Prove that the probability of no particle on the positive axis tends to $0$.

Question

There is a particle located at the origin of the real line. At time $t=1$, the particle splits into two, and each of the resulting particles either jumps to left by $1$ or right $1$ with equally probability $1/2$, independently of the other particle. At time $t=2$, each particle splits into two and do the exactly same random jump as before, independently of other particles and of what happened in the past. This process continues endless. Let $A_t$ denote the event that there is no particle on the positive axis immediately after time $t$, and we want to prove that $P(A_t)\to0$ as $t\to\infty$.

We can denote the maximum coordinate of all particles at time $t=n$ as $Y_n$, and coordinates of two particles generated by the original one at $t=1$ as $x_1$ and $x_2$. Hence, $$P(Y_n\le x)=(P(Y_n\le x|x_1=1)P(x_1=1)+P(Y_n\le x|x_1=-1)P(x_1=-1))\cdot$$ $$(P(Y_n\le x|x_2=1)P(x_2=1)+P(Y_n\le x|x_2=-1)P(x_2=-1))$$ Note that $P(Y_n\le x|x_1=1)=P(Y_{n-1}\le x-1)$ and $P(Y_n\le x|x_1=-1)=P(Y_{n-1}\le x+1)$ and same for $x_2$. Hence, $$P(Y_n\le x)=(\frac{1}{2}P(Y_{n-1}\le x-1)+\frac{1}{2}P(Y_{n-1}\le x+1))^2$$

Then, we want to show that $\lim\limits_{n\to\infty}P(Y_n\le 0)=0$.

As $P(Y_n\le x)\le(P(Y_{n-1}\le x+1))^2$, and so $P(Y_n\le 0)\le(P(Y_{n-1}\le 1))^2\le(P(Y_{n-2}\le 2))^4\le\cdots$

Then I am not sure how to bound it with something tend to $0$.

Is this a reasonable way to deal with this problem? Is there any other way to do this?

Thanks for any hint and suggestion.

Answer 1

Basic approach. Suppose we track just a single particle that does not split up, but simply walks one discrete step at a time, with $1/2$ probability of going left one unit or right one unit. Let $B_t$ be the event that immediately after its $t$th step, it has a positive coordinate.

By symmetry, using the binomial distribution, what is $P(B_t)$? What simple bound does it have for any $t$?

Now note that the original process is the superposition of many such particles, with $2^t$ particles after the $t$th step. These are all independent walks, so the probability that none of the $2^t$ particles has a positive coordinate after the $t$th step is simply

$$ P(A_t) = [1-P(B_t)]^{2^t} $$

Using the above bound for $P(B_t)$, what can you say about $P(A_t)$?