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I have to prove whether the next statement is true or not:

'if {Xn} for n>=1 to infinitive it is such a martingale that for everything n>=1, Xn>=0 and E|Xn|=1, then the sequence {Xn} for n>=1 to infinitive converges on L1'.

I think I have to use Doob's martingale convergence theorem but I don't know how to use it to get convergence on L1, because as I understand with that theorem you get almost sure convergence.

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A positive martingale $(X_n)$ converges almost surely toward a random variable $X_{\infty}:= \lim_{n \rightarrow \infty}X_n$. But in $L^1$, we do not have automatically a convergence, even if $(X_n)$ is bounded in $L^1$.

Here is a counter-example:

Let $(Z_n)$ be a non-symetric random walk on $\mathbb{Z}$ and $(X_n)$ a process defined by $X_n = (\frac{q}{p})^{Z_n}$ where $p = 1-q$, $p\in]0,1[$, $p\not= 0.5$, representing the probability of up and down of the random walk.

Then $(X_n)$ is a positive martingale converging almost surely toward $X_{\infty}=0 \,a.s.$ . If your statement was true, we would have $E[X_0]=E[X_{\infty}]\Leftrightarrow 1=0$ which is false.

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