Consider $X_1,X_2,...,X_n$ the i.i.d. Bernoulli random variables, with $P(X_i = 1) = p$ and $P(X_i = 0) = q$. Take the process $S=(S_n)_{n\geq 0}$ where $S_n= X_1 +...+ X_n$ for $n = 1,2,\dots$ and $S_0= 0$: For which $p$ is $(S_n)$ a martingale? sub martingale? super martingale?

Basically I let $F_n=\sigma(X_1,...,X_n)$ be the filtration $F$ at $n$. To find $p$ such that $X_1,..,X_n$ is a sub martingale/martingale I found

$E(S_{n+1}|F_n)\leq S_n \implies E(X_{n+1}|F_n)+E(S_n|F_n)\leq S_n \implies E(X_{n+1})+S_n\leq S_n \implies p\leq 0 $. Thus $p=0$. Similarly when $p\geq 0$ then $X_1,...,X_n$ is a super martingale

The result seems trivial, but I am not sure if this correct. Thanks

  • $\begingroup$ Most likely your Bernoulli variable has support $\{-1,1\}$ instead. $\endgroup$
    – A.S.
    Nov 16, 2015 at 22:07
  • $\begingroup$ The support is $\{0,1\}$. I think its pretty much in all cases of a random variable follows a Bernoulli distribution. $\endgroup$
    – user60887
    Nov 16, 2015 at 22:49

1 Answer 1


Yes, it is correct. So the sequence $(S_n)$ is never a martingale unless in degenerated cases. We could note that the conclusion would not be the same if we replaced $P(X_i = 0) = q$ by $P(X_i = -1) = q$.


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