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1

$begingroup$

Define $S_n = Sigma_{i=0}^n X_n$, where $X_n = pm1$ with probability $1/2$ for each case.

I am trying to show that for a walk of length $2n -2k$ starting at $0$, the probability that it does not hit $0$ at all (except its start point) is the same as the probability that it ends at $0$.

Formally, I am trying to show that
$$
P(S_1, S_2, ...S_{2n-2k}neq 0) = P(S_{2n-2k} = 0).
$$

I have a "hand-wavy" proof. Call the set of walks of length $2n-2k$ that do not revisit zero $G$, and the set of walks that end at zero $H$. We can form a bijection $G rightarrow H$ by taking a walk in $G$ and reflecting it across the horizontal line $y=S_{0.5(2n-2k)} = S_{n-k}$, since doing so will guarantee that such walk ends at $0$. Since such bijection exists, the probabilities that members of these sets occur must be equal.

First of all, is this logic correct, and secondly, is there a way to make this more formal?

probability proof-verification random-walk

share|cite|improve this question

edited Mar 12 at 3:53

lithium123

asked Mar 11 at 22:04

lithium123lithium123

637316

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Proving that the function you describe really is a bijection is a good start!
  $endgroup$
  – Greg Martin
  Mar 11 at 22:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $X_n = pm 1$ with probability $1/2$ each, or $0$ or $1$ with probability $1/2$ each?
  $endgroup$
  – Brian Tung
  Mar 11 at 22:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's the first. I've made the appropriate edit in the question.
  $endgroup$
  – lithium123
  Mar 12 at 1:55
  

  
  
  
  

add a comment | 

1

$begingroup$

Define $S_n = Sigma_{i=0}^n X_n$, where $X_n = pm1$ with probability $1/2$ for each case.

I am trying to show that for a walk of length $2n -2k$ starting at $0$, the probability that it does not hit $0$ at all (except its start point) is the same as the probability that it ends at $0$.

Formally, I am trying to show that
$$
P(S_1, S_2, ...S_{2n-2k}neq 0) = P(S_{2n-2k} = 0).
$$

I have a "hand-wavy" proof. Call the set of walks of length $2n-2k$ that do not revisit zero $G$, and the set of walks that end at zero $H$. We can form a bijection $G rightarrow H$ by taking a walk in $G$ and reflecting it across the horizontal line $y=S_{0.5(2n-2k)} = S_{n-k}$, since doing so will guarantee that such walk ends at $0$. Since such bijection exists, the probabilities that members of these sets occur must be equal.

First of all, is this logic correct, and secondly, is there a way to make this more formal?

probability proof-verification random-walk

share|cite|improve this question

edited Mar 12 at 3:53

lithium123

asked Mar 11 at 22:04

lithium123lithium123

637316

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Proving that the function you describe really is a bijection is a good start!
  $endgroup$
  – Greg Martin
  Mar 11 at 22:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $X_n = pm 1$ with probability $1/2$ each, or $0$ or $1$ with probability $1/2$ each?
  $endgroup$
  – Brian Tung
  Mar 11 at 22:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's the first. I've made the appropriate edit in the question.
  $endgroup$
  – lithium123
  Mar 12 at 1:55
  

  
  
  
  

add a comment | 

$begingroup$

Define $S_n = Sigma_{i=0}^n X_n$, where $X_n = pm1$ with probability $1/2$ for each case.

I am trying to show that for a walk of length $2n -2k$ starting at $0$, the probability that it does not hit $0$ at all (except its start point) is the same as the probability that it ends at $0$.

Formally, I am trying to show that
$$
P(S_1, S_2, ...S_{2n-2k}neq 0) = P(S_{2n-2k} = 0).
$$

I have a "hand-wavy" proof. Call the set of walks of length $2n-2k$ that do not revisit zero $G$, and the set of walks that end at zero $H$. We can form a bijection $G rightarrow H$ by taking a walk in $G$ and reflecting it across the horizontal line $y=S_{0.5(2n-2k)} = S_{n-k}$, since doing so will guarantee that such walk ends at $0$. Since such bijection exists, the probabilities that members of these sets occur must be equal.

First of all, is this logic correct, and secondly, is there a way to make this more formal?

probability proof-verification random-walk

share|cite|improve this question

edited Mar 12 at 3:53

lithium123

asked Mar 11 at 22:04

lithium123lithium123

637316

$endgroup$

share|cite|improve this question

edited Mar 12 at 3:53

lithium123

asked Mar 11 at 22:04

lithium123lithium123

637316

share|cite|improve this question

edited Mar 12 at 3:53

lithium123

asked Mar 11 at 22:04

lithium123lithium123

637316

asked Mar 11 at 22:04

lithium123lithium123

637316

asked Mar 11 at 22:04

lithium123lithium123

637316