Sample once from the Uniform(0,1) distribution. Call the resulting value 
iters = 200 locations = rep(0,iters) top = 1 multiplier = 2 for(i in 1:iters) { locations[i] = runif(1,0,top) top = locations[i] * multiplier } windows() plot(locations[1:i],1:i,pch=20,col="blue",xlim=c(0,max(locations)),ylim=c(0,iters),xlab="Location",ylab="Iteration") # Optional save as movie, not a good idea for more than a few hundred iterations. I warned you! # library("animation") # saveMovie(for (i in 1:iters) plot(locations[1:i],1:i,pch=20,col="blue",xlim=c(0,max(locations)),ylim=c(0,iters),xlab="Location",ylab="Iteration"),loop=1,interval=.1)
Deriving the distribution of the r.v.s gets messy after t=2 or so, but you can derive the first two moments of this process by inducting over conditional expectations:
E[x(t)]=(1/2)*(c/2)^(t+1)
E[x(t)^2]=(c^(2(t-1)))/(3^t)
so the expectation ->0 if |c|infty if |c|>2, and ->1/2 if c=2.
the second moment ->0 if |c|infty if |c|>3^.5 and ->1/c^2 if |c|=3^.5
Apologies if I made a typo but I think the idea is more or less correct.
So just a comment on your simulation– the expectation should converge to 1/2, but as you can tell, there is some volatility that will become infinitely bad as t->infty.
I think it is possible to derive the mgf of the process in terms of the mgf of the Unif(0,1) which is just 1/(n+1).
I think my questions was too vague.
Take a look at the final value of the sequence for high values of “iters”. See how often you “end up” at a high number, let alone marginally above 0. There is a particular multiplier that seems to guarantee that you will end up with a big number instead of tending to 0.
To make the testing easier I bundled things up in a function that works like a distribution you can sample from:
http://www.statisticsblog.com/code/multiUni.r
Try running this:
z = multiUni(1000,1000,2.1)
mean(z)
That’s an estimator of the expectation after 1000 trials with 1000 iterations each, using multiplier 2.1.
While conditional expections is the way to go for deriving a closed expression for expected value, I’m not sure that Eric’s answer is the correct one. The limits of integration in the expectation is a little strange, and I think you end up with an odd product series for the expectation, but I need to check my algebra.