Identity for simple 1D random walk

The question is to find a purely probabilistic proof of the following identity, valid for every integer n, 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

\frac{n}{2^{2n-1}}\,{2n-2\choose n-1}.

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

E[(S_n)^2;S_{2n}=0] = E\left[\left(\sum_{i=1}^nX_i\right)^2 ; S_{2n}=0\right]=E\left[\sum_{i=1}^nX_i^2 + \sum_{i\neq j}X_iX_j ; S_{2n}=0\right]
E[(S_n)^2;S_{2n}=0]=E\left[\sum_{i=1}^nX_i^2 ; S_{2n}=0\right]+ E\left[\sum_{i\neq j}X_iX_j ; S_{2n}=0\right]

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_is are i.i.d and S_{2n} is a symmetric function of (X_i). So we arrive at

E[(S_n)^2;S_{2n}=0] = nP[S_{2n}=0] +n(n-1)E\left(X_{2n-1}X_{2n}; S_{2n}=0\right)

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
E\left(X_{2n-1}X_{2n};S_{2n}=0\right) = -1\cdot q + 1\cdot(P[S_{2n}=0]-q) = P[S_{2n}=0]-2q

Note that if X_{2n-1}X_{2n}=-1 then X_{2n-1}+X_{2n} = 0 and S_{2n-2}=S_{2n}, therefore
q= P[S_{2n-2}=0,S_{2n}=0]= P[S_{2n-2}=0,X_{2n-1}+X_{2n} = 0]= \frac 12 P[S_{2n-2}=0]
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.
E\left(X_{2n-1}X_{2n}; S_{2n}=0\right) =P[S_{2n}=0]-P[S_{2n-2}=0]

Inserting into the main expression and multiplying out we have

E[(S_n)^2;S_{2n}=0] = nP[S_{2n}=0] +n(n-1)\Big (P[S_{2n}=0] – P[S_{2n-2}=0]\Big)

=n^2P[S_{2n}=0] – n(n-1)P[S_{2n-2}=0] \tag{A}

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
P[S_{2n}=0] = P[E_0 \cup E_1 \cup E_2] = P[E_0] + P[E_1]+P[E_2]

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

P[S_{2n}=0] = \frac 12 P[S_{2n-2}=0] + \frac 12 P[S_{2n-2}=2]

Inserting this into equation [A] we obtain

E[(S_n)^2;S_{2n}=0] = n^2\left( \frac 12 P[S_{2n-2}=0] + \frac 12 P[S_{2n-2}=2]\right) – n(n-1)P[S_{2n-2}=0]

\Rightarrow E[(S_n)^2;S_{2n}=0] = \frac n2 (2-n)P[S_{2n-2}=0] + \frac {n^2}2P[S_{2n-2}=2] \tag{B}

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},

P[S_{2n-2}=0] = P[Y_{2n-2}=n-1],\;\; P[S_{2n-2}=2] = P[Y_{2n-2}=n]

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

G_Y(z) = \left (\frac 12 +\frac 12z\right)^{2n-2}
while we have

P[S_{2n-2}=0] = P[Y_{2n-2}=n-1] = \frac 1 {(n-1)!}G_Y^{(n-1)}(0)
P[S_{2n-2}=2] = P[Y_{2n-2}=n] = \frac 1 {n!}G_Y^{(n)}(0)

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

G_Y^{(n)}(0) = (n-1)G_Y^{(n-1)}(0) = (n-1)\cdot[(n-1)!]P[Y_{2n-2}=n-1]

and so

P[S_{2n-2}=2] = P[Y_{2n-2}=n] = \frac {n-1}{n} P[Y_{2n-2}=n-1] = \frac {n-1}{n}P[S_{2n-2}=0]

Inserting this into equation [B] we obtain

E[(S_n)^2;S_{2n}=0] = \frac n2 (2-n)P[S_{2n-2}=0] + \frac {n^2}2\frac {n-1}{n}P[S_{2n-2}=0]

\Rightarrow \frac n2P[S_{2n-2}=0] \cdot (2-n+n-1) = \frac n2P[S_{2n-2}=0]

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

E[(S_n)^2 \mid S_{2n}=0] = \frac {n^2}{2n-1}

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,
P[S_{2n}=0] = P[Y_{2n}=n],\qquad P[S_{2n-2}=0] =P[Y_{2n-2}=n-1]

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

P[Y_{2n}=n] = \frac 1{2^{2n}}{2n \choose n} = \frac 12 \frac
{2n-1}{n}\frac 1{2^{2n-2}}{2n-2 \choose n-1} = \frac

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

E[(S_n)^2;S_{2n}=0] = n^2\frac {2n-1}{2n}P[S_{2n-2}=0] –

=\frac{n}2\,P[S_{2n-2}=0]\Big (2n-1 – 2(n-1)\Big) =

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.

Source : Link , Question Author : Did , Answer Author : Alecos Papadopoulos

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