Sample once from the Uniform(0,1) distribution. Call the resulting value . Multiply this result by some constant . Repeat the process, this time sampling from Uniform(0, ). What happens when the multiplier is 2? How big does the multiplier have to be to force divergence. Try it and see:

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.