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If we have $X_k$ random variables with average $0$ and independent, why is the $\sum_{k=1}^n X_k$ a martingale for the sigma algebra $\mathcal F_n$ generated by $\{X_1,\ldots, X_n\}$?

I basically only have to prove that the expected value of $X_{n+1}$ knowing $\mathcal F_n$ is $0$, but somehow this isn't intuitive to me at all. Could anyone give me pointers on this?

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  • $\begingroup$ If the expected value of $X_{n+1}$ is $0$ but the expected value of $X_{n+1}$ knowing $\mathcal F_n$ is not $0$ then $X_{n+1}$ is not independent of $X_1,X_2,\ldots,X_n$ $\endgroup$
    – Henry
    Jun 14, 2012 at 22:01

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Just check the definition. Let $S_n:=\sum_{k=1}^nX_k$.

  • We have that $S_n$ is integrable since so are $X_j$.
  • Let $n$ an integer. We have by linearity of conditional expectation that $$E[S_{n+1}\mid\mathcal F_n]=E[S_n+X_{n+1}\mid\mathcal F_n]=E[S_n\mid\mathcal F_n]+ E[X_{n+1}\mid\mathcal F_n].$$ Since $S_n$ is $\mathcal F_n$ measurable, we have $E[S_n\mid\mathcal F_n]=S_n$ and since $X_{n+1}$ is independent of $\mathcal F_n$, we have that $E[X_{n+1}\mid \mathcal F_n]=E[X_{n+1}]=0$.
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  • $\begingroup$ Alright thanks, now I have another question: if I change the definition of F(n) so that it's any sigma-algebra that can measure X1, ..., Xn, do I still have this result? It seems that now we don't have X(n+1) independent of F(n), since F(n) can't be defined with just X1, ..., Xn $\endgroup$
    – lezebulon
    Jun 14, 2012 at 22:12
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    $\begingroup$ The result may be not true, for example if you take $X_n$ non-zero and $\mathcal F_n=\mathcal F$ which makes all $X_n$ measurable. $\endgroup$
    – Davide Giraudo
    Jun 14, 2012 at 22:14

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