# Identity for simple 1D random walk

The question is to find a purely probabilistic proof of the following identity, valid for every integer $n\geqslant1$, where $(S_n)_{n\geqslant0}$ denotes a standard simple random walk:

Standard simple random walk is defined as $S_0=0$ and $S_n=\sum\limits_{k=1}^nX_k$ for every $n\geqslant1$, where $(X_k)_{k\geqslant1}$ is an i.i.d. sequence such that $P[X_k=+1]=P[X_k=-1]=\frac12$.

Of course, the RHS of the identity is

For a combinatorial proof, see this MSE question and its comments.

For an accessible introduction to the subject, see the corresponding chapter of the Chance project.

I don’t know if what I will write is a “purely probabilistic proof” as the question requests, or a combinatorial proof, but Did should decide that. At the end I do use combinatorial identities (UPDATE 12-1-2014: an alternative final step of the proof has been found that does not use the identities. The initial final step is preserved at the end, in grey).

The LHS of the identity can be re-written as

hence

But $X_i^2=1$ almost surely for every $i$. Likewise, the distribution of $(X_i,X_j)$ conditionally on $S_{2n}=0$ does not depend on $i\ne j$ since the $X_i$s are i.i.d and $S_{2n}$ is a symmetric function of $(X_i)$. So we arrive at

The random variable $X_{2n-1}X_{2n}$ takes on the values $\{-1,1\}$. Let $q \equiv P[X_{2n-1}X_{2n}=-1; S_{2n}=0]$. Then $P[X_{2n-1}X_{2n}=1; S_{2n}=0]=P[S_{2n}=0]-q$ hence

Note that if $X_{2n-1}X_{2n}=-1$ then $X_{2n-1}+X_{2n} = 0$ and $S_{2n-2}=S_{2n}$, therefore

the last equality being due to the fact that $S_{2n-2}$ and $X_{2n-1}+X_{2n}$ are independent and that $P[X_{2n-1}+X_{2n} = 0]= \frac 12$.
So

Inserting into the main expression and multiplying out we have

Guided by what we want to prove, the issue now is to express $P[S_{2n}=0]$ in terms of $P[S_{2n-2}=0]$. We note that the event $\{S_{2n}=0\}$ can only occur and so has strictly positive probability, if and only if one of the following events occur:

1) $E_0 \equiv [\{S_{2n-2}=0\}\; \cap \{X_{2n-1}+X_{2n}= 0\}]$
2) $E_1 \equiv [\{S_{2n-2}=2\}\; \cap \{X_{2n-1}+X_{2n}= -2\}]$
3) $E_2 \equiv [\{S_{2n-2}=-2\}\; \cap \{X_{2n-1}+X_{2n}= 2\}]$

The events inside the brackets, if we look at them as elements of a $3\times 2$ matrix, are (pardon my language) … “horizontally independent and vertically mutually exclusive”. Combined these imply that

while the “horizontal” independence implies that the joint probabilities can be separated as the product of the probabilities of the two events involved in each case. More over, by symmetry we have that $P[S_{2n-2}=-2]= P[S_{2n-2}=2]$, while
$P[X_{2n-1}+X_{2n}= 0]= 1/2$ , $P[X_{2n-1}+X_{2n}= -2]= 1/4$ , $P[X_{2n-1}+X_{2n}= 2]= 1/4$. Using all these results we arrive at

Inserting this into equation $[A]$ we obtain

Equation $[B]$ is an improvement over eq. $[A]$ because, through probabilistic arguments, only $S_{2n-2}$ is now present in the RHS. Now, if we denote by $Y_k$ a binomial random variable with probability of success $1/2$, $Y_k = Bin(k,1/2)$, then, the probabilities related to $S_{2n-2}$ can be expressed in terms of $Y_{2n-2}$,

It is then a natural step to use the probability generating function of $Y_{2n-2}$, which is

while we have

where the superscript parentheses denote the order of derivative taken. But

and so

Inserting this into equation $[B]$ we obtain

which is what we wanted to prove. One could argue that the use of the probability generating function is too “algebraic” a step, and that in general, the moment we enter binomial distribution territory, combinatorics rage in the background -but at least now they rage only in the background.

And don’t forget, on the side, rearranging the relations above we can obtain the conditional expectation

INITIAL final step of the proof (using combinatorial identities)

We have arrived at equation $[A]$.

Using $Y_k$ to denote a binomial random variable with probability of
success $1/2$, $Y_k = B(k,1/2)$, the probabilities related to $S_n$
can be expressed as probabilities of binomial random variables,

Since we want at the end to have only $P[S_{2n-2}=0]$ present, we must
somehow express $P[Y_{2n}=n]$ in terms of $P[Y_{2n-2}=n-1]$, and here
is where I use the combinatorial identities, since one can arrive
through them at

Reverting back to $S$– notation and inserting in the main expression,
we have

which is what we wanted to prove. Maybe there are other ways to relate
$P[S_{2n}=0]$ and $P[S_{2n-2}=0]$, but they currently escape me.