How does one prove that a simple (steps of length $1$ in directions parallel to the axes) symmetric (each possible direction is equally likely) random walk in $1$ or $2$ dimensions returns to the origin with probability $1$?

Edit: note that while returning to the origin is guaranteed $(p = 1)$ in $1$ and $2$ dimensions, it is not guaranteed in higher dimensions; this means that something in a correct justification for the $1$- or $2D$ cases must fail to extend to $3D$ $($or fail when the probability for each direction drops from $\frac{1}{4}$ to $\frac{1}{6})$.

  • $\begingroup$ Can you think of a reason why it should hit points that are not at the origin? $\endgroup$ Jul 23, 2010 at 4:13
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    $\begingroup$ @Jonathan: I'm not sure what you mean--in 1-d, the random walk hits every integer with probability 1, and in 2-d, the random walk hits every lattice point with probability 1. $\endgroup$
    – Isaac
    Jul 23, 2010 at 4:16
  • $\begingroup$ Then I guess I am confused by the question :( $\endgroup$ Jul 23, 2010 at 4:18
  • 2
    $\begingroup$ I think it will hit all points a infinite number of times. $\endgroup$ Jul 23, 2010 at 4:55
  • 2
    $\begingroup$ @Isaac: If you know you will hit every point with probability 1, then you will return to each point with probability 1 $\endgroup$
    – Casebash
    Jul 23, 2010 at 5:04

10 Answers 10


See Durrett, Probability: Theory and Examples (link goes to online copy of the fourth edition; original defunct link). On p. 164 Durrett gives a proof that simple random walk is recurrent in two dimensions.

First find the probability that simple random walk in one dimension is at $0$ after $2n$ steps; this is clearly $\rho_1(2n) = \binom{2n}{n}/2^{2n}$, since $\binom{2n}{n}$ is the number of paths with $n$ right steps and $n$ left steps.

Next, the probability that simple random walk in two dimensions -- call this $\rho_2(2n)$ -- is at $0$ after $2n$ steps is the square of the previous probability. Consider the simple random walk which makes steps to the northeast, northwest, southeast, and southwest with equal probability. The projections of this walk onto the x- and y-axes are independent simple random walks in one dimension. Rotating and rescaling gives the "usual" SRW in two dimensions (with steps north, east, south and west) and doesn't change the probability of being at $0$.

So $\rho_2(2n) = \left(\binom{2n}{n}/2^{2n}\right)^2$. This is asymptotic to $1/(\pi n)$, and the expected number of returns to the origin is the $\sum_{n \geq 1} \rho_2(2n)$, so the expected number of returns to the origin is infinite. It's not hard to show (and is in fact true, but the proof is unenlightening so I'll leave it to Durrett) that in this case the probability of eventually returning to the origin is $1$.

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    $\begingroup$ Though it is not explicitly stated here, note that this argument also tells you exactly why you should expect the statement not to be true in 3 and higher dimensions. $\endgroup$ Aug 4, 2010 at 4:14
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    $\begingroup$ Perhaps someone could edit this to connect the dots and demonstrate what happens at d>2? $\endgroup$ Oct 10, 2012 at 23:33
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    $\begingroup$ For $d>2$ the probability goes like at least $n^{-3/2}$. One can now apply the Borel-Cantelli lemma to show that with probability $1$ the random walk does not return to the origin an infinite number of times. $\endgroup$ Apr 3, 2014 at 20:20

I'll try it for 2D and then we can get 1D as a corollary [excercise!]...

This is the only proof I know of, there may be a more intuitive (and less messy without tex!) proof out there, but I like this one- it uses generating functions in a really nifty way.

