, 1, .95)
> > + y1[i] <- z1[i]*x1[i]
> > + }
> >> mean(y1)
> > [1] 0.605
> >> y2 <- 0
> >> for (j in 1:200){
> > + z2[j] <- rbinom(1, 1, .6)
> > + x2[j] <- rbinom(1, 1, .8)
> > + y2[j] <- z2[j]*x2[j]
> > + }
> >> mean(y2)
> > [1] 0.515
> >>
> > and this I want to simulate 200 time, in order to see the
> > distribution of y1
> > and the distribution of y2.
> > How can I repeat it and see all the results? (all the 200 means of
> > y1 and
> > y2)
> > Thank you,
> > Sigalit.
> >
> >       [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-
> > guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>

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