I know that the process $X_t$ is a martingale when:

  • $\mathbb{E}[|X_t|]<\infty$ for each $t$,
  • $\mathbb{E}[X_t|\mathcal{F_s}] = X_s$ for $s<t$.

Using properties of the conditional expectation we can show that:

$\mathbb{E}[X_s] = \mathbb{E}[X_t]$.

My question is:

Can we use this property ($\mathbb{E}[X_s] = \mathbb{E}[X_t]$) to prove that the process is a martingale?

I noted that in many sources people show that $\mathbb{E}[X_t|\mathcal{F_s}] = X_s$ to prove that the process is a martingale.

I think that these conditions are equivalent and showing the first condition should be easier. Am I right?

  • $\begingroup$ It is not possible to prove this way. Note that $X_S$ is a random variable, while $E[X_S]$ is a constant. Just as an example, supose $X\sim Gamma(\alpha,\beta)$ and $Y \sim Gamma(\beta,\alpha)$. Both have same mean $E[X]=E[Y]=\alpha \beta$, but its false that $X=Y$. $\endgroup$ – A.T May 28 '18 at 14:42

Fractional Brownian motion with $H\neq 1/2$ is not a martingale yet $E(B_s)=E(B_t)=0$ for all $s,t$.

  • $\begingroup$ Thank you. I think that my assumption that conditions: $\mathbb{E}[X_t|\mathcal{F}_s] = M_s$ and $\mathbb{E}[X_t] = \mathbb{E}[X_s]$ are equivalent is not correct. It should be an implication between them. Hence we cannot use the second condition to prove that $X_t$ is a martingale. Am I right? $\endgroup$ – MathMen May 28 '18 at 15:10
  • $\begingroup$ @Jonas Al-Hadad yes you are right. $\endgroup$ – user223391 May 28 '18 at 16:16
  • $\begingroup$ @Jonas Al-Hadad, just to add onto your claim, while what you proposed is not true, there is a similar equivalent statement. $X_t$ is a martingale iff for every bounded stopping times $S \leq T$, $E(X_S) = E(X_T)$. $\endgroup$ – James Yang May 30 '18 at 16:38
  • $\begingroup$ I'm confused why there's a delete vote on this answer. $\endgroup$ – user223391 May 30 '18 at 17:39

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