Consider the probability of being at the origin after 2n steps (notice we cannot return in an odd number of steps):


$u_{2n} = p(n,n,0,0)+ p(n-1,n-1,1,1)+...+p(n-k,n-k,k,k)+...+p(0,0,n,n)$ (when $n\neq0$)

Here $p(u,d,l,r)$ is the probability of the first 2n steps being u up, d down, l left and r right in any order. Each order has probability $\frac{1}{4^{2n}}$, and there are $\frac{(2n)!}{u!d!l!r!}$ distinct orders, giving $p(n-k,n-k,k,k)=\frac{1}{4^{2n}} \frac{(2n)!}{(n-k)!(n-k)!k!k!}$

Now, since $(2n)!=n! n! \binom{2n}{n}$ we have $p(n-k,n-k,k,k)=\frac{1}{4^{2n}} \binom{2n}{n} \binom{n}{k}^2$ giving

$u_{2n}= \frac{1}{4^{2n}} \Sigma_k \binom{2n}{n} \binom{n}{k}^2$ which, by one of those silly binomial results, can be contracted to

$u_{2n}= \frac{1}{4^{2n}} \binom{2n}{n}^2$

Let us put that in our back pocket for now and consider instead the probability of first returning after 2n steps $f_{2n}$ --- this is rather difficult to tackle directly but we can make ourselves a cunning little formula involving it, jazz out some generating function fun and seal the deal with some.

The formula in question is: $u_{2n}= f_2 u_{2n-2}+ f_4 u_{2n-4} +....+f_{2n-2}u_{2} + f_{2n}$

Which we shall not prove so much as explain: to return to the origin after 2n steps (LHS) you must either first return after 2 steps and do a 'return to the origin in 2n-2 steps walk' (first term RHS) or first return after 4 steps and do a 'return to the origin in 2n-4 steps walk' (2nd term) or... or first return after 2n-2 steps and do a 'return to the origin in 2 steps walk or first return to the origin after 2n steps.

We shall now tweak our formula just a tiny bit so it has the right symmetry properties for what is to come, we do this by adding $f_0=0$ and $u_0=1$ to give

$u_{2n}= f_0 u_{2n}+ f_2 u_{2n-2}+ f_4 u_{2n-4} +....+f_{2n-2}u_{2} + f_{2n}u_0$

Which is secretly a statement about generating functions (this is the sweet bit!), see look at the generating functions of $u_n$, $f_n$:

$U(x)= \Sigma_m u_{2m} x^{2m}$, $F(x)= \Sigma_m f_{2m} x^{2m}$

We see:

$U(x)= 1+ U(x)F(x)$

(where the '1+' is to compensate for the fact that $u_0$ does not appear in the product) Rearranging to:

$F(x)= \frac{U(x)-1}{U(x)}$

Observe that the probability of return is $\Sigma f_n= F(1)$, which comes out as 1 because $U(1)$ diverges by some tedious stirlings formula bounds that I forget.

Edit: Until Tex comes online, this is pretty unreadable, so here's a link to some lecture notes I found with the same proof (and, fortunately, the same notation!). Enjoy!

Edit$^2$: Hooray, Tex has come online!!! Enjoy.

  • $\begingroup$ Hi, this is a great answer but can you explain your statement (or provide a proof/source) that $2n! = n!n! {{2n} \choose {n}} $ $\endgroup$
    – analystic
    Apr 13, 2015 at 0:58
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    $\begingroup$ @analystic $\binom{n}{k}$ is always equal to $\frac{n!}{k!(n-k)!}$. $\endgroup$
    – Wojowu
    Nov 30, 2015 at 21:06
  • $\begingroup$ @TomBoardman I was able to manipulate the proof to get the result that in case the probabilities of moving in 4 directions are unequal(2D case), the man can return to the origin eventually with probability 1 regardless of the probabilities. Is this result false? I might be missing a trivial clause. $\endgroup$
    – reyna
    Jul 19, 2020 at 5:34

Let $X_n = 1$ if at the $n^{th}$ step we are back at origin else it is $0$

$\mathbb{E}(X_n)$ is the probability of return after $n$ steps = $P_n$

Let $X = \displaystyle \sum_{n=1}^{\infty} X_n$ counts the number of returns

$\mathbb{E}(X) = \displaystyle \sum_{n=1}^{\infty} \mathbb{E}(X_n)$

Expected number of returns = $\mathbb{E}(X) = \displaystyle \sum_{n=1}^{\infty} P_n$

Note that $P_n = 0$ if $n$ is odd

Let $\rho$ be the probability that the random walk returns (at-least once) to $0$

