# equivalence of $E[X_\infty]=1$ and $X$ is a u.i. martingale on $[0,\infty]$

Let $(X_t)$ be a strictly positive supermartingale on $[0,\infty)$. Hence $X_t$ covnerge to $X_\infty$ a.s. Now how can I show the following: $E[X_\infty]=1$ is equivalent to $(X_t)$ is a uniformly integrable martïngale on $[0,\infty]$.

hulik

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This result does not hold, as the following classical example shows. Let $U$ be uniform on $[0,1]$. For every $n\geqslant0$, let $U_n$ denote the integer part of $2^nU$, $\mathcal F_n=\sigma(U_n)$, and $X_n=2^n\cdot[U_n=0]$.

Then $(\mathcal F_n)$ is a filtration and $(X_n)$ is a nonnegative martingale with respect to $(\mathcal F_n)$, starting from $X_0=1$. Furthermore, $X_n\to X_{\infty}=0$ almost surely, hence $\mathrm E(X_{\infty})=0$, and $(X_n)$ is not uniformly integrable since $\mathrm E(X_n)=1$ does not converge to $\mathrm E(X_\infty)=0$.

For a strictly positive example, add a positive constant to each $X_n$.

About the revised version: Assume first that $(X_n)$ is in fact a martingale and that $\mathrm E(X_0)=1$.

In one direction, if $\mathrm E(X_\infty)=1$, then Scheffé's lemma ensures that $X_n\to X_\infty$ in $L^1$. Any sequence in $L^1$ which converges in $L^1$ is uniformly integrable hence $(X_n)$ is uniformly integrable. In the other direction, if $X_n\to X_\infty$ in $L^1$, $\mathrm E(X_n)\to \mathrm E(X_\infty)$. Since $\mathrm E(X_n)=1$ for every $n$, $\mathrm E(X_\infty)=1$. QED.

Finally, if one only assumes that $(X_n)$ is a supermartingale, what could impose that $(X_n)$ is in fact a martingale? To wit, consider the deterministic example $X_n=1+\frac1{n+1}$.

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@ did: I'm so sorry! as you posted your answer I was starting to edit my question. It should be $E[X_\infty]=1$ Again, sorry! – user20869 Jun 11 '12 at 7:58
@ did: Thank you for updating your answer. However, there still some questions around. First why is it true that, $E[X_n]\to E[X_\infty]$ (this is necessary to apply Scheffé's lemma.). Second, the generalization to a martingale: It would be clear if $E[X_0]=1$, otherwise I do not see how to adapt the case of supermartingale to martingal. – user20869 Jun 11 '12 at 9:04
(1) The convergence of the expectations is a part of the conclusion of S lemma. (2) There is no generalization to a martingale. Rather the result holds for martingales and cannot hold for supermartingales. – Did Jun 11 '12 at 9:15
@ did: Thanks again: to Scheffé: I know this result: $f,f_1,f_2,\dots \in L^1$ and $f_n \to f$ a.e. Then $E[|f_n|] \to E[|f|] \Rightarrow E[|f_n-f|]\to 0$. To obtain $L^1$ convergence I have to verify $E[|f_n|] \to E[|f|]$, right? Why is this true? – user20869 Jun 11 '12 at 9:48
?? $E(X_\infty)=E(X_n)=1$ for every $n$. – Did Jun 11 '12 at 10:46