# problem about almost sure and L1 convergence

• Nov 11th 2009, 09:58 AM
yavanna
problem about almost sure and L1 convergence

Let $\lim_{n \to \infty}X_{n}=X$ a.s.

And let $Y=\sup_n|X_{n}-X|$.

• Proove that $Y<\infty$ a.s.
• Let $Q$ a new probability measure defined as it follows:
$\displaystyle Q(A)=\frac{1}{c} \mathbb{E}\!\left[1_{A} \frac{1}{1+Y}\right]$, where $\displaystyle c=\mathbb{E}\!\left[\frac{1}{1+Y}\right]$.
Proove that $X_{n} \rightarrow X$ (in $L_{1}(Q)$).
• Nov 11th 2009, 10:28 AM
Laurent
Quote:

Originally Posted by yavanna

Let $\lim_{n \to \infty}X_{n}=X$ a.s.

And let $Y=\sup_n|X_{n}-X|$.

• Proove that $Y<\infty$.
• Let $Q$ a new probability measure defined as it follows:
$\displaystyle Q(A)=\frac{1}{c} \mathbb{E}\!\left[1_{A} \frac{1}{1+Y}\right]$, where $\displaystyle c=\mathbb{E}\!\left[\frac{1}{1+Y}\right]$.
Proove that $X_{n} \rightarrow X$ (in $L_{1}$).

It should be pointed out that the first question is just a question about real sequences (i.e., not probability) : prove that a convergent sequence is bounded, and explain why it is sufficient to conclude.

As for the second question, be very careful! It is not $L^1$, but $L^1(Q)$ that you probably mean; this is very different. The only advice then is to use the bounded convergence theorem; that should do the trick pretty neatly.
• Nov 11th 2009, 10:49 AM
yavanna
Thanks!

Quote:

Originally Posted by Laurent
It should be pointed out that the first question is just a question about real sequences (i.e., not probability) : prove that a convergent sequence is bounded, and explain why it is sufficient to conclude.

As for the second question, be very careful! It is not $L^1$, but $L^1(Q)$ that you probably mean; this is very different. The only advice then is to use the bounded convergence theorem; that should do the trick pretty neatly.

Yes, you're right... It was $L_{1}(Q)$.

But the convergence to be verified in the first question is a.s.

Quote:

Originally Posted by yavanna

• Proove that $Y<\infty$ a.s.

• Nov 11th 2009, 11:40 AM
Laurent
Quote:

Originally Posted by yavanna
But the convergence to be verified in the first question is a.s.

Yes. The assumption means that for $\omega$ in an event of probability 1, the sequence $(X_n(\omega))_n$ of real numbers converges. However, this convergence implies that the sequence is bounded. Thus, you have the inclusion of events : $\{\lim_n X_n= X\}\subset\{Y<\infty\}$. If the first one has probability 1, then the same holds for the second one.