$\rho_k$ be the probability that we return exactly $k$ times to the origin

$\rho_0 = 1 -\rho$ and in general, $\rho_k = \rho^k (1- \rho)$

$\mathbb{E}(X) = \displaystyle \sum_{k=0}^{\infty} k \rho^k (1-\rho) = \frac{\rho}{1-\rho}$ if $\rho < 1$

Hence, from the above we see that,

If $\rho < 1$, then $\mathbb{E}$(Number of returns) is finite i.e. $\displaystyle \sum_{n=1}^{\infty} p_n < \infty$

If $\rho = 1$, then $\mathbb{E}$(Number of returns) is infinite i.e. $\displaystyle \sum_{n=1}^{\infty} p_n = \infty$

In one dimensions, we have $p_{2n+1} = 0$ and $p_{2n} = \frac1{2^{2n}} \binom{2n}{n}$. Note that the middle binomial coefficient is the largest and hence $\frac1{2^{2n}} \binom{2n}{n} \geq \frac1{2n+1}$ and hence the $\displaystyle \sum_{n=1}^{\infty} p_{2n}$ diverges

Or we could use the stirling's formula to get a better estimate and get $p_{2n} \approx \frac{c}{\sqrt{n}}$ and hence the sum diverges

In two dimensions, we have $p_{2n} = \frac1{4^{2n}} \displaystyle \sum_{k=0}^{n} \binom{2n}{k} \binom{2n-k}{k} \binom{2n-2k}{n-k} = \frac1{4^{2n}} \displaystyle \sum_{k=0}^{n} \frac{(2n)!}{k! k! (n-k)! (n-k)!}$

Note the maximum is at $k = \frac{n}{2}$ and by Central Limit Theorem you can approximately say that intervals of length $\sim \sqrt{n}$ around $\frac{n}{2}$ where this is of largest size

Hence, each such $k$ contributes $\sim \frac{\sqrt{n}}{(\sqrt{n})^4} = \frac1{n^{3/2}}$ (By stirling's formula).

Number of $k$ is of the order of $\sqrt{n}$ and hence $p_{2n} \sim \frac1{n}$ and so again $p_n \rightarrow \infty$ since the harmonic sum diverges

In general, $p_n$ is about the size $\frac{c}{n^{d/2}}$ so $\displaystyle \sum_{n=1}^{\infty} p_n < \infty$ if $d > 2$. This is from the convergence and divergence of the p-series. $\displaystyle \sum \mathbb{E}(X_n) < \infty \iff \mathbb{E}(X) < \infty \iff \rho < 1 \implies$ Probability(return infinitely often) = 0

  • $\begingroup$ how can i prove: ρk=ρ^k * (1−ρ) $\endgroup$
    – Shreyans
    Aug 12, 2022 at 18:26

$P =$ probability that if you are at $1$, you will return to $0$.

$Q =$ probability that if you are at $2$ you will return to $0$.

Assuming you start at $0$, you will either go to $1$ or to $-1$, so finding the probability $P$ for $1$ is the same as the probability if you go to $-1$.

From the point $1$, there are $2$ possibilities:

1) you can go immediately back to $0$

2) you can go to $2$.

So, $P = \frac{1}{2} +\frac{1}{2} * Q$

What is $Q$:

From the point $2$, you have $2$ possibilities:

1) You can go even further out.

2) You can return to $0$. What it the probability you will return to 0?

Well, the probability that you will return to $1$ is $P$ because returning to $1$ from $2$ is the same probability as returning to $0$ from $1$. And, you will have to return to $1$ before you can return to $0$.

\begin{align} P & = \frac{1}{2} + \frac{1}{2} P^2\\ 2P & = 1+P^2\\ P^2 – 2P + 1 & = 0\\ (P – 1) (P – 1) & = 0\\ P & = 1.\end{align}

  • $\begingroup$ Well, my formulas didn't paste well. But, I'm sure you can sort it out. $\endgroup$
    – user8115
    Mar 11, 2011 at 15:14
  • $\begingroup$ P = 1/2 + 1/2P^2 ---- 2P = 1 = P^2 ---- P^2 - 2P +1 = 0 ---- (P-1)^2 = 1 ---- P = 1 $\endgroup$
    – user8115
    Mar 14, 2011 at 11:58
  • $\begingroup$ What if we add R=probability that if you go to 2 you will return to 0. Doesn't it lead to P = 1/2 + 1/2P^2+1/2P^3? $\endgroup$
    – asmani
    Dec 22, 2020 at 11:01

Since my reputation is less than 50 and my suggested edit has been rejected. I am adding explanation to why $ U(x) = 1 + U(x)F(x) $ . This answer assumes background from Tom's answer.

