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I can´t prove that a martingale has constant expected value. $$ \mathbf{E}[M_t]=\mathbf{E}[M_0] $$

Thanks people.

A.P.
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albino
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1 Answers1

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It holds for any sigma-algebra $\mathcal{F}$ that

$$\mathbb{E}[ \mathbb{E}(X \mid \mathcal{F}) ] = \mathbb{E}(X).$$

Note that this does not require that $\mathcal{F}$ and $X$ are independent. Since a martingale satisfies

$$\mathbb{E}(M_{n+1} \mid \mathcal{F}_n) = M_n,$$

we get

$$\mathbb{E}(M_{n+1}) = \mathbb{E}[\mathbb{E}(M_{n+1} \mid \mathcal{F}_n)] = \mathbb{E}(M_n)$$ for all $n \in \mathbb{N}$. Hence, $\mathbb{E}(M_n) = \mathbb{E}(M_0)$.

saz
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  • Sorry for the pedantic question, but what exactly is a sigma-algebra $\mathcal{F}$? It seems in some notes on Martindale, $\mathcal{F_n}$ is defined as ${ X_n, X_n-1,\cdots X_1 }$. Is it just some succinct way to write the large set of variables we're conditioning on? – 24n8 May 10 '20 at 01:01
  • @Iamanon $\mathcal{F}_n$ is typically the $\sigma$-algebra generated by $X_1,\ldots,X_n$, i.e. $\mathcal{F}_n=\sigma(X_1,\ldots,X_n)$. – saz May 10 '20 at 09:30