$L^1$ convergence of this martingale

61 Views Asked by At

Let $(X_n)_n$ be a martingale, with $X_0=1$ and $X_n=2U_nX_{n-1}$ such that $(U_n)$ is i.i.d and follows a continuous uniform distribution on $[0,1]$

Does $X_n$ converge in $L^1$ space ?


I showed that $X$ converges a.s towards $0$ (using the logarithm and the law of large numbers) and that it does not converge in $L^2$.

For this question does this have to do with uniform integrability?

1

There are 1 best solutions below

0
On BEST ANSWER

As you showed $X_n\rightarrow 0$ almost surely.

If $X_n\rightarrow 0$ in $L^1$, then $E(X_n)\rightarrow 0$ which is not true since $E(X_n)=1$.

From Jensen's inequality $E(X_n^2)\geq E(X_n)^2$, hence $L^2$ convergence implies $L^1$.