$ U(x) = \sum_{m=0}^{\infty} u_{2m} x^{m} $, $F(x) = \sum_{k=1}^{\infty} f_{2k} x^{k} $

It is trivial that $ u_0 = 1 $.

$ U(x)F(x) = (\sum_{m=1}^{\infty} u_{2m} x^{m}) (\sum_{k=1}^{\infty} f_{2k} x^{k}) $

$ U(x)F(x) = \sum_{n=1}^{\infty} (\sum_{m=1}^{n} f_{2m} u_{2n-2m}) x^n $

$ U(x)F(x) = \sum_{n=1}^{\infty} u_{2n} x^{n} $

$ U(x)F(x) = U(x) - 1 $

NOTE: Proof of $ U(x) = 1 + U(x)F(x) $ is cited from here Go to page 3.


I'll do 1D.

1D walks are building binary strings, 010101, etc.

Say take six steps. Then 111111 is just as likely as 101010.

However, how many of the possible sequences have six ones? 1. How many of the possibly sequences have three ones and three zeros? Much more.

That number is called multiplicity, and it grows mighty fast. In the limit its log becomes Shannon entropy.

Sequences are equally likely, but combinations are not.

In the limit the combinations with maximum entropy are going dominate all the rest. So the walk is going to have gone an equal number of right and left steps...almost surely.

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    $\begingroup$ Conceptually, I agree (and there's a fairly similar expected-value argument on Wikipedia), but I'm looking for a rigorous theoretical proof. (I know it's doable and I know I have it buried in a notebook somewhere, but I can't for the life of me find the notebook or reconstruct the proof at the moment.) It's also interesting that while true in 1 and 2 dimensions, it is not true in higher dimensions. $\endgroup$
    – Isaac
    Jul 23, 2010 at 4:39
  • $\begingroup$ I don't really find writing rigorous proofs fun, but if you would like a derivation of how multiplicity becomes entropy in the limit check out: en.wikipedia.org/wiki/… $\endgroup$ Jul 23, 2010 at 4:46
  • $\begingroup$ Hmm, I definitely find it surprising that the same line of argument would fail in higher dimensions. Do you have a link? $\endgroup$ Jul 23, 2010 at 4:47

I can prove the 1 dimensional case a bit more formally than Jonathan. First we only look at the absolute values. Let us try to calculate the probability of this never exceeding x. The probability of 2x+1 consecutive moves all being the same is some p>0. If this ever occurs, then the absolute value will exceed x. Consider n groups of 2x+1 moves, the probability that at least one of these is all the same is 1-(1-p)^n, which approaches 1. So the probability of reaching each absolute value x (other than 0) is 1.

Now, lets consider the probability of reaching 0 again. Without loss of generality, suppose our first move is +1. We have a 100% chance of reaching a point a distance of 1 from this. There is a 50% that the first such point is 0 and 50% that it is 2. From 2, we have a 100% chance of reaching a point two away from this. 50% chance that this is 0, 50% that this is 4. Repeating, it is easy to see that we have to reach 0 again.

Furthermore, after we have reached an absolute value x, the probability of reaching it again is 1. So we reach each absolute value an infinite number of times. Due to symmetry, we can expect to reach each x an infinite number of times.

EDIT: I originally tried to apply it to the 2d case as you can see below. This is incorrect, as even though there will be no maximal absolute value, there is no reason we will necessarily come back down.

Similar reasoning will work in the 2d case. Instead of using symmetry, we simply note that each point with the same Manhattan distance from 0 has an expected value that is a non-zero proportion of the expected value all points a particular Manhattan distance away.

  • $\begingroup$ I'm having trouble following the tail of the first paragraph--I'm missing something between going to groups of 2x+1 moves and still hitting x, but it may just be me at this point. $\endgroup$
    – Isaac
    Jul 23, 2010 at 5:04
  • $\begingroup$ @Isaac: I am trying to show that there can be no maximal absolute value achieved. If there was a maximal absolute value, x, then there could never be 2x+1 moves which were all the same, which is clearly not going to be the case if you wait long enough. I just broke it into groups to prove this a bit more formally. $\endgroup$
    – Casebash
    Jul 23, 2010 at 5:07
  • $\begingroup$ At the moment, what I'm thinkings is that once you're looking at the groups, you might have the all-the-same group happen in a late enough group that it's offset by earlier groups and doesn't force you past x. $\endgroup$
    – Isaac
    Jul 23, 2010 at 5:17
  • $\begingroup$ @Isaac: In that case we would have exceeded the absolute value earlier. That is why I choose the value 2x+1. The 1d proof is sound - the 2d proof is not however $\endgroup$
    – Casebash
    Jul 23, 2010 at 5:20
  • $\begingroup$ To clarify: I refer to 2x+1 consecutive moves $\endgroup$
    – Casebash
    Jul 23, 2010 at 5:27

If you'd need only the 1d case, you can get this statement from the Law of the iterated logarithm.

This does not give much combinatorial insight, and well... requires said law, which is a little more work to proof.


For the one dimensional case

Assume there is a non-zero probability p that 0, the starting point is the leftmost point of the walk. Then there is a 1 - p chance the walk lands on -1 at some point. Given that it lands on -1 there is a p chance that -1 turns out to be the leftmost point. Continuing in this way we end up with a geometric distribution for the leftmost point of the walk over 0, -1, -2 ... which sums to 1

So if there is a nonzero probability that 0 is the leftmost then there is a 1 probability that there is a leftmost point. Same goes for rightmost point. This implies a probability 1 that the walk is bounded on the left and right.

But the probability that the walk bounded on the left and the right must be zero. Because after trimming down the boundaries we would find an upper and lower limit which the walk lands on infinitely many times and bounces back in the same direction every time. This has a probability of 0.

So it is impossible to have a nonzero probability that the origin is the leftmost point. From there is follows easily that the walk hits every point infinitely many times with probability 1.

For the 2 dimensional case just consider the "shells" of walking distance n around the origin. Every step moves to an adjacent shell either inward or outwards with bounded probabilities.


For $k\ge 1$ let $p_k$ be the probability of returning to the starting point in exactly $k$ steps, and not earlier. Note that the probability of ever making it back equals $$p=\sum_{k\ge 1} p_k$$ (the $p_k$ measure disjoint events). Clearly $p\le 1$.

Let $b_n$ be the probability of returning in $n$ steps. We have $$b_n = \sum_{k_1 + \cdot + k_s = n} p_{k_1}\cdot\ldots \cdot p_{k_s}$$ Indeed, being back at time $n$ might not be the first time, there might have been $s$ returns, and $s$ varies too (from $1$ up to possibly $n$).

Now, from the above we conclude that $$\sum_{n\ge 1} b_n = (p_1 + p_2 + \cdots) + (p_1 + p_2 + \cdots )^2 + (p_1 + p_2 + \cdots)^3 = \sum_{s\ge 1} p^s$$

(both sides can be $+\infty$, but since we are dealing with positive numbers all is OK).

Conclusion: $$\sum_{n\ge 1} b_n = \infty \Leftrightarrow p=1$$

In some cases it easy to check whether $\sum b_n$ is finite or not.

Note: if we denote by $B(t) = \sum_{n\ge 0} b_n t^n$ ($b_0 \colon = 1$) and $A(t) = \sum_{k\ge 1} p_k t^k$ then the equalities above for $b_n$ mean for formal power series $$B(t) = \frac{1}{1- A(t)}$$

Now, plugging in $t=1$ does the trick. However, if you feel uncomfortable, consider the alternate $$B(t) = \sum_{s\ge 0} A(t)^s$$ where $t=1$ is allowed because of the positive coefficients.


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