Proof of Martingale

[Dhamnekar Win,odd]

Let [math]X_i, i \geq 1[/math] be independent with [math] P( X _i = 1) =p, P(X_i =-1) =q = 1-p[/math]  with [math]S_n= \displaystyle\sum_{i=1}^n X_i[/math] Show that [math](\frac{q}{p})^{S_n}, n \geq 1[/math] is a martingale with mean 1.

Solution:

Let's prove this step-by-step:

1) The [math]X_i[/math] are independent random variables with:
[math]P(X_i = 1) = p[/math]
[math]P(X_i = -1) = q = 1 - p[/math]

2) Let [math]S_n = \displaystyle\sum_{i=1}^n X_i[/math] be the partial sums.

3) Consider the stochastic process [math]Y_n = (\frac{q}{p})^{S_n}[/math]. 

4) To show this is a martingale, we need to prove:
[math]E[Y_{n+1} | Y_1, ..., Y_n] = Y_n[/math]

5) We have:
[math]E[Y_{n+1} | Y_1, ..., Y_n] = E[(\frac{q}{p})^{S_{n+1}} | S_1, ..., S_n][/math]
[math]= (\frac{q}{p})^{S_n} E[(\frac{q}{p})^{X_{n+1}}][/math] 
[math]= (\frac{q}{p})^{S_n} (q + p) = Y_n[/math]

Where we have used the independence of the [math]X_i[/math] and the definition of conditional expectation. 

6) Clearly [math]E[Y_n] = E[(\frac{q}{p})^{S_n}] = 1[/math] since [math]S_n[/math] has a symmetric distribution about 0.

Therefore, [math](Y_n)_{n≥1} = ((\frac{q}{p})^{S_n})_{n≥1}[/math] is a martingale with mean 1.

In my opinion, this solution is correct. Do you have any doubts? 

 

Edited July 17, 2023 by Dhamnekar Win,odd