Prove that a process with independent increments and a constant mean is a martingale

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It is a basic property of conditional expectation. See https://en.wikipedia.org/wiki/Conditional_expectation#Conditional_expectation_with_respect_to_a_.CF.83-algebra

If a random variable $X$ is independent from a $\sigma$-algebra $\mathcal{F}$, then $E[X|\mathcal{F}]=E[X]$. Here the increment are independent, which means that $X_t-X_s$ is independent from $\mathcal{F}_s$. Thus $E[X_t-X_s|\mathcal{F}_s]=E[X_t-X_s]$.

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Dave ddd
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Dave ddd

Updated on October 24, 2020

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  • Dave ddd
    Dave ddd about 3 years

    how to prove that a process with independent increments and a constant mean is a martingale?

    in a solution to this problem i found this :

    $X_t - X_s$ is independent from $F_s$ hence :

    $$E[X_t-X_s|F_s]=E[X_t-X_s]=E[X_t]-E[X_s]=0$$

    why $E[X_t-X_s|F_s]$ = $E[X_t-X_s]$ ?

    if you have any other way of proving this i'm also interested. thank you

  • Dave ddd
    Dave ddd over 7 years
    great thank you!