[sixhigh]

Whops - that's probably wrong. The entire post.

E[x_n] = 1 for all n,

but still x_n --> 0. (p-almost sure)

Say you walk n steps and k times the coin lands on heads. The value of x_n then will be 1 * 1.1^k * 0.9^(n-k).
For this value to be greater than (say) x_0 = 1 we get the following expression:

1.1^k*0.9^(n-k) > 1

Solving this expression for k (and using Derive) we get

k > n * ln(1/0.9) / ln(1.1/0.9)

or

k > n * 0.525

Now since k has binomial distribution (with the parameter p=1/2) we know by the law of large numbers that it will (p-almost surely) not be in this region for large n.

The EV of x_n being 1 although x_n almost surely approaches 0 follows for nearly the same reasons the St.-Petersburg paradox does.

Now since the random variable approaches zero almost surely it will almost surely never hit a certain value above 0 infinitely often. Thus (i guess) it will hit no value above 0 with probability 1.