How Can I Prove the Second Equation from the First in a Random Walk Probability?

AI Thread Summary
The discussion centers on deriving the second equation from the first in a random walk involving increments X_i that can take values -1 or +1. Participants emphasize the importance of the independent and identically distributed (i.i.d.) nature of the X_i's in proving the relationship rigorously. A key insight is that the distribution of pairs (X1, X2) is the same as (X2, X1), which stems from their i.i.d. property. This property allows for the exchange of indices without affecting the overall probability statements. The conversation highlights the foundational role of i.i.d. in establishing the desired proof.
jakey
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Hi guys,

I was reading about random walks and i encountered one step of a proof which i don't know how to derive in a mathematically rigorous way.

the problem is in the attached file and S is a random walk with X_i as increments, X_i =
{-1,+1}

I know that intuitively we can switch the indices to obtain the second equation from the first but how do we prove it rigorously?

EDIT: btw, I am just looking for hints, not the entire solution. i think one of the possible hints is that the X_i's are i.i.d. but i can't think of a way to use this
 

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Since all X_i have the same distribution renumbering the indices makes no difference.
 
hi mathman, thanks btw! so there's no rigorous proof for this?
 
I don't what you need to make it rigorous.
 
Mathman's lemma: (X1,X2) has the same distribution as (X2,X1).

Proof: P[X1<=x1,X2<=x2] = P[X1<=x1]P[X2<=x2] = P[X2<=x1]P[X1<=x2] = P[X2<=x1,X1<=x2]
 
You started with the assumption that the Xi's were iid. Part of the definition of iid is that they are identical - that is, every marginal probability statement for one variable can be exchanged for any probability statement about another. The other bard of the definition of iid is that they are independent. This fact allows us to extend the above from marginal probability statements to any arbitrary joint probability statement